By way of Raymond Johnson, the best statistics multiple choice question ever written on a chalkboard. Try not to think too hard. [via]

By way of Raymond Johnson, the best statistics multiple choice question ever written on a chalkboard. Try not to think too hard. [via]

Obviously that’s a nice question which deserves that we shouldn’t be thinking on it…not even too hard… :P

The answer is C because most people will think they are correct and its the highest number.

B

Read the question again – it never implies the answer is one of the multiple choice options, if you rephrase it then it’s more understandable.

Picture there being 3 light bulbs and 4 doors, 2 doors lead to the light bulb on the left, and the other 2 doors lead to 1 of the light bulbs each. If the bulbs light up randomly, what’s the probability of walking through a door into a lit up room? If the left one lights up you’ll be right 50% of the time, if the middle one is on then 25% of the time, and if its the right one then 25% of the time as well. Each of these light bulbs randomly will light up 33% of the time, therefore –

[33% * 50%] + [33% * 25%] + [33% * 25%] = Probability of being right = 33%

And there we go, the answer to this question is 33%.

The question is not multiple choice, the problem that the question is about is. It’s not a paradox, there is obviously an answer, and it’s clear as day once you think about it.

By that logic, Luke, if the question doesn’t imply that one of the multiple choice options is correct, then you can choose /any answer/ to the question at random. Which means the real probability of being right is closer to 1/infinity.

For any question, given 4 choices, 2 of them being the same, all the choices equally probable of being correct, then the probability of selecting the right answer at random is 33%. My point was that the problem that [ 25% / 50% / 60% ] is the answer to isn’t stated, and doesn’t need to be. I think it can either be interpreted this way, or it’s just a novelty with no real meaning, but not a paradox.

The choice of the answer is random so you do not get to think you are correct

Good analysis Luke but it begs the question: What if the four choices were:

a. 33%

b. 50%

c. 60%

d. 33%

But Luke, what part of the chalkboard question is analogous to the random on/off state of the light? The correct answer to the question won’t turn on or off. It will simply be the correct answer. I don’t think there’s any actual probability at work there, in other words–a given answer is, or is not, correct. The only question is our probability of choosing it.

I don’t know what I would think, though, if A & D were both 50% and B were 25%.

Once you found out that no answer is right, we can make the problem more fun by changing C to C = 0%.

I wonder what you think about it…

If and only if one chooses 0% to be a possible answer the problem can become a paradox.

I don’t think that’d necessarily be a paradox either (but it makes you think a lot more!)…. There can be no ‘correct’ answer in this instance … so the probability of selecting it is zero. I don’t think that’s actually the same thing as saying zero is the correct answer, though…… You’d need to be very careful how you worded the question….. I’m leaning towards thinking it can’t be done….

the question is: what is the probability that you randomly choose the right answer to this question.

we have to possibilites:

1. there is a right answer to this question

2. there is no right answer to this question

to see weather or not this is a paradox one can do by testing this two possibilites.

if we suppose that there is a right answer, we have three possible answers:

1.1 the probability that we randomly chose the right answer is 25%

1.2 the probability that we randomly chose the right answer is 50%

1.3 the probability that we randomly chose the right answer is 0%

we did soppose that we have a right answer, so one of this three must be correct.

if we find contradictions in all the three cases, the hypothesis that there is a right answer must be rejected.

if the first one is correct ( A) and D)) we have a chance of 50% percent to choose this answer at random. This is in contradiction with the answer itself since it states that we have a chance of 25%.

the same reasoning applies for the other two answers; we always get contradictions.

We therefore must conclude, that there can not be a right answer.

Now let us suppose that there is no right answer.

If there is no right answer, the probability to randomly choose a right answer is 0%, (since there is none). But if the probability to randomly choose the right answer to this question is 0% and 0% is a possible answer to the question, we get a contradiction, since C becomes a correct answer .

You’ve nailed it, dedeco — …but… there IS a solution!

The question specifies that the answer is chosen at random — but it does NOT specify that the distribution of the random selection is uniform! :-D

a paradox is not a well defined concept, it’s just an _apparent_ contradiction, but actually is a conceptual error by he who describes the paradox

I think Dedeco you are right!

sorry, Fela — a paradox is most certainly NOT a conceptual error.

Chesterton put it well when he said that a paradox was “truth standing on its head to gain attention,”

0% because there is no specific question. The question could be any question, so the chances of a, b, c, or d being correct are effectively 0.

Well, we need to separate here 2 things: answer to the question and a chance to get this answer. Consider the same problem but instead of percentages you have balls. It is still the same problem.

What is the chance that you choose the blue ball if you randomly choose from the basket containing: blue, red, black, blue?

The answer is 50%. The same answer is is for the above question.

Who told you that blue is the correct answer to the question?

Or in the question above: you are saying that 25% is the correct answer and because 2 out of 4 answers are 25%, 2/4 makes 50%. That´s obviously non-sense, because you don´t know if 25% is the correct answer!

To make your answer correct, the question must have been: what is the chance to get 25% as answer at random choice.

In my opinion you could say, that there are 3 different answers, so the chance to get the right one at random choice is 33,33%, but actually there is no question and so there can´t be a correct answer to a not asked question!

Luke: “My point was that the problem that [ 25% / 50% / 60% ] is the answer to isn’t stated, and doesn’t need to be.”

Yes, it is stated. I understand that you want very much for that not to be the case, but there it is. You want for the question to say “if you are given a standard multiple-choice question with four answers, two of which are identical, and there is a correct response, what is you chance of picking it randomly?” And yet that is not what the question says. It is plainly a multiple choice question, and it plainly refers to itself.

You don’t get to forbid just one part of our shared reality. A question on a blackboard with four answers labeled A through D beneath it is a multiple choice question. If you’re allowed to deny that, then you might as well deny that the question is written in English at all. Maybe it’s a secret code! Maybe the only correct answer, according to that secret code, is to quack like a duck! Maybe we’re all hooked up to the Matrix, and there is no blackboard!

But I’m betting you didn’t want this to turn into an argument about Rene Descartes and David Hume. Well, you don’t get to take just one step down that rabbit hole. Either there is no spoon, or that is a multiple choice question, it sets up self-referential paradoxes, and there can be no suitable answer.

“Maybe the only correct answer, according to that secret code, is to quack like a duck!”

Hahahaha! You just saved my day! Thank you!

Actually the answer to the question of choosing the right answer is B), but if that is the correct answer to this question, then there is only a 25% probability of choosing at random correctly, so then A or D is the correct answer. Which then in turn gives you a 50% probability of guessing correctly, which would make the correct answer B… Which brings me to a parallel

Vizinni: All I have to do is divine from what I know of you: are you the sort of man who would put the poison into his own goblet or his enemy’s? Now, a clever man would put the poison into his own goblet, because he would know that only a great fool would reach for what he was given. I am not a great fool, so I can clearly not choose the wine in front of you. But you must have known I was not a great fool, you would have counted on it, so I can clearly not choose the wine in front of me.

Man in Black: You’ve made your decision then?

Vizzini: Not remotely. Because iocane comes from Australia, as everyone knows, and Australia is entirely peopled with criminals, and criminals are used to having people not trust them, as you are not trusted by me, so I can clearly not choose the wine in front of you.

Man in Black: Truly, you have a dizzying intellect.

Vizzini: Wait till I get going!

the princess bride, great scene, great reference

wonder if it is the last kid action film with a lecture on logic in it

(or for that case any real reference to math)

Any question can be suitably answered with a quote from Princess Bride, if one puts enough thought into it.

Well done!

I totally agree, and it could be seen as a bit of a joke about students trying to guess their way through multiple choice exams, by giving a multi choice question that is contradicted by guessing and thus unanswerable. I’m just amazed at how so many people want to see this as a puzzle rather than a joke. The clue is in the OP’s statement: “try not to think too hard about this”!

brilliant! great to know that others use Vizinni’s logic/illogic thinking as reference!

No answer in the choices.

Case 1: A is correct, then D is also correct. The probability will be 0.5. But the answers there are 0.25. So, not right.

Case 2: B is correct. The prob. of choosing B will be 0.25. But, the answer there is 0.5.

Case 3: C is correct. The prob. of choosing C will be 0.25. But, the answer there is 0.6.

So, all of the choices are not correct.

First assume its a multiple choice question if so:

Since there are two 25% then 25% cannot be the right answer. So, we are left with two choices and a probability of 1/2 of getting the right answer. So based on this scenario the answer would be B 50%.

Second assume its a theoretical question if so:

Then in this case the answer can be any probability and we are not just limited by the 4 options. You could justify 33% because the chance of 25% being right is 1/2 * 1/3 (25%, 50%, 60%) = 1/6, chance of 50% being right is 1/4 * 1/3(25%, 50%, 60%) = 1/12, and chance of 60% being right is 1/4 * 1/3 (25%, 50%, 60%) = 1/12. We can find the expected value by summing the probabilities (1/6) + (1/12) + (1/12) = 33%.

wow… im pretty good at math and logics and all tht considering im in the gifted program….. but wat the heck! im confused!!!!!

i think there exists no correct answer.

probability of getting 25 =50%, ….50=25%,…..50=25%

by the question, the probability of the correct answer should be the answer itself…which is not so for any of the 4 choices…

Luke’s analogy is faulty at best. The random lighting up of the light bulb implies that the correct answer to this question can change at random. It is not the correct answer to the question that changes at random, rather it is our answer choice that is random. Thus, there is only one element of randomness, and therefore only one element of probability to consider. The question is, in fact, seemingly paradoxical, based on the way it is worded. The probability of choosing the correct answer randomly from any given 4-option multiple choice question is 25%. Since this question refers to itself, it changes things. At first we would say the answer is 25%. Since there are two options that say 25%, that means that the probability of choosing the correct answer to this question is 2/4, or 50%, so we would choose B. However, if the answer truly is 50%, and there is only one option for that answer, the probability of choosing it at random is 25%, which would make the answer either A or D, which brings us back to 50%, and so on forever.

As per me answer is 25% … Here is how…. prob of ans being A or D is 25% + 25% i.e 50% if A or D is correct then we deduct the prob of either B or C from it .. i.e 50% – 25% so total prob is 25%. In other way say A or D is wrong then we remove 50% from total 100% so remaining is 50% for B or C and again prb of either B or C being correct is 25% so we remove 25% from 50% so it again comes out to be 25% :)

That is clearly not the bestir question ever written because it is badly written. What does it mean to “choose an answer at random?” If I have to randomly choose from A, B, C, and D, the answer is 1/4. But if I have to choose between 25%, 50%, and 75%, the answer is 1/3.

Please relieve me from this anxiety, I absolutely must see my reasoning validated:

The question is a paradox, and no answer is correct, right?

My logic is:

If the answer were 25% (A or D), random choice would be 50% right.

If the answer were 50% (B), random choice would be 25% right.

If the answer were 60%(C), random choice would be 25% right.

So none can be correct.

I’m with Paulo on this

Yeah, I like your logic, I got to the point of considering 50% being right when you have the two 25%’s there, and just shuddered to a stop in my mind.

+1

Paulo im with you.

Now the real treat would be if he added 0% as answer C instead of 60%. As it would imply none could be correct, but by answering it you would be correct (and you have chosen 0% which is none is correct)… my head hurts… statistiception

Actually, there are only 3 possible options, because choosing A is mathematically the same choosing D, so the random choice would be 33% right for A and D (as 1 option only), B and/or C.

Is this some kind of self-mind-fuck i have, or what?

@Nacho That’s incorrect. Choosing A and D are not mathematically the same. You cannot combine them into one option. Although they have the same value associated with them, and are two separate choices.

If a question is: What does 2+2 equal? And the choices are a) 4, b) 4, c) 4, and d) 5. What is the chance of choosing the correct answer at random? You wouldn’t say 50% just because a, b, c all have the same value of 4, and the other value choice is d as 5. Here, three of the four have the correct answer of 4, to make it 75% chance.

Actually, though there is 4 options, there is only 3 possible answers, the question states answer… and A and D are the same answer…, he has to be awesomely right, but still I think it’s 0% given the paradox. :S

Most people reading this question appear to make two fairly assumptions, which are fairly natural, but are not explicitly stated:

(1) That you are randomly selecting from answers (A), (B), (C) and (D) only.

(2) That the method of random selection should give each of the above options (A,B,C and D) an *equal chance* of being selected.

The question is insufficiently specified, and so strictly speaking it has no ‘correct’ answer.

If we replace (2) with:

(2b) That my method of random selection should give each *unique answer* (from the list) an equal chance of being selected.

…then it is equivalent to Nacho’s understanding of the problem.

Neither assumption (2) nor assumption (2b) is more or less correct than the other, because no information is given by which we could distinguish between these two possibilities.

jajaajaa… Paulo your answer is the best!!! I was stady statistics and you are totally right!!!!

Exactly. It’s self-referential.

I believe there is a correct answer: 0%. The question is “what is the chance you will be right”? The hypothetical question is, if you choose one of A-D, will it be the right answer? The actual question is, what is the chance it will be the right answer. 0%.

Bobby you are 100% correct.

But the question says IF you pick the answer randomly, which you are not. So it is still 50%.

think this way. The probability of you chosing B is 25%. But there are two 25% in four answers. So, your chance is still 50%!

I’m with Paulo: 50% can not be the answer because you only have 25% chance of picking it; neither can 25%, because you have 50% of chance of picking it…

I completely agree with Helena. Most of the posters here are looking at the question as if it says “Choose an answer at random. Now that this is done, what is the chance that you were right?”. These are assumptions that you cannot start with.

The reason this is an assumption is through the use of the “proverbial you” which exists in the english language. The question may not refer to “you” the reader. If the author of the question meant the “proverbial you” then the question could be accurately rephrased as “Suppose someone chose a question at random (maybe you, maybe someone else). What is the chance that that person would be right?”.

Marking this as a statistics question is a red herring to distract people from this. And, if the question were not written in english, then this “proverbial you” situation might not exist (I can’t speak for all languages).

The word “If” adds to the mayhem. If you did, then what? But the question is not commanding you to make that choice. Which is where most people are getting snared.

If the author of the question meant the “proverbial you”, then the answer is clearly B. The fact that the author is not commanding you to pick the random answer, but is merely placing it as a hypothetical idea, then the answer is still B. The problem only becomes a contradiction if you are forced to pick an answer which is clearly not the case.

I can also see the case that it is a trick question because the answers below are not necessarily the answer to the question, which other readers have suggested. Readers of douglas adams may remember, “42” and “how many roads must a man go down”. The “answers” listed below the question may not be related to the answer of the question you seek.

Ultimately this fun little exercise only exists because of its vagaries, its vagueness if you will. And thats why its says not to think about it very hard. A little clarification about what the author meant makes all the difference.

Well presumably there is a correct answer, it’s either a) b) c) or d), and if the actual value is irrelevant since the correct answer is one of those letters (keys) rather than its value, then it’s 25%.

If the values matter, then the question is faulty and thus invalid.

That’s as far as I can go without chemical influence.

Similar to what I concluded. Ignore the values themselves since you were asked to choose at random. The nature of this form of question is that there is one and only one correct choice. The creator of this particular multiple choice question has apriori determined what they deem to be the correct choice. Whatever you choose at random may or may not match what the question creator deemed the correct answer. As it turns out I believe the probability that your random choice matches what the question creator chose is 25%, but note that doesn’t mean my choice or the correct choice was either (a) or (d).

+1

25% chance of being correct.

The values are relevant because that’s how multiple choice questions work. You look at the values and choose which one is correct. Here it is not asking you to choose at random, it is asking you to choose the correct value.

The fact that two of the values are the same implies that there may be more than one correct answer in this format of multiple choice, hence 25% goes out of the window.

That was my original thought, too. But it’s really a syntax-versus-semantics kind of issue, as follows. If there’s simply one assigned correct answer, either (a) or (b) or (c) or (d), then it’s a syntactic issue and the other three answers are just plain wrong, regardless of what values the answers denote. But suppose the question were, “How much is 1+1?”, and the four possible answers were (a)2 or (b)3 or (c)4 or (d)2. Then the correct answer must be decided semantically, so to speak, with both (a) and (d) correct, and with random guesses correct half the time. And the actual question’s wording suggests a semantic interpretation to me, so I’d say 50%.

Contrary to my earlier comment, which I realized was wrong as

soon as I re-read it, I now think:

There’s a question and a meta-question (a question about

the question). The question itself is multiple-choice as

indicated. But, despite first appearances, the meta-question

is outside that multiple choice context, and you simply need

to answer it with a percent number, not by the (A)-(D) choices.

And then the numerical answer’s obvious, and is, in fact,

denoted by one of the (A)-(D) choices, though the choice

that answers the meta-question isn’t the same choice that

answers the question itself. … Or something like that:)

@BILL:

m with ur answer..

@Snackdaddy, Since when, during a multiple choice exam or question, is it okay to ignore the answers attached to the letter choices? That is the whole point of this question. The key is that this question refers to itself, therefore the values attached to the answer choices must be considered, which in turn gives rise to the endless loop of logic one enters when trying to answer this question. Again, that’s the whole point.

Yep, it’s a paradox. It’s a terrible question.

I actually thought it was quite ingenious.

The version I gave to my class had 0% instead of 60%… so that the “none can be correct” was covered, making it a proper paradox. I broke at least one kid’s brain with it :)

I don’t see that. ‘This question’ isn’t even a question so we must assume that it is referring to ANY multiple choice question. Therefore the answer is 25%. BUT that answer is there twice so its 50%. There is only one answer stating ‘50%’ so that is the right answer. But it is still wrong. So what do we do? Common sense. There is a 25% chance you’ll get the answer right to a problem with four possible answers. If two of those answers are the exact same thing, then your professor is an asshole and you should answer E. none of the above: you are an asshole. Then you would be right.

There isn’t a 25% chance if 2 of the answers are right (which is possible here).

the chance of choosing correct answer or incorrect answer is constant! then whether you choose A & D or not, you have chance of 50% for choosing correct option. Then B is the answer, although it is not correct

Paulo ANY answer would be correct. Because “IF YOU CHOOSE TO ANSWER AT RANDOM” the way the question asks ….any answer you choose “AT RANDOM” will be correct. SO AS LONG AS YOU DONT THINK and JUST CHOOSE RANDOMLY” you’ll have choosen the correct %

yeah I agree with this, most definitly a paradox

I think the answer is None because given the answers available, 1 out of 4 is normally 25%, but it’s listed twice, so in this instance, you would have a 50% chance of getting the answer correct, answer B, however if answering randomly, you’d only have a 1 in 4 chance of answering B (50%)… but by that virtue, the answer is A or D… 25%… , but because it’s available twice to choose from the answer is B 50%… it goes round and round. I don’t believe there is an answer.

But then what if option C was 0%?

exactly. That’s what i think.

different person has different idea

Answer is B. There are 4 choices. Random choice means 1 of 4, so you have 25% chance…..but 25 appears twice, therefore a 50% chance when random selected.

That’s not a paradox. None of the choices is correct, so, if you choose an answer at random (assuming you are randomly choosing from A, B, C, and D, giving equal weight to each one), then there is a 0% chance of getting the correct answer. This is not a paradox because there is, in fact, a 0% chance of picking 0% (under the same assumption as above).

I agree JW! TOTALLY!

Right, so the probability is 0%…

I think you are wrong Paulo.

My logic is :

The answer to the question is unknown to us … so it could be 25%(x2), 50%,60%. Tree unique options. Thus 33.33(3)%

I totally agree with you.

But I also love the KISS principle, so….

Oh yeah… paulo watz ur face…… ur right!!!!

Yep. I like this logic.

Paulo Ferreira’s logic, that is…

“So none can be correct.” Exactly. Therefore the chances of getting the answer correct is 0%.

The phrased question is just a hypothetical – it doesn’t say you have to answer it ramdomly, just that if one did what would be the chance that you pick the correct answer.

b) 50%

Ah, but if b) is the correct answer, then the chance of picking the correct answer randomly is 25% – and therefore b) cannot be the correct answer.

You’re assuming that the actual value in the answer is to be self-referentially the same as the chance of picking an answer randomly. The question does not state this. You are adding a layer of complexity that does not exist into the question.

just because you say “Ah” before your statement, doesn’t mean its enlightening.

Alan is correct. The question specifically references itself. It states “this question” not “a multiple-choice question with four possible answers”. It is purposely being paradoxical.

the answers are not the probabilities… to make this easier, simply replace the answers with something like a) green, b) blue, c) red, d) green. Then it’s obvious… the chance of getting the right answer depends on which answer is correct. Simple, right?

Patricio, your idea is typical of a generation that grew up on multiple-choice tests. You begin by assuming that one of the answers must be correct–or at least, must be in the answer key. But that misses the point here. It is possible that 25%, 50%, 60% and 25% are all incorrect, and we need to find the correct answer from first principles. As, incidentally, all students should have to on their exams….

actually pretty enlightening

b is the correct answer to the question, but the important issue is the intelligence behind the question. The text moves you into the worl of randomness while the answer is quite straight and direct…

This is clever

It’s not clever, it’s pretentious. You can’t remove the context from a self-referential question; changing the options is not appropriate, as Psyphi noted.

It’s a poor question unless the answer is fill-in-the-blank, in which case I think that Bobby’s suggestion (0%) deserves some philosophical consideration.

Actually the answer to the question of choosing the right answer is B), but if that is the correct answer to this question, then there is only a 25% probability of choosing at random correctly, so then A or D is the correct answer. Which then in turn gives you a 50% probability of guessing correctly, which would make the correct answer B… Which brings me to a parallel

Vizinni: All I have to do is divine from what I know of you: are you the sort of man who would put the poison into his own goblet or his enemy’s? Now, a clever man would put the poison into his own goblet, because he would know that only a great fool would reach for what he was given. I am not a great fool, so I can clearly not choose the wine in front of you. But you must have known I was not a great fool, you would have counted on it, so I can clearly not choose the wine in front of me.

Man in Black: You’ve made your decision then?

Vizzini: Not remotely. Because iocane comes from Australia, as everyone knows, and Australia is entirely peopled with criminals, and criminals are used to having people not trust them, as you are not trusted by me, so I can clearly not choose the wine in front of you.

Man in Black: Truly, you have a dizzying intellect.

Vizzini: Wait till I get going! Now, where was I?

The answer is B

The odds of picking b) at random are 25%.

The answer is B – 50%

The probability of picking 1 right answer out of 4 choices is 25% (thats 1/4). Since there are 2 25%, and you are only allowed to pick one then the 25% choices are not the answer.

That leaves you with 2 choices left (B and C). … Therefore the right answer is 50%

Change of picking a,b,c or d = 1 in 4 (1/4)

Change that 25%, 50% or 60% is the correct answer is 1 in 3 (1/3)

Change you pick the correct answer is 1/4 * 1/3 = 1/12 = 8.3%

or… perhaps it implies that 25% is the correct answer as with 4 possible answer the change of picking the correct one (provided that there is only 1 correct answer) is 25% and as you have a 50% change of picking A or D the answer is 50%.

Chance that 25%, 50% or 60% is the correct answer is 0 in 3 because all three possibilities can be refuted. If the chance that you pick the correct answer were 8.3%, then the chance that you pick the right answer could not also be 25%, 50% or 60%.

Another problem with this approach is that 8.3% is not an answer choice : )

Not possible to calculate. If you answer random, you can’t see result to calculate real chance.

This break space-time and demostrates 11th dimension on string theory.

Best comment in the thread

The answer cannot be B. To figure out why, just imagine if all the answers were 25%. Got it?

it’s 100%, since they are all wrong, whichever you choose is the correct one, bitcheesssszzzzzzzzz, it’s a nice metaphor for life, whatever you do, you’ll never get it right, the only way is to fool yourself into thinking that you have done things right, motherfucking.

the only way one can be right is to be an electron, which is, to not think is to be right.

don’t think, feel!

I agree. It’s 100%. It only asks “if you pick an answer at random”. We can all pick an answer at random, so the answer is 100%.

Apparantly, most can’t.

The answer is 50%.

The chances for a good answer is 1/4 but there are two right answers so it’s 50%

But David, if the right answer is 50%, then there are not two right answers — there is only one.

Right. There is only one answer to the problem as stated in the problem. “Choose an answer” specifies we are to choose only one answer. We all know that means the probability of getting the answer right is 25%. But there are two choices for 25%. And you can only choose one based on the problem as stated above. So if we know the answer is 25% then the correct answer has to be A or D. Meaning we effectively only have two choices. So wouldn’t that mean you have a 50% chance at choosing the right answer? Or the probability to choosing 1 answer among two choices? Maybe my math is wrong.

@Jeff S

Yes 50% is the answer but you will not choose between A and D since there is only 1 right answer. A and D is already out to become the only answer therefore what remains is B and C… The answer is B.

The answer is B – 50%

The probability of picking 1 right answer out of 4 choices is 25% (thats 1/4). Since there are 2 25%, and you are only allowed to pick one then the 25% choices are not the answer.

That leaves you with 2 choices left (B and C). … Therefore the right answer is 50%

When Jeff S writes “We all know that means the probability of getting the answer right is 25%. ” this statement interpolates something that is not in the question, i.e., that there is a “right” answer (which implies that there is also a ‘wrong” answer.) But there is nothing in the question that provides any clue as to what is right or wrong. The predicate is, “if you choose an answer at random”. If that is the only condition, there is no “right” or “wrong” result, just a random selection. To determine whether a selection is right or wrong, you need some more information in the question. The question just fails.

Think of it this way. A person goes into a store with 4 items. If s/he buys one item at random, what is the probability that the item s/he buys will be the correct one? You can’t answer it. You need to know what is correct, which is outside the question.

David, I’m with your reasoning. Nicely explained by Jeff S, by the way.

it’s self-referencing, meaning as soon as you pick one it becomes wrong…

so the chances of you getting it right are 0%

Con Obs is mostly correct. I would amend it slightly to say, because it is self-referential, no extrinsic standard to determine “right” or “wrong” exists. Thus, there is no “right” or “wrong” answer. The question fails because it provides nothing intrinsic and specifies no extrinsic basis to answer it/

The answer is 0%. It would never be correct

But if there was a 0 choice that would have been correct, and so you would have 25% probability of choosing it at random and therefore it would be incorrect again. :)

No, because 0% is not one of the options, so there is a 0% chance that you will pick it.

Is a fuching loop.

The answer is B 50%, because if you answer to the question at random, with 4 posibility answers you have a 25% of posibility of answer the correct one, so if you have two answers (A and D) with this number, the probability of put the correct answer is of a 50% (answer B)

so… if the answer is B the probability is 25%… loop.

50%

Your answer can be correct or incorrect. So 50%.

Bonus.

What is the chance to win the lottery? 50%. You win or not.

Fail. That’s like saying “what are the chances that I will be struck by lightning?” and answering “50% because either you’re hit or you’re not.”

That’s not how probabilities work.

you should switch on your sense of humor…

Yeah, Nilo. It’s like saying your chance of winning the lottery is 50%, because you either win or you don’t.

Damn Bayesians.

+1 to Paulo.

If someone says 50% please give an explanation :P

Because as Paulo says, if 50% is the correct answer, you have a 25% of choosing that one if you take one randomly.

But, the question stipulates that you pick the answer at random, therefore you should not be thinking about it at all – just answer at random. The percentages are inconsequential since you pick at random. If you answer “B”, then… you haven’t picked at random at all, and have not followed the guidelines.

“If”! It’s hypothetical.

why all you agree on the question being valid? I think its formulation is wrong and it should be eliminated from the test.

it’s not a test, it’s written on a chalkboard

I totally agree, no student would accept such a question in a test !

I would say it’s B, explanation as follows.

Which answer is correct is not in question, so we assume each answer is possibly correct.

“If you choose an answer [..] at random [..]”

This reflects a dice roll with four sides. So basically to get one answer the chance is 25% on each possible answer.

However two of those answers are the same, so you have doubled the base chance, which yields 50%.

Which means that A and D are no longer the correct answer, if the correct answer is now 50% rather than 25%. This logic then loops back, since you have a 25% chance to select B, which would be a wrong answer (because you need a 25% chance to select 25%, which is not possible in this example).

There is no correct answer here.

The answer is: A and C are correct.

I think we are mixing up two issues. The question doesn’t ask about the chance of picking the right answer. It asks about the probability of being the right answer one chosen at random. There is a slight difference and in this second case there’s no link between the options and the right answer.

So, once picked one at random, the probability of being the right one is 1/3. Imagine other example, A), B), C) are 25% and D) 75%.

It’s not a paradox, just a really funny riddle.

Well, it depends on what you mean by “at random”, and “correct”. If by “at random”, we mean that each of {A, B, C, D} is selected with equal probability, and “correct” answers are some pre-selected subset of {A, B, C, D}, then it follows that the probability of a correct answer has to be 0.25 times the number of elements in the “correct” subset. So C is right out. For the reasons Paulo specified above, A, B, and D are wrong. So none of the answers are correct (the subset of right answers is the empty set), and the probability, obviously, of picking a correct answer is just 0.

I agree with this one. You assume you have to pick from one of the options, but none of them are correct, so the answer is 0%. Alternatively, even if there WERE a correct answer, if you randomly picking from an infinite number of answers, the probability of picking the correct one is 1 divided by infiinity, which is 0%.

My name is Miguel, which is my name?

A) Miguel

B) Bob

C) Andy

D) Miguel

The right answer is in two options, so, the answer is A and D.

It’s not asking which one is correct, it is asking what your chances are of choosing the correct answer. In your example, there is a 50% chance of choosing Miguel, but in their example, there is a 50% chance of choosing 25%, which invalidates the answer.

Think about what the question is saying again.

That is not a parallel to this question. You have inserted a non-random premise that provides a standard for evaluating the answers’ correctness. The chalkboard does the opposite. it just has randomness as its sole premise. So there is no “right” or “wrong” answer. You could have one or 1000 choices to select from. It doesn’t matter.

My name is Tom, which is my name?

A) Miguel

B) Bob

C) Andy

D) Miguel

The right answer is in no options, so, I can pick randomly as much times as I like. My chance to pick the right answer is 0

Paulo is correct, this is a logical paradox. Out of statistic, this is a logic question and has no answer. There are thousands of riddles of this type out there and the simple answer is that there is no answer. This is why they are called paradoxes.

First, we think about first sentence:

P(A) = 25%

P(B) = 25%

P(C) = 25%

P(D) = 25%

if A = 25% and D = 25% then:

P(25%) = 50%

P(50%) = 25%

P(60%) = 25%

Second, we can resolve the question:

Our chance is 50%, then we should check B :)

Are you sure? yesss XD

In that case, only B) is the right answer. What’s the chance of picking B out of 4 options? well it’s 25%. So you have a loop.

The question anyway it’s just funny. No answer is correct, hence is not a valid choose-an-option question.

I think it would be much more interesting with possible answers:

A) 25%

B) 50%

C) 0%

D) 25%

This is the one that makes my head spin is circles. In this case, I think that there will always be a “TRUE” answer in the option set, however I am pretty sure that it can not be selected because of the very rules of the “If one is chosen by random” bit…

It is almost so beautiful that I want to cry.

I don’t think this one is that bad either if you approach it in a sound way. Observe…

If you choose A then you are incorrect

If you choose B then you are incorrect

If you choose C then you are incorrect (because the correct answer would have to be 25% if there is one occurrence of it in the set of choices)

If you choose D then you would be incorrect

It doesn’t matter that 0% is a choice because you can’t choose it and be correct at the same time. So the probability of choosing the correct answer remains at 0%

I think the answer is 33.33% because they are asking about the chance. Replace 25%, 50%, 60% for A,B,C and answer again.

You read the answers and dont know the correct, then you answer at random. Yo have 3 options for choose. A(25%) B(50%) and C(60%) then 1/3 = 33.33%

I’m with you.

I choose NOT to choose; therefore, 0%.

I have another question. I have a dice with four sides. The dice has the number two writed in two sides, and the number 1 and 3 in the others.

I chose a random number between 1 and 3. I roll the dice. What are the chances for get the numer i choose at random in the dice.

33.33%

No, just because there are only 3 possibly outcomes, does not mean there is a 33% chance of landing on each side. It is still a 4-sided die, meaning you actually have 4 possible outcomes:

1, 2, 2, or 3.

Look at it this way, I will assign ‘sides’ to each number:

1 = top

2 = left

2 = right

3 = bottom

You have a 50% chance to land on either the left or right side (a 2, or a 2) of the die, a 25% chance to land on the ‘top,’ or a 25% chance to land on the ‘bottom.’

25%

The right answer to the question is 25% so… it is 50%

because there is only 1 option with 50%….

The answer is 25%.

To explain it better… there are 2 questions in the question.

1 is the answer of the question and the second one the probability of choosing it.

The probability of get an answer in 1/4 so it is 25%

Then, the question 2 is that you have 2/4 posibilities of choosing 25% so 50% is the second right answer.

Then, the pobability of choosing 50% is 25%…

I dont know if i am explaining myself.

I’ve seen paradoxes like this before. They are very funny. I was inclined to choose A at first because I assumed 25% right when I saw the four choices. Then I realized that appeared twice so I then thought the answer was B. I then realized 50% appeared once so the chance of choosing it would be only 25%, so there is a paradox.

One similar back-and-forth paradox that I have always liked is:

1. Statement 2 is true.

2. Statement 1 is false.

Your example of two statements that refer to one another, is a paradox — a form of the liar’s paradox. But this question, though it contains a “loop” in the choices, is not a paradox. It has a simple answer.

It’s B. You choose either the right or the wrong answer. 50% :)

There is no question at all.

The answer is 100% because there is no question to be answered.

i agree. no question is ever actually asked, so there can be no answer.

There is no spoon

No matter what answer is correct, I love this answer.

Not enough information was given, it never articulated an actual question so we can’t infer that the answers listed in A,B,C,D was an exhaustive list.

For example, if I write 4 numbers on a board (1,2,3,4) and ask someone to pick a number, most people would instinctively pick one of these 4 numbers even though it was never explicitly stated that this was the entire domain to pick from (heck, e or pi would be reasonable answers).

It’s never articulated what “correct” means either. “Correct” might be the answer that’s written on a piece of paper that’s in a sealed envelope somewhere, ie “correct” does not mean statistically correct, but just that it matches the pre-selected value. Similarly, perhaps some scantron machine has been programmed to know that some answer is “correct”, regardless of the statistical merit of the answer. Then guessing randomly would net you a 25% of being scored “correct” by the scantron. Modulo mechanical failure.

Francisco is right!

There’s no specific question posed. The text just refers to “this question” but it doesn’t state what the question is.

Exactly! It’s not a statistics problem–it’s philosophy, or tongue-in-cheek linguistics.

It’s a trick question because it isn’t really a question at all.

The question is there, it’s simply self-referential.

Hey guys, just divide the things in the right direction.

This is the question:

What is the chance you will be correct?

They ask for AN ANSWER, they are not saying choice one option. So whatever a,b,c,d contains, the answers to that question is 33.33%. (because two options have the same value)

The confusion is because the options has percents, and you think you need to choice one option, for clarify just switch your mind and think in this way:

Having the next options, if you choose one option as random, what is the chance you will be correct?

a) red

b) blue

c) white

d) red

Answer to the question is 33.33%, because we do not know the right answer but we know that there are three options in the universe of options.

Again, they ask for the probabilities not said, choice one option.

I agree with this

With that question, how do you know what is “correct” to be able to assess the probability of answering it correctly?

The answer to a multiple choice question must be A, B, C, or D, not 33%. You must have failed your SATs.

33% chance of hitting

Best said. 33,33% chance of hitting (Mejor dicho un 33,33% de acertar)

I’m only posting this because I want to receive future comments — and the only way you can get on the subscription list is to leave a comment! This is a fascinating dialogue — a truly great mental floss! Keep it up… and would one of you please resolve the issue (if there IS such a solution) so I can get back to work?

OK, my explanation for this issue is:

– If I have chosen randomly A) or D) then I’ve failed, because A) and D) have the same result, and this result is 25%. Then, I have had 50% to choose A) and D) and other 50% for chose another.

– If I have chosen randomly B), then I’ve failed, because I have only a 25% probabilities to choose B) and this makes this answer wrong, but makes option A) and D) correct. Then, if I would have choosed A) or D) then I would be right, and I had 50% probabilities to chose A) or D).

– If I have chosen randomly C), then I’ve failed, because I have only 25% probabilities to choose C) and make this answer wrong, but makes option A) and D) correct again. Then, if I would have choosed A) or D) then I would be right, and I had 50% probabilities to chose A) or D).

That means, if I have chosen randomly, I have had 50% to answer it wrong.

Then, the correct answer is: b) 50%

Thats what i said but the resolution to me is 25% not 50%. the correct answer is 50% because of A and D so… if 50% is the correct answer, it is 1 option of 4!

Well, we can see in other way.

The time for do the exam are close to finish, and you don’t know which answer is correct.

Then, you answer randomly. You have a 25% to choose A), B), C) or D). Only one is correct, therefore the probability to answer correctly is 25%, but wait, if 25% is correct, then you have 50% not 25% to be right. Therefore, if you answer the question thinking, you will know that you have 2 options with 25%, A) and D). Then, if you choose randomly, you would have 50% to be right, and 50% to be wrong (because A) and D) are correct).

But the question was what is the chance to be right choosing randomly. If you choose A) or D) you’re right, but, you dont are answering the question. The question would be, what is the chance to choose 25% randomly? The chance to choose 25% randomly is A) + D) = 50%. Therefore the correct answer is B) 50%.

:)

It is a 33.33% because you choose between three valid responses (25, 50 and 60). Another thing is that you had asked that the possibility of choice is 25% (which would be 50%) or that the choice is 50% or 60% (in these cases would have been a 25%).

In definitif correct result is 33.33%.

In your response, if the ‘answer’ is 33%, then neither A nor B nor C nor D are correct, meaning there are no valid responses, meaning the correct answer would actually be 0%.

I think you’re right and I’m wrong, because in this particular case there is no valid answer. I like this kind of question.

This is what computer scientists call an “undecidable” question. There are many examples of undecidable problems. Wikipedia has a small sampling: http://en.wikipedia.org/wiki/List_of_undecidable_problems

Paulo Ferreira is right!

P(A) = P(D)= 50%

P(B) = 25%

P(C) = 25%

But A=D=25% therefore different from the 50% above

B=50% different from 25%

C=60% different from 25%

It is never correct!

I think the options haven’t to be read it until the end. How to do it? Simple, I suggest this: your random choice is 1/4 and an answer is true with probability 1/2 (bernoulli distribution). Then (4 choices)*(1/2 to be true)*(1/4 random)= 50%. Your final answer wouldn’t be a random choice. B).

I actually get .375. Here’s my math:

P(picking A) = P(picking B) = P(picking C) = P(pickingD) = .25

P(A is correct) = P(D is correct) = .5

P(B is correct) = .25

P(C is correct) = .25

So the probability of you guessing A and A being the correct answer is .5*.25 = .125

The probability of you guessing B and B being right is .25*.25 = .0625

” ” C and C being right is .25*.25 = .0625

” ” D and D being right is .5*.25 = .125

Sum those events and you get .125+.0625+.0625+.125 = .375

The trick is that the question isn’t really multiple choice – it’s short answer asking you to evaluate a multiple choice problem.

Jay is correct; the answer is 37.5%.

The key is to not assume that one of the four multiple choice answers is the answer to the question. Framed differently, the question actually implies “If there were a multiple choice problem with the following four answers, and you picked an answer to that multiple choice problem by random chance, what is the probability that you picked the right answer?”

That assumes there is at least one right answer.

You are fundamentally changing the question, so any answer you provide does not apply to the question in this actual post. The question in the image is self-referential, so you should not infer that it is referring to anything other than itself.

You are assuming that one of the answers is correct and there is an equal chance that the answer is 25%, 50%, 60%, or 25%. But it is more reasonable to assume there is an equal chance that 25%, 50%, or 60% is the correct answer. So it’s more like:

probability of you guessing A and A being the correct answer is 1/3*1/4 = 1/12

The probability of you guessing B and B being right is 1/3*1/4 = 1/12

” ” C and C being right is 1/3*1/4 = 1/12

” ” D and D being right is 1/3*1/4 = 1/12

Sum those events and you get 1/12+1/12+1/12+1/12 = 4/12 = 1/3 chance of being correct with a random choice.

right

i think that the correct answer is c), the reason is kind of crazy…let’s imagine that we got a sample, this sample is 25%,50%,60%,25%. so the probability asociated to each one are:

25%—-) 1/2

50%—-) 1/4

60%—-) 1/4

if we have the mean, it would be 0,25*0,5+0.5*0.25+0.6*0.25= 40%

if we consider that we have done all this thing not randomly, the random answer would be 100%-40%= 60%

any body agree?

What you need is to ask the question to 4 people and hope that two of them will say answer B.

Let’s run a simulation. Ask 1 person. He says A. A could be correct, but if A is the correct answer, we now have 100% of the answers in our sample correct, whereas A predicted only 25%. This is a bad result due to small sample size, of course.

So we ask another person. The second person guesses D. Now we have 50% of the people guessing A and 50% of the people guessing D. They could both be the correct answer, but in both cases, they say that only 25% of the random answers would be correct, not 50%.

Now things get interesting in our simulation if the next two people answer B. In this sample of 4, we now have two people answering B (50%) and one person each for A and for D. They could all be the correct answer. If A is the correct answer, we now have 1 person guessing this answer correctly. However, if A is correct, the person guessing D is also correct. So that would mean that we have 2 people guessing the correct answer at random out of a sample of 4. Whereas if A and D would have been correct, we would have predicted only 1 person to get it right.

On the other hand, the two people that guessed B represent 50% of our sample. And answer B does indeed say that you would have a 50% chance. Our two random people got it right.

Unfortunately, if we run this simulation a million times, it is unlikely that half a million people would at random choose B, so in the long run, this doesn’t work, but in the short term there will be a moment where it is possible that B is the correct answer and that 50% of the sample guessed it at random

B)50%

Because the correct answe is 25%, and you have 50% to hit it

Exactly. I Agree.

Se you are saying that the correct answer is 50% because the correct answer is 25% ?

Hmmm…..

100%. See my detailed answer on Quora, and add your own answer for posterity: http://b.qr.ae/tfIo2g .

Nathan, is this some kind of record for the most number of comments received per word of your original post? Nice work.

100%. I am always correct!

I think people are missing the obvious. Everybody seems to agree that A, B and D can’t be right as they contradict each other…therefore the only possible correct answer is C 60%. It doesn’t matter why as long as you pass the test :-)

I think there is a 0% chance. It is impossible to choose the correct answer. All the various reasoning above is correct to a point, but fails to consider that the answer changes when the question is answered. You would, as noted, randomly have a 1/4 chance, but there are two 25% answers, so you have 1/4+1/4=50%, which is now the answer. So, as noted, you now have a 1/2 chance is the answer is 25% and a 1/4 chance if the answer is 50%, which (I’m gonna guess here) would work out to 60%. That would futher affect the odds, but it doesn’t really matter. The answer is a moving target, its always changing as soon as a guess is made. That means there can never be a correct option (unless you restructure the question to bound it in some way, say within 1% difference), so the answer becomes 0%.

There’s two 25%, one 50% and another 60% . Picked randomly, the chance of the problem is recursive. In fact if we pick 50% is right ’cause we are on {25%, 25%} U {50%,60%} = 100/2. But 20 percent they’re correct too, in every case A and D are the same [ {25%} {25%} {50%} {60%} ] = 100/4. I quess the quiz are on 75% – X% where X is a variable between 60% and 75%. it is correct? Think there is more of 60% of probability of you choosed the correct one… programmed at a computer could truncate the whole machine.

Mathematically, the answer is 1/3 .. if we make the assumption that there actually exists a right answer on this list. I remove any ‘numerical’ meaning from the answers, treating them as discrete events. So we need the probability that an event is correct, AND the probability that we actually pick the correct event. We have 3 different answers (25, 50, 60) and the chance of any one of those being ‘the correct one’ is 1/3. The chance of picking 25 is 1/2 (because there are 2 ot those instances) . Chance of picking 50 or 60 is 1/4 each. So we multiply the (chance of each event being correct) with the (chance of picking it) and take the sum.

Event Chance correct Chance of picking

25 (A+D) 1/3 1/2

50 (B) 1/3 1/4

60 (C) 1/3 1/4

Sum (1/3 x 1/2 + 1/3 x 1/4 + 1/3 x 1/4) = 1/3.

Philisophically .. this question is false. The answer is 42. ;-D

If you follow the instructions and just choose at random with no further thinking, then wouldn’t it be 33.3% since two answers are identical (effectively there are only 3 answers)?

no.

The answer is 50%.

1 out of 4 answers = 25%

Because 2 answers are the same && the answer is 25%, the random selection of 25% as the correct answer is 2 out of 4 which resolves as 50%.

Simplified:

The question is: “If (hypothetically, you don’t must to do this) you choose an (an = one) answer to this question at random, what is the chance (The chance is 1/4 = 25%) you will be correct?”

A) 25% —————— Correct chance

B) 50%

C) 60%

D) 25% —————— Correct chance

Then, do you have a 50% to be correct. Choose B) 50%.

I’m astonished. “Randomly” is NOT interchangeable with “With equal probability”. For example, the values in craps (a game involving rolling two six-sided dice and summing the values on each) are determined randomly, but not all values are equally likely; rolling a 7 is 6 times more likely than rolling a 2.

The proper answer to this question is “What’s the distribution?”

i like your thinking but ultimately I think its just you answering the question with another question.

Because the original question didn’t give enough information.

It’s like me saying “If you pick a number at random, what are the odds that you picked 1?”

Picked a number from what set of numbers? The question needs more parameters that have not been given.

The assumption is that the distribution is flat.

Good idea, but craps isn’t random – it’s the result of two random events combined. This is just one. Random is by defintion that all possible results are of equal probability. a “fair” coin is random, a 60% coin is not. Random describes the distribution.

The only way to win is not to play.

It is just an endless loop haha. At first you think 25% is the correct answer. But there are two of them. So it is 50%. But that means that B is the correct answer which means you only have 25% change you got it right, where you can conclude it is either A or D, and the 50% came back again……

aka, there is no answer to this question.

This is a trick question. It’s not about statistics, it’s about quantum mechanics.

If you observe the answer, the answer changes states.

+1

Most excellent!

Like, as they say.

But, for those that are trying to solve it under the assumption that you select randomly with equal probability, you would do better to write the question more clearly.

If you were to select a letter randomly from {A,B,C,D}, with equal probability, what is the probability P that the number associated with that letter is equal to P? The last portion is what “that you will be correct” means.

With equal probability means 25% chance of getting A, 25% chance of getting B, etc. The chance of getting A OR D is 25%+25%=50%.

There are exactly three possible values you can get from your selection: 25%, 50% and 60%. The probability of getting 50% is the probability of getting B which is 25%. Same with 60% and C. The probability of getting 25% is the probability of getting A OR D which is 50%.

Thus there is no correct answer.

Those advocating that “you aren’t forced to choose it randomly” are actually removing one of the elements (that the number is equal to P) of the question that is not removable. We know you aren’t actually choosing it at random (if you were, then 1/4 of the people on here would be choosing 60%), but the question is “If you did”. If I were to ask “If tomorrow is Friday, what is today?” You don’t have to wait until Thursday to answer it, but Thursday is still the only correct answer.

Here’s a better wording, that removes “you” entirely:

If a letter is selected randomly, with equal probability, what is the probability P that the number associated with that letter would equal P?

P(“25%”)=P(A OR D) = P(A) + P(D) = 50%; “25%” != 50%

P(“50%”)=P(B)=25%; “50%”!=25%

P(“60%”)=P(C)=25%; “60%”!=25%

I think the more important aspect of your wording is that you make explicit what “correct” means. Now it should be rather obvious that no answer is correct, and that we get a true paradox if we associate choice C with 0% instead of 60%, just as dedoco has pointed out right in the beginning of the comments.

Simply none,

If A or D is the correct answer then you have a 50% (can’t be A or C then)

If B is the correct answer you have a 25% (but B marks 50%)

If its C same logic as B.

If its D same logic as A.

So no one is the answer.

The question is invalid. It’s asking a hypothetical question with your hypothetical self as the person picking an answer at random. However, since you’ve seen the answers, and your hypothetical self is an extension of you, your hypothetical self cannot be trusted to be unbiased and pick a random number. :)

You’ve just destroyed John Rawls’ life work.

I’m disappointed, it’s really simple. B(n=4,p=1/2)=1/2. https://en.wikipedia.org/wiki/Binomial_distribution

Sum(P(A)+P(B)+P(C)+P(D))=(1/4(1/2(1)+1/2(0))+(1/4(1/2(1)+1/2(0))+(1/4(1/2(1)+1/2(0))+(1/4(1/2(1)+1/2(0))=1/2

Let’s think about this very simplistically. The question is: “If you pick an answer to this question at random what are your chances of being correct?” Let’s not think about the multiple choice at the moment, I think thats the red herring used to confuse you. So, instantly the answer to the question on it’s own is going to be either correct or incorrect. So I believe the answer is B) 50% because I don’t believe this is a math problem it’s a logic question and your chances of being right or wrong is going to be 50%. Granted I’m just a layman and could be completely wrong but it’s the answer that makes the most sense to me.

Logic is math. Math is logic. If you take a (proper) college Logic course, I promise you’ll be doing math and using symbols.

The answer is 50%, but it is NOT a letter. It does not ask you to do anything with the answer; rather it asks you what is the answer. As serveral of you have said if different words if you have 4 choices with two correct you have 50% chance of getting it. Where I am confused is why are you trying to reapply the answer to get a new answer.

I think the confusion comes from the fact that while probability of selecting 25% is 50% (1/2), the probability of selecting 50% is still 25% (1/4)

Because the problem requires you to do so. See my comment above.

Rolling with dice is uniform distribution. Rolling with 2 dice is sum of two uniform distribution. We select 1 of 4 options at random in this case so it means equally. You didn’t select the D answer more then the A. So we can agree its Uniform Distribution.

I got a D in statistics in college – anybody let me see their paper?

The number of answers to this question is infinite. Therefore “i feel” every answer will be not correct.

You are so dumb. The ir is no answer, is a LOOP or also called as a PARADOX.

This question is a PARADOX. If you pick A ( %25). That will lead you to B as an answer, but picking B as the answer will lead you again to (%25) ’cause B is just a 1/4 of the answers, so you will pick A or D again, and again and again and again.

Read something about PARADOX and Go to school again plz.

:O you are right!

you have the definitive answer.

37.5% b) and c)

The question starts: If you choose an answer to “this question”

This question in the structure of the sentence represents the question being posed. Although we have a reference to the question we do not know the contents of the question. Like in programming I can have a reference to an object but I may not know what is in that object.

For our purpose “this question” could be one of all the possible questions that can be asked. So for example if the question being referred to was ‘how many atoms are there in the universe?’ then none of the probabilities provided would be correct.

This is my logic. If “this question” could refer to any question that could possibly be asked and the answer I give could be any answer that I could ever give then the chances that the answer I give will be correct will be

1 ( Answer I chose) / all possible answers I could choose

In the question it does not specify that we need to choose from answers A, B, C or D so I think the above answer would still be valid.

: )

possibilities

a – 25%, b – 25%, c – 25%, d – 25%

If correct answer is set [a,d] = [50% right, 50% wrong] 50% chance of this being correct set = .5 * .5 = .250

If correct answer is set [b] = [25% right, 75% wrong] 25% chance of this being correct set = .25 * .25 = .0625

If correct answer is set = [25% right, 75% wrong] 25% chance of this being correct set = .25 * .25 .0625

.250 + .0625 + 0.0625 = 0.375

37.5% chance

———–

Programatic Proof Just For Sanity Check:

public class Main {

public static void main(String[] args) {

int iterationsTop = 1000;

int iterationsBelow = 100000;

double pctsTtl = 0;

for (int y = 0; y < iterationsTop; y++) {

String[] values = {"25%", "50%", "60%", "25%"};

int won = 0;

int iterations = 100000;

for (int i = 0; i < iterationsBelow; i++) {

String correctAnswer = values[(int) (Math.random() * 4)];

String checkValue = values[(int) (Math.random() * 4)];

if (correctAnswer.equals(checkValue)) {

won++;

}

}

//System.out.println("won=" + won + " / " + "iterations=" + iterations);

double wonPct = (double) won / iterations;

pctsTtl += wonPct;

}

System.out.println((double)pctsTtl / iterationsTop + " is the average percentage over " + iterationsTop + " trials.");

}

}

OUTPUT:

0.3750187399999998 is the average percentage over 1000 trials.

(I’m just going to go with that this is a trick question though, just as a hunch lol)

I pick E – All of the above.

Obviousness is learned. I think most of you need to go read Godel Escher Bach by Douglas Hofstadter, immediately.

Ian, there is 4 options to pick up, 2 of them are the same. But this 50% value has his own counting on the possibilities. (25%, 25%) ar just a part of the group that we can pick up, I think that the problem is asking us which percent of succesful we can achieve on these options. So there is a parallel question, cause more of them can be true. I hope clear up something.

LOL … it is indeed a trick question that I will venture stumps every statistician! Look carefully: the question does not ask what percent of a population of querrants would be right, it asks for the chance that YOU will be correct. That chance is 50%, you will be right, or you will be wrong, there are NO other options! If YOU try again, again you will have a 50% chance of being right, because the outcome set has not changed and the two trials are not dependent, they are unique; even once you have found the right answer, if the answers were hidden (they say choose at random, so we don’t want to be influenced by the contents ergo double-blind us to the answers behind the alternatives) it says no where that two answers could not be right, so it is still 50:50!

“That chance is 50%, you will be right, or you will be wrong, there are NO other options!”

And you have a 50% chance to win the lottery. You either win it or you don’t.

And there’s a 50% chance you’ll understand where your logic is wrong.

N8 is right on this one, mrG. The two outcomes are NOT EQUALLY LIKELY. Imagine a jar with 99 one-dollar bills and 1 100-dollar bill. The probability of picking the 100-dollar bill at random is 1/100, not 1/2, even though there are only two outcomes, viz., you pick the 100-dollar bill or you don’t.

Eduardo is Right.

is a Paradox. http://en.wikipedia.org/wiki/Paradox

Is funny how many people try to get the answer even whit codes.

sorry for my english.

e) 20%

A winner.

commenting, maybe I’ll win a book

Here is the original problem:

If you choose an answer to this question at random, what is the chance you will be correct?

a) 25%

b) 50%

c) 60%

d) 25%

In order to understand this problem, it is best to frame the question and its possible answers differently. The question is really asking: If you choose an answer to a question from four possible answers, of which two answers are the same value, and any answer may be correct for each instance of choice, what are the odds that you will chose the correct answer? For example:

What is the answer to this question?

a) AA

b) AB

c) AC

d) AA

There is a 2-in-4 (50%) chance of choosing AA. There is a 1-in-4 (25%) chance of choosing AB. There is a 1-in-4 (25%) chance of choosing AC. So, do not treat the question or the answers in a literal way. Look at them hypothetically. You have to imagine that a correct answer may be present in each of the possible answers in order to calculate the odds. For example:

What is the capital of France?

a) Paris

b) London

c) New York

d) Paris

What is the capital of France?

a) London

b) Paris

c) New York

d) London

What is the capital of France?

a) New York

b) London

c) Paris

d) New York

Now, I am not a mathematician, nor a statistician. I am just a dumb raconteur. So, I cannot actually do the math. But logically, this seems to make sense as an explanation for how this question works. Although, to be honest, I’m not sure what the answer is. If this were on a standardized test, I would chose b) 50%. But I would be guessing.

I retract my guess. The odds for randomly choosing the correct answer does not have to be one of the answers presented. According to what I wrote (above), you can still calculate the answer and come up with a percentage. The answers listed in the original problem are illustrative of what I outlined: Simply that two answers are the same and the remaining two answers are different. Again, I suck at mathematics, so someone here has probably already calculated the correct odds.

Dis town ain’t big enuf fer da 2 of us, padna

There is something wrong there. With 4 possible aswer: the chance i will be correct can only be: 0 , 0.25, 0.50, 0.75 or 1. There aren’t any 33% or 37.5%.

The answer to the question being asked is 0.

The question asks “If you choose an answer to this question AT RANDOM, what is the chance that you will be correct?” (emphasis mine). If you answer any 4-option multiple choice question randomly, there is a 25% chance you are correct. In this case, there are two ways to select 25% RANDOMLY, so this question has a 2/4 = 1/2 = 50% chance of RANDOMLY choosing 25%.

Now, if you were to answer RANDOMLY, that means there would be a 50% chance of your being correct. However, you’re answering it with logic, NOT RANDOMLY, and thus select B and get full credit. QED

It’s a kind of paradox. It has not right answer.

The chanse of being correct is 0%. Since that is not one of the options in the question, this is not a proper multiple choice question.

Like this:

What is the Capitol of France:

A: London

B: Berlin

Now how did I get 0%?

Well just do what the question asks:

Choose one at random and see if you where correct.

The propability of picking 25% (A andD) is 50%

So if you picked a or d you where wrong.

The propability of picking B or C is 25% so if you picked one of those you would also be wrong.

Hence, which ever answer you pick you will be wrong.

The propability is 0% and the question does not have the correct options.

If you replace the C with 0% however..

If you pick C you will still be wrong and hence be right.. :)

The answer is 33.33% It’s NOT saying one of the choices is a correct answer.

Since there are only three choices to make

-25%

-50%

-60%

You have a 1/3 chance of getting the answer right.

You choosing from 4 options. How the hell did you lost the 4th? You cannot just dump it even if its the same.

You are correct that there are 3 different options but there are 4 choices still: A,B,C,D. The teacher wont tell the student that the A and D the same so you cannot choose the D. You can still choose both even they are the same so you have double the chance you are correct . Again the possible chances: 0,25,50,75,100.

It’s 50%. If you choose randomly an answer, the probability is 25% of getting the right answer. Then, you see how many options are saying 25%, there are two of them, so you got a 50% chance of getting the right answer. So, the option is B.

Initially I was supporting the 0% option, I confess… ;)

Bruno – What you said is correct, and it is one of many conditionals (i.e. if you pick X, then probability of being correct is Y). But. remember the question has asked you to choose at random. So, imagine that the correct answer is one in this set {25%, 50%, 60%} and there are four possible way to pick randomly {A,B,C,D}. Make a list of all combinations, such as I picked A and the answer is 25%, therefore I am right. These look like this:

P (choice A & answer A) = correct

P(choice A & answer B) = incorrect

P(choice A & answer c) = incorrect

P(choice A & answer D) = correct

…..

Add all of these up (16 total combinations) and see how many are ‘correct’. The answer is 6/16, or 37.5%. This makes sense intuitively – your probably of landing on the right answer is a little better than 1/3 when picking A,B,C, or D randomly.

Yeah, I see your point Pete. But in that case, you’re working with multiple correct answers scenario. I mean, you can’t assume all are right and play probability on that.

I’m on the side of “0%” for this one. It is the only answer that matches its own frequency in the options presented.

But that suggests another variation. Same question, but the given choices are:

a) 25%

b) 50%

c) 50%

d) 75%

Now, a), b) and c) all appear correct at first blush, but if they are all correct, doesn’t that make d) correct instead? Here we go again…

Cant be answered!!

Hmm.. I believe is that this question is clearly a paradox…..It is a question in a question.

And to answer a question we must have the correct answer.

:

So if you look at the options: 2 are similar 2 unique so totally 3 unique (4 in all)…So now if you assume that one of the options is correct:

your chance to randomly guess the correct option will be 1/3(0.5)+1/3(0.33)+1/3(0.33)=38.888%

Here 1/3 is the chance of that unique option being correct…. and the value in the bracket is the probability of picking the 3 unique options

But no option equals 38.888% so our initial assumption is wrong. None of the options present the correct answer.So if we have to CHOOSE then we will always be wrong. So for the probability to exist there should be a right answer but as there is none the probability is NULL.(0%)

If one of the options was 0% then the options would also be paradoxical :D

I’m going to insert my own junk math here.

It doesn’t matter if there are 2 choices or 2,000 choices. There are only two possible outcomes: You are either correct, or you are not correct.

Therefore, the answer is 50%.

Clearly it is 99% right now, at least according to the news.

There is more than one way to answer the question, depending on how you read it.

To me, “answer” means “the stuff next to a letter choice.” So if I were to choose one of the 3 possible “answers,” the odds of being right SHOULD be 1/3. So, strictly reading the question, “the chance you will be correct” is 1/3, even though this isn’t listed in the choices.

You could also interpret “answer” as “one of the letter choices, regardless of the stuff next to it.” Say, for example, I program a computer to accept only A as the “correct” answer. In that case, you have a 1/4 chance of answering correctly – you’re wrong if you pick B, C, or D because I was looking for A.

Or you could read “answer” as “any letter choice that has the right stuff next to it.” Selecting at random, you should have a 25% chance (the right stuff) of answering correctly. Two letter choices have the “right stuff” next to them, so now we are looking at a 50% chance of answering correctly, which corresponds to letter choice C, which takes us back to a 25% chance of answering correctly because C is one out of four choices, and so on . . . In this case, the question is paradoxical and thus faulty.

Most tests say “choose the BEST” answer. By that logic, 0% is the best answer in my opinion.

Furthermore if I can “choose an answer at random” who’s to say I can’t choose “Martin Luther King” or “Teenage Mutant Ninja Turtles”? The question doesn’t say “from the choices below.” Zero percenters FTW.

If randomly choosing the correct answer is 50% (which I agree), what is the chance of randomly choosing the incorrect answer?

nobody said you have to select one of the options below, just that your answer most be random, so whatever you answer is correct, so the change is 100%…

There is no question, so there’s no answer. But if there were a question, the answer would be e) “Saskatoon

This one reminds me the best philosophical question ever: “What do you think about this question”… of course one student got it right … he just answer: “What do you think about this answer!” xD

I think B is the correct answer.

The tests, normally, only have one correct answer.

Option A and D are the same. You can only choose one option. So A and D are not correct.

The options that have we have to choose randomly are to B and C.

You have only 2 answers, the chance will be 50%

Luckily, there are some tests, if graded by a human and not a machine, in which the correct answer is to be smart enough to recognize the invalidity of the question and write down “none of the given answers can be correct and here’s why….” followed by a good explanation. This could be one such test, since if the answer is 50% it’s really 25% and if the answer is 25% it’s really 50% and you spin forever in this loop.

What would really make this a mind-blowing mess is if answer (C) had been 0%. Once you figure out that (A),(B), and (D) cannot be correct then the right answer would be (C). But then because there’s one possible right way out of four to answer it, it wouldn’t be 0%, it would be 25% and you’d be back to the answer being A or D again and restart the loop that way.

The problem is ambiguous : do we select one or more answers ?

If we can only chose one answer, then the correct answer is 0% since none of the answers are correct (100/3 = 33.333… %), essentially we consider A and D to be a single answer.

However with multiple answers allowed, there is a paradox that is created :

The right answer is A AND D, for which the probability is 25% * 25% = 1/4 * 1/4 = 1/16 = 0.0625. The answer is therefore 6.25 %. But since the choice of answers doesn’t show 6.25 %, we don’t have any opportunity to answer that so the answer is 0%.

In any case, the answer is 0% and we can’t answer this question.

That said, I’m not a mathematician and I’m probably wrong (it’d be interesting to calculate the likeliness of that ?).

After 10 minutes of thinking and wasting my time

I’m gonna go on a limb and say C:60%

I may be wrong, but here’s my reason why.

Randomly pick one = 50% chance it’s the A or D, if that is the right answer, most people would then choose B:50% not randomly. It is a sort of loop, but I guess both answers are right. Or should i say, all three of them! Therefore answer is 60% because it’s closer to 75% =D I win.

Seriously! No one considered 60%? If the guy was smart enough to make this question, why would 60% be there randomly? Hmm? ok maybe it is random. This has gotten too far out of hand! Time to play video games.

Its 25%

You cannot assume that the value of the answer has anything to do with its correctness. Maybe correct is D) 25% and not A) 25%

Let me revise my thinking here.

B) is the correct answer and you have a 25% chance of picking it.

100%. The answer is outside the namespace of this question. All you are asked to do here is to pick between A B C or D at random. If you accomplish that task, then no matter which letter you chose you got it right. The percentages have nothing to do with it. Could be fruits, or vegetables. Boom.

Agreed! You did it, so you’re correct!

This is an incomplete question. The first part of the question says “if you choose an answer to “THIS” question at random.” What is the “THIS” question? Is it the second part of the question, which is, “what is the chance you will be correct?” If so, first, we need to know the correct answer to original “THIS” question. Then only we can determine whether the correct answer is present in the answer choices.

An interesting variant of this problem would have the answers choices of:

a) 50%

b) 25%

c) 75%

d) 50%

Why does everyone assume the answer chosen randomly must come from the set below the question? That’s certainly not how the question’s worded. I choose randomly from the uniform distribution [0, 1], and my probability of getting it right is 0%. Now, if I answer it, I might write 0%. If the problem asks me to show my answer by circling one of the potential answers from below, I’d go to the teacher and show my work and say that my answer is not in the set.

If the wording were changed, I’d probably go a little loopy like everyone else and mention the liar. But the wording is a perfectly good question with a misleading blob of unrelated information below it.

The question is phased wrong.

Probabilities are expressed as fractions (1 / 2, 1 / 4, 3 / 4), as decimals (.5, .25, .75), or as percentages (50%, 25%, 75%) between 0 and 1. Chance is only expressed as a ratio, such as a 50/50 chance.

And my comment was phased wrong too… ;-)

Assuming one of the four options is correct, the random chance is 25%. Since there are two choices of 25%, the chance of picking one of these is 50%. The odds of choosing 0% at random is 25%. Hence the odds of choosing the correct answer at random (50%) is 25%.

33%

Let me elaborate, the question states that an answer is randomly selected, then asks the probability of that randomly selected answer being correct. There are only three viable options: 25%, 50%, and 60%. The probability of picking one of these three options is 33%. People need to stop focusing on the 50% chance of choosing 25%. That is not the question.

u r wrong brigid. its 25 or 50

There are only three viable options, but not the same chance to pick each randomly, because you have 2 possible options with 25% A and D therefore have more chances to be piked than 50% (B) or 60% (C )

Come on peoples, this question is easy.

Ignoring what the answers are, there are 4 of them, and randomly answering a 4 answer question would give you a 25% chance of getting it right.

Looking at the answers, both A & D are 25%. This means the odds of you randomly picking 25% (the correct answer) would be 50%.

So the answer has to be B, 50%.

I should rephrase that 3rd paragraph.

From the answers, the chances of selecting the possibly of randomly answering a 4 answer question (25%) is 2 in 4, or 50%. So the answer is B, 50%

1/3 – There are three distinct answers, the chance of randomly picking the right one is thus 1/3. The fact that one of them is stated several times is irrelevant. Imagine each of the three distinct responses had been written down 1,000 times – it would still remain just 3 distinct answers, hence, the chance of being right is 1/3.

Actually no – it depends: if “25%” is the correct answer then there is a 50% chance of randomly picking the right one. If the correct answer is one of the other two options, the chance of randomply chosing the correct one is 25%.

It’s an invalid question therefore it can’t be answered.

isn’t it 50%? As you have a 50% chance of a right answer and a 50% chance of a wrong answer. So if you picked an answer at random you would either fall into the half that is right or the half that is wrong. 50%?

Of course B. You have a 50% chance of choosing the correct answer which is 25%.

It’s 0%… If you choose at random, your chances of getting the correct answer are 0%

Because two of the choices are the same, I believe we all are in agreement that A or D can be the correct answer. So the correct set of answers are [A OR D], [B], AND [C]. The correct answer is the probability of picking [A OR D], [B], OR [C] effectively reducing your out come to 3 choices [A OR D], [B] OR [C] instead of [A], [B], [C], [D]. So the chances of you being correct is 33% (1 out of 3 possible out come).

This is similar to flipping a trick coin with both sides being the same (let say heads). Even though the coin has two sides, the possible outcome is only one. So the chances of getting head is 100%.

Oh this is so sweet!

All the spinning is due to the “Best statistics question ever” framing.

But that’s not on the board!

What if instead, the title above the photo was “Best social psych question ever?”

Then you’d be in the realm of Zimbardo and all the other peer pressure experiements …

As such, it’s a brilliant piece of misdirection which everyone (I think, I haven’t read every answer) has falling into.

Here’s another option which gets you out of the knot:

Hide the “answers” and the “title”.

Now read what’s left on the board.

With this framing you are no longer the test taker you are the test creator.

As such, it becomes a statement rather than a question:

“The questions you ask colour the world you experience.”

Let’s say you choose the answer “BLUE”.

Is that the correct answer?

Since in this framing, you are the test creator only you know if it is the correct answer, a correct answer, a random answer, the most evocative answer …

Once again, the question tuns into an introspective “opening”.

As such, this is precisely the operating mode of a koan.

What is the nature of choice?

Can you actually make a random choice?

Why do so many folks feel constrained to “options” presented below the “?” ?

If you had to put together a team to solve really hard, complex, policy questions it might be interesting to put everyone in a room with this one the board a see what happens.

I’ve only read a few comments and most people say the obvious answer is 25% but since there are 2 of them it is the same answer as 50%, and so on. Well since two 25% are 50% then the one 50% is 2 choices at 25% thus giving us 5 options. After that math we only have the ability though to choose between what’s listed but since 25% or 50% is correct the actual answer to this question is C) 60%

C) 60% is the answer. Read above

It’s an invalid question :

1) we can’t behave “randomly”

because as we read question and choices, we acquire data about the probability distribution so it’s a bias for randomization.

2) choices are not independent rather depending on conditions.

but these conditions are contrary with each other. so question built a paradox.

though if we can an option to determine an answer out of question/choices logic then the answer would be %0 because of contrary conditionals.

What is confusing is the difference between the selection (A,B,C,D) and the value. (25%, 50%, 60%).

Correct answer is 50%. You have a 50% chance of selecting A or D which is the probability of randomly selecting the correct selection B.

“The chance of being correct” means that you must know the right answer from the start, in order to solve this problem. But no one knows what the correct answer is. Some people assume it’s written in one of given options; some don’t. All probability reasoning above is done with some right answer already in mind (even if unconsciently). At the end, one fundamental truth remains: in order to reason about chance, you must know the answer beforehand. And you don’t. If you put a) Tomato, b) Carrot, c) Lettuce, d) Tomato could you answer the question? IMHO no, because I don’t even know which vegetable is the right one. They could even be all wrong, AFAIK. That piece of information hasn’t been given, and therefore, I cannot answer the question. But once I know that, then I can reason about chance to some extent (and I say to some extent because the set of possible answers hasn’t been given either; although some assume the set if based on all the available options, this might not be the complete set, and that’s what leads some people to state other values beyond the ones written).

I know the answer, but don’t want to spoil it for everyone. The only thing I can say is that the question is valid, all information is given, and it is not a paradox. Good luck!

P(T|A)P(A)+P(T|B)P(B)+P(T|C)P(C)+P(T|D)P(D)=5/12

(Teacher’s rant)

I’m going to argue that this is the worst statistics question ever…for a class exam. Teachers often make the mistake of trying to test for cleverness instead of testing for hard work. “Trick” questions often have too much of a disconnect from content delivery and learning that actually happened during the course. Teachers who fail to keep track of their own content throughout the semester will fall back on trick questions to challenge students, instead of testing them for how well they studied the content that was actually delivered.

I agree Karmella! However, I don’t know if this was in fact an exam question, or maybe just a bonus question, or simply an illustration? I find that in statistics courses it is already so difficult to get students’ interest, that resorting to such clever “riddles” is the only way to get them to pay attention. Hopefully this wasn’t really on any exam for marks!

it’s just a normal question. as i said.just apply complete probability formula and get the answer.

this question is hard for people,because they don’t totally understand the meaning of that formula.

I agree i with Karmella as well, but any question that can stir this much discussion is a good one! I think that mental engagement for all students is the key challenge in any math class, and this one certainly makes my brain hurt just a little.

Not an exam. God forbid we tested people for their ability to think and apply the knowledge they acquire in classes to novel situations. We definitely should stick to making sure they can mindlessly execute the tasks thought to them. All hail rote learning!

When I first read the problem, I was laughing out loud. I saw the paradox right away.

Nonetheless, I believe there are four correct answers to this problem, but only 3 of them are given.

A and D are both correct answers, because choosing any answer at random among four answers, regardless of their values, assuming there is at least one correct answer, will always give a probability of 25%. In any event, if A and D are both correct, than that also makes B a correct answer, because, among 4 answers, if 2 are correct, then the probability of picking either of those two answers is 50%. It is possible, however, that none of the given answers is correct, in which case, there are three possibly correct answers which are given, and a fourth possibly answer which is not given. In this case, there would be 5 possible answers, only 4 of which are given, and only 3 of the given of which are possibly correct. That means that any four out of 5 possible answers could be correct depending on the circumstances, so the probability of guessing a correct answer would move up to 80%. The problem does not state, after all, that you must choose only from the available choices for a correct answer.

The real problem with this problem, of course, is that it is poorly written. No well-designed math or logic problem should depend upon true/false statements about the data itself specified to alter the problem. Such a problem lacks adequate data sufficiency, and is therefore not a valid logic problem. This is such an example.

The chance is 1 out of 4. We don’t know the question so we don’t know the answer. Any of them could be correct, so like any other question with 4 answers, we have a 1 in 4 chance of picking the right answer.

Riddle me this: What the difference between “If you choose an answer from one of the following choices” and what’s actually written on the blackboard? Thusly. we can choose any answer. But if we chose 0% then our choice isn’t random, as is written on the same blackboard. It also clearly and undeniably says “to *this* question” not “to a question.”

Therefore, I choose towel day.

“If you choose an answer to this question at random, what is the chance you will be correct?

A) 25%

B) 50%

C) 60%

D) 25%”

The question is in natural language, which is somewhat more fuzzy than symbolic logic, so let us state some assumptions.

Assumptions:

A1. The question referred to as “this question” is the question written on the blackboard. That is, the question is self-referential.

A2. The set of answers that may be chosen is the set of answers given: [25%, 50%, 60%, 25%].

A3. The mechanism for random selection is uniform, (as rolling a fair die, or pulling tiles from a bag).

A4. The set of answers contains the correct answer.

A5. There is one correct answer.

A6. We are given enough information to solve the stated problem

By assumption A2, we make the following inference:

I1: There are 4 possible answers.

With that, and assumption A3, we infer the following:

I2: The probability of selecting any answer randomly is 25%.

Now, with that, and assumption A5, we infer:

I3: The correct answer is 25%. (Assuming there is one correct answer (A5), and there are 4 answers (I1), each equally probably (A3), then the chance of selecting the correct answer is 1/4.)

So far, this is pretty uncontroversial.

But, when we examine the answers, must either discard inference I3, or assumption A5, because 25% occurs more than once in the set.

If given that 25% is the correct answer, then it cannot be the case that there is only one correct answer.

If given that there is only one correct answer, then it cannot be the case that 25% is the correct answer.

Let’s look at the assumptions one at a time:

If we discard assumption A1, we are cannot proceed unless the question referred to is provided. A non-starter, and contrary to assumption A6.

If we discard assumption A2, we must assume some other set of answers. Which is the correct set of answers to pick from? This violates assumption A6. (For the sake of argument, let’s assume it’s the set of all numbers. The chance of picking any one number from that infinite set is a limit of 1/N as N approaches infinity, which is 0.)

If we discard assumption A3, we cannot assign a reliable probability to any answer. This violates assumption A6

If we discard assumption A4, we may as well assume !A4 (The set of answers does not contain the correct answer.). With that assumption, we arrive at an answer of 0%. (Chances of selecting an answer not in the set from the set is 0%; this follows from the definition of sets.)

If we discard assumption A5, we can proceed as follows.

From A2, and A3, we find

I4. The chance choosing an answer of 25% is 50% (1/4 + 1/4). Now, we might infer:

I5: The correct answer is 50% (which is based on I3, 25% is the correct answer, and I4, the chance of selecting “25%” is 50%). BUT, I3 is based upon A5, which we have discarded. We should therefore discard I3 as well, and having discarded I3, discard I5, which is based upon I3. (Also, there seems to be a contradiction: we have inferred that both 25% and 50% are the correct answer. The chances of selecting 25% or 50% is 75%.)

If we discard assumption A6, the answer could be anything. We cannot proceed.

I think the most reasonable answer is 0%, or not enough information.

Dammit … I totally agree. (Hate it!)

In my opinion this was the most clarifying answer so far.

There is no correct answer to this question because the question asks you to choose the answer at random.

There is no right answer.

If the correct answer was 25%…

then you’d have a 2 in 4 chance of randomly picking 25%…

then the correct answer would be 50%…

then you’d have a 1 in 4 chance of randomly picking 50%…

then the correct answer would be 25%…

then you you have a 2 in 4 chance of your brain melting…

a fish. or 42.

The correct answer is 0%.

The question never really said you *had* to pick one of those answers, it only says *if* you did!

In any case, it’s a great catalyst for debate! It would be a great question to post in an online course, but you may need a dedicated server and adjunct to manage it…

The question is asking the chances you have to pick the right answer given 4 specific values; it is not asking you to select an answer from the list. So this is just a matter of probabilities.

To simplify this and clarify it, I’ll use A, B, C and D

if we have:

1)A

2)A

3)A

4)A

then chances of picking the right answer is 100%

.25*1 + .25*1 + .25*1 + .25*1 = 1

If we have:

1)A

2)A

3)B

4)B

then chances of picking the right answer is 50%

.25*.5 + .25*.5 + .25*.5 + .25*.5 = 0.5

If we have:

1)A

2)B

3)C

4)D

then chances of picking the right answer is 25%

.25*.5 + .25*.5 + .25*.5 + .25*.5 = 0.5

and now…. the answer to our question

If we have:

1)A

2)A

3)B

4)C

then chances of picking the right answer is 37.5%

.25*.5 + .25*.5 + .25*.25 + .25*.25 = 0.375

So, the answer to the question above is 37.5%

This does not make sense. Are you saying that A is the correct answer? If so, the second case and the last case are equivalent. You are also assuming that the right answer is in the set [A,B,C,D].

The answer is not any of the options. Do not assume it is a multiple option question. The question is what is the probability of picking the right answer for the values shown, not select the right answer from the options. I just used letters instead of number to avoid confusion.

In other words, in my example above, I replaced 25%, 50%, 60% with A, B, C.

true

Here’s one more

If we have:

1)A

2)A

3)A

4)B

then chances of picking the right answer is 62.5%

.25*.75 + .25*.75 + .25*.75 + .25*.25 = 0.625

This is the same as rolling an unfair 3 sided die. 50% is as equally likely as 60%. 25% is twice as likely as either 50% or 60%. If p is the probability of getting a 50% then P(50%) = P(60%) = p and P(25%) = 2p. All probabilities sum to 1, p + p + 2p = 1. That means 4p = 1 so p = 1/4 or 25%. Out of four choices the right answer occurs twice so your chance of getting that is 50% if you picked randomly.

No, you can only get 50% if you have 2 different values each one occurring with the same probability, in my example above that is:

1)X

2)X

3)Y

4)Y

then chances of picking the right answer is 50%

.25*.5 + .25*.5 + .25*.5 + .25*.5 = 0.5

not if you already know the answer is 25% and after that randomly select from (a,b,c,d). Each is as likely at that point but we know 2 out of the 4 are correct and 2 out of 4 are incorrect.

Where can I get a 3 sided die? :)

I think a 3 sided die could be made in the shape of a cylinder, so that the probability of landing on either end and the probability of it landing on its side would all be equal.

I do like the theory of 42……but I think it’s B) for the following pseudo-mathematical reasons:

First way to get to (B), state your assumptions. Assume that a) b) c) and d) are all the “guesser” sees. You see the values associated with those choices. You know that there is a 25% chance of blindly guessing one out of four. You see that two choices have that as a possible answer, therefore the unnamed dummy has a 50-50 shot at getting the answer right.

Proof the Second: believe it or not, involves even less math….. So back in high school, I was taught how to take tests well (thank you, no child left behind, or the NYS regents exams, or the SAT’s, yadda yadda yadda……). So how would you solve this assuming you’re the guy guessing, but you can see the answers you’re guessing at. You see that two are mathematically equivalent, which either means you can eliminate both of them, or ask the proctor and try to get her to tell you which one is RIGHT right, as opposed to just right. Lets go with option #1, that means that there are only two to guess from, one divided by two is 0.159154* pi, round up, you get B) 50%

At least according to my years of SAT test prep in the Guessing-the-Correct-Answer-on-Multiple-Choice-Tests-Even-if-you-have-Little-or-No-Understanding-of-the-Topic-in-Question Academy.

My actual rationalization for this answer is that when confronted with a paradox, if you’re forced to choose an answer, choose the most consistent -ish one, and usually, the answer that your gut instinct likes (on first glance) is wrong. The one you know has to be right after you spend an hour thinking it over is also wrong. It’s usually the one with the perfect mix of unlikeliness without being too outlandish. I am aware how very depressing the whole process is…..I guess the idea that a series of multiple choice questions administered in a high stress environment can accurately asses an individual’s true intellect is the real paradox

I had to repost as my chosen values (A B C D) seem to create confusion due to fact that the multiple options in the question above are using the same letters as identifiers. So here it is with different letters:

The question is asking the chances you have to pick the right answer given 4 specific values; it is not asking you to select an answer from the list. So this is just a matter of probabilities.

To simplify this and clarify it, I’ll use W, X, Y and Z

if we have:

1)W

2)W

3)W

4)W

then chances of picking the right answer is 100%

.25*1 + .25*1 + .25*1 + .25*1 = 1

If we have:

1)W

2)W

3)X

4)X

then chances of picking the right answer is 50%

.25*.5 + .25*.5 + .25*.5 + .25*.5 = 0.5

If we have:

1)W

2)X

3)Y

4)Z

then chances of picking the right answer is 25%

.25*.5 + .25*.5 + .25*.5 + .25*.5 = 0.5

and now…. the answer to our question

If we have:

1)W

2)W

3)X

4)Y

(think W is “25%”, X is “50%” and Y is 60%”, and remember these are just values, *not* any of the possible answers to the question)

then chances of picking the right answer is 37.5%

.25*.5 + .25*.5 + .25*.25 + .25*.25 = 0.375

So, the answer to the question above is 37.5%

This is incorrect. The probability of picking 1 of 4 answers is 25% (assuming a uniform distribution). If you get to pick 4 times, you get .25+.25+.25+.25=1, or a 100% chance of getting the correct answer. If all the answers are the same, (and correct!) then you have a 100% chance of getting the correct answer. That’s your first case, as far as I can tell.

The second case, you have a 50% chance of selecting W, and a 50% chance of selecting X. If one of those is the correct answer, you have a 50% chance of getting the correct answer by chance.

In the third case, you have a 25% chance of selecting any of the answers. If one is correct, you have a 25% chance of getting the right answer by chance.

The last case, now the chance of getting the right answer depends upon which of the answers is correct. If it’s W, you have a 50% chance of getting it. Otherwise, you have a 25% chance of getting it right. I think you are trying to apply the probability that an answer is correct, but I don’t think that’s valid. There is (we assume) a single, correct answer. There’s no 50% chance that W is the actual answer. The answer is the answer, and it has a 100% chance of being the answer. The other answers have a 0% chance of being the answer. So, the .25 terms in your last equation should be 1 for the correct answer, and 0 for the others. Calculating the final probability depends upon the correct answer.

And you are still assuming that the correct answer is one of the 4 answers given, and don’t think it actually is one of the givens.

Sorry but this is not what I’m saying. Re-read my answer. The chance to get the right answer if all 4 values are different is 25%, that would be when the options (in my example) are W X Y and Z. And the chance to get the right answer when 2 of the 4 values are the same is 37.5. That is the correct answer. And nowhere in my reply it says that the answer is one of the given options (A B C or D) but exactly the opposite, that the answer to the question is *not* one of the multiple options.

Please explain, in detail, how you are coming up with this 37.5% number. What are these terms meant to represent?

I think it’s clear from the question both that 1) the set of answers that must be picked from are the given answers, and 2) we are supposed to interpret the question to mean that the correct answer is in the set of given answers. Under that interpretation, the correct answer should be found in the list of answers.

If the choices were 25%, 50%, 60%, and any other number, then each number has a 1 in 4 chance of being selected. If one is correct, there is a 25% chance that the correct number will be pulled at random, and the correct answer would be 25%.

If you assume that the answer to the problem is in the list, then one or more of the answers is correct. And every other answer is incorrect. The correct answer has a probability of 1 of being correct. The incorrect answers have a probability of 0 of being correct. So, there is no place to put a probability that a given answer is correct. It would be like saying what is the probability that true is true? What is the probability that false is true?

You seem to be calculating the probability of picking, at random, the answer to some question, unspecified, which has multiple choice answers:

A) 25%

B) 50%

C) 60%

D) 25%

Then, it would make sense to calculate the probability that a given answer is correct, because you don’t know anything else about the question. But that’s not this question.

And, this is all built upon the assumption that _the correct answer is in the list_. If it’s not in the list, then by definition, there is 0% chance that you find it in the list, and the answer to the question is 0%.

Yes, I agree. The answer to the question is close to Zero.

The challenge with multiple-choice questions is not to identify the correct answer, but rather, to identify that answer that the teacher believes to be correct. Any moron knows that the odds of selecting the correct answer from a four-question multiple choice problem is 25%. (The values of the answers as written are irrelevant, the probability is still 25%.) Consequently, either (A) or (D) is the truly correct answer. The teacher, who believes himself to be quite clever but who otherwise fails to grasp the necessity of writing well-worded and intuitively understandable multiple choice questions, reasons thus: “If ‘25%,’ which is the correct answer, is written twice, then that adds up to ‘50%’!” According to the teacher’s reasoning, then, the correct answer is (B) and that is the answer that will be graded as if it were correct. In the end, rather than illustrating an interesting problem pertaining to probabilities, this question provides an illuminating example of how multiple-choice questions often gauge a student’s ability to psychoanalyze the teacher and to navigate astoundingly poorly worded multiple-choice tests rather than assess his mastery of the subject material.

The answer probability is A or D 25% and your chances of being correct is 50% But if you choose randomly then chance of being correct is 33.3%.

Its pretty obvious that there is no real answer. First of all, it says “at random”, which means that there would have to be a pre determined answer for you to get the answer right AT RANDOM. Secondly, if the chance of you getting the answer right is 25% because there are four options then as there are 2 25%’s the total chance of getting it right is 50%, which includes option b so there are 3 options that could be correct but each one of them debunks the other one. So this does not have an answer, a bit disappointing really

Helloooo…Simplify the question. If you choose an answer to THIS question, what is the chance you will be correct? Well, the answer to THIS question is that you can only be right or wrong. You have a 50/50 chance.

well, the fact that this post has 200+ comments…

/joke

This question was posted on Math StackExchange site:

http://math.stackexchange.com/questions/76491/does-this-question-even-have-an-answer

I think this is not a test. The question do not say we have to choose one of the four options, it ask what chance we have to do right if we choose a ramdom option. The solution, i think, it is not choose neither A, B, C or D; the solution will be to write: “If we choose one ramdom of 4 (if one of then its correct) there will be 25% but there are two answer with 25% so if we throw up a four side dice, the percent of take 25% will be 50%”.

PD: sorry for my english

A, D, AD, or DA are equally right, my explanation:

http://x.thexs.us/2011/10/29/probability-of-choosing-the-right-answer/

“This question” refers to itself. But it has no content regarding what it seeks. Since it could be any arrangement of words- or for that matter- any arrangement of pictures, objects, numbers or things or whatever, it is infinitely broad in scope. Therefore, the “answer” if given, is tantamount to like trying to throw an object into the air and hitting an asteroid that just happens to be there at that moment. The odds approach zero.

As the question is non-specific as to what is required then it isn’t possible to give a random answer and therefore there is no chance of getting it right.

Its a paradox between answers (A and D) and answer B. There is no right answer. Stop trying to read so far into it!

The answer is B, and it’s not a paradox.

You have a 25% chance to randomly choose a standard multiple choice question correctly, so the hypothetical answer is 25%, which is both A and B.

However, since the proposed correct answer appears twice out of four possible outcomes, that means you have a 2/4 chance to answer correctly. That’s 50%.

The answer is neither A nor D because that’s only the answer to the HYPOTHETICAL SCENARIO that the real question you’re answering has put forward.

Either the correct answer is 25%, 50%, or 60%.

If the correct answer is 25%, you have a 2/4 chance of picking it randomly.

If the correct answer is 50%, you have a 1/4 chance of picking it randomly.

If the correct answer is 60%, you have a 1/4 chance of picking it randomly.

E(X) = 1/3(2/4) + 1/3(1/4) + 1/3(1/4) = 2/12 + 1/12 + 1/12 = 4/12 = 1/3

I agree!

P(Answer=25%)*P(Choosing25%) + P(Answer=50%)*P(choosing50%) + P(Answer=60%)*P(choosing60%)

P(Answer=25%)*(0.5) + P(Answer=50%)*(0.25) + P(Answer=60%)*(0.25)

Since we are running in circles around between 25% and 50% continuously, the probability at any time that 25 or 50 is the right answer is 50%.

(0.5)*(0.5) + (0.5)*(0.25) + (0)*(0.25)

0.25 + 0.125 + 0 = 0.375 = 37.5%

The answer is A. here’s why:

When you read the question, as soon as you hit the end of the question, by “correct?”, immediately your brain knows choosing random means 1 out of however many possible answers. You then look at the NUMBER of possible answers, not the answers themselves. You see A,B,C,D, for a total of four. Without looking at the answers, you already know the answer is 1 in 4 = 25%. You know the answer is 25%. You start reading the possible answers, you hit A =25%. You mark down A. The rest of the questions should not matter, if you know 100% you are right. B,C and D shouldn’t matter at all. End of discussion.

Unfortunately, we changed the answers just by observing them.

Is anyone reminded of The Sicilian from The Princess Bride when reading this discussion?

IMHO , You need to reference the quiz maker or quiz giver what is the question’s expected answer, is the % the answer expected (25%,50%, or 60%)? or the letter of the answer expected (a,b,c,d)? and if there is only one correct answer or more than one, as that is not clear if [“a) 25%” = “d) 25%”] .

The question is: The answer to w h i c h question, and given w h i c h choices? ;)

The trick is that it is easy to confuse the values and the answer. The proper selection is b. That is because there are two of four ways to pick 25% which is the correct answer.

Said another way, two of four choices provide the correct probability so you have a 50% chance of picking it at random.

i get tired of saying this, NO 50% is not the answer. You are forgetting that the question must have a predetermined answer for you to be able to pick the right answer at random. i know that a multiple choice with 4 answers has a 1/4 chance of being randomly correct, but since two options are same that makes it 50% BUT no one said that 25% HAS to be the right answer, at random you could as easily pick 50% or 60%

Wow, I can’t believe I read the whole thing. If you’re reading this, you’re probably thinking the same thing. :)

Words first, then the math. The central issue here, as many have pointed out in different ways,”what is the question *actually* asking, as worded.” Because the question contains ambiguity, you have decide how you are reading the question and then answer accordingly.

I personally don’t read this as a multiple choice question, but rather an open-ended question that asks about a hypothetical answering of a multiple-choice question. I’m willing to assume the question is possible, that there is random distribution, and (key) I’m interpreting the meaning of “this” in the question in such a way that we are being asked “If you answered the question below, tell me the probability you would get it right.” (Nothing in the sentence says your *final* answer has to be one of the below. I read the sentence as saying that the multiple choice part is data needed to be able to come up with the correct answer. That just makes the most sense to me given the exact info we have.)

Many are instead interpreting “this” in such a way that the question becomes self-referential, IE trying to answer the question changes the answer etc. And there are other ways to read it and choose assumptions, such as “why are we assuming there is a correct answer listed?”

The language is imprecise (in particular what “this” exactly refers to in the question), and you have to make some choices about what assumptions you are willing to use. which is why so many knowledgeable people are coming up with different answers. There are different ways to read this, and each of those ways of reading it has a correct answer.

But I digress. For me personally, given exactly what we were given, I see this question as asking:

“A random number generator with equal distribution must randomly generate one of the following colors:

A) Red

B) Green

C) Blue

D) Red

– The correct choice for that question is either Red, Green, or Blue.

– “A)”, “B)”, “C)”, and “D” are each equally likely.

– For the scenario where “Red” is correct, either of answers “A)” and “D)” count as correct.

Given all that, calculate the probability that the random number generator will yield the correct answer.”

Note that the *actual question* is that last line. There’s nothing saying it’s multiple choice. The multiple choice was part of the information we base our calculation off of, not a set from which we must select our own answer.

In that scenario, which makes the most sense to me personally given the way the question is worded, the final answer is 37.5% because there are the following scenarios:

1) The correct answer is “Red.” [There is a 50% chance of the random number generator getting this scenario right, since A or D would count.] [Let’s call this scenario “P1” when we get to the math.]

2) The correct answer is “Green.” [There is a 25% chance of the random number generator getting this scenario right, since only one of the four choices it chose from was correct.] [P2]

3) The correct answer is “Blue.” [There is a 25% chance of the random number generator getting this scenario right, since only one of the four choices it chose from was correct.] [P3]

Three scenarios, each equally likely to happen. Three possible answers, two of which happen 25% of the time and one of which happens 50% of the time.Thus we get the following calculations for how likely each scenario is to happen and how likely the random number generator is to get the correct answer in that scenario:

P= (chance of scenario happening) * (chance of random number generator choosing correct answer in that scenario)

Calculating for each scenario:

P1= .25*.50=.0625

P2= .25*.25=.0625

P3= .25*.25=.125

Put that all together for your overall probability and you get .125+.0625+.0625+.125 = .375 (which is 37.5% when expressed as a percentage.)

Thus for the overall question, the computer will randomly generate the correct answer to the given hypothetical problem 37.5% of the time.

I like how Jay put this: “The trick is that the question isn’t really multiple choice – it’s short answer asking you to evaluate a multiple choice problem.”

If you read the ambiguous parts of the question otherwise, then there are correct and incorrect answers for your own version. Put another way, this is best evaluated as a two-part question:

1) “What do you think the question is actually asking?” [The language part of the question, particularly what does “this” in the sentence mean and what will your assumptions be?)

2) “Given your answer to 1), what is the correct calculation? (The actual number crunching.)

Thank you OP for causing me to lose an embarrassing amount of my morning. It was all worth it even if nobody at all reads this. :)

You say:

—–

P= (chance of scenario happening) * (chance of random number generator choosing correct answer in that scenario)

Three scenarios, each equally likely to happen.

——-

Why then you attach 25% probability to each scenario when calculating?

Why the

Maybe this is a better way to say it. The correct answer is 25% and the chance that you will select it is 50% (2/4). Nothing more complicated than that.

When all answers are wrong unless you pick a,b,d, which would equal 100% since no factual basis for an answer is required, so you would have to assume you would be right which would equal 100% . Therefor one must assume the theory of asking one’s Boss if he worked hard today, knowing full well the answer you get will be a lie. It is akin to the question of if a tree falls in the forest and no one is around to hear it does it make a sound when everyone knows it will be screaming all the way down.

kind of what I was trying to say then I got off onto percentages which i should not have done.

The answer has to be 50%. If you chose 25% as your random answer, you had 2 choices out of 4 .. A and D which is 50% and if you decided 25% was NOT you choice you still had 2 choices out of 4 … 50%

so what’s the answer?????

There is no chance that you will be correct, because there is no real question.

Justr because there is a 50% chance of choosing 25% in one random guess does not make it correct. Certainly, the answer cannot be 50% because you have only a 25% chance of choosing it. 25% is the correct answer because there is a one in four chance you will choose any one of the letters. The percentages are irrelevant.

The right answer is 33,33%, because there is just 3 diferent options.

Sorry if i’m making an already existing point, but there are to many answer to read them all, got bored.

In the computer engineering world, similar problems occur, the dreaded stackoverflow! :)

It went on an crazy experiment using a programming language. My opinion that this is an incorrect question, it may even “compile” but if reaches runtime it will always crashs and burns with an error at runtime.

This question seems to me to be impossible to answer. It is recursive without an stop condition!

Geeky me.

// Meta-code (inspired on C++) of an simplistic experiment to make my point… :)

int main( ){

for( int i=0; i<100; ++i ){

unsigned int pick = random()%4;

if( pick == answerToQuestion() )

++correctGuesses;

else

++LogIncorrect;

}

LogExperimentalResult( correctGuesses/ (correctGuesses+incorrectGuesses));

}

usinged int answerToQuestion(){

return answerToQuestion(); // bug – recursive :)

}

“The most common cause of stack overflows is excessively deep or infinite recursion”

Tony is right. The chance that a random answer is right must me close to zero base on two facts:

1. There is no question to answer (check it out, all it says is “to this question”)

2. A random answer is anything: (i.e. 1, q, blue, yes, elephant, Pi……)

The multiple options in there are just there to make you go on a wild chase after nothing, aka diversion.

I’ll re-word it:

How likely it is that you will be correct by picking a random access to a question?

*”answer”, not “access”

Ioana [Oct/28) is correct…I’m quite confident:

* The question is: “if you choose an answer to this question [below] at random, what is the chance you will be correct?”

* This is not the same as: “What are the odds if you choose one of the answers below [a, b, c, d] at random that the value selected will be the same as the odds of having chosen it?”

Said differently, the question is: what are the odds that if you choose one of the numbers below [25%, 50%, 60%] it is correct (whatever that value might be or mean; and whether you choose (a) or (d) it is the same value or answer: 25%).

If you’re still with me:

Your professor poses a problem on the blackboard along with three possible solutions to that problem [25%, 50%, 60%]. S/he hands out index cards and instructs the class to write their names on the card along with the answer. S/he also specifies that you must put one answer on the card, assures you that there is indeed an answer and therefore ‘none of the above’ / ‘all of the above’ / ‘not enough information’ are all off the table.

Sadly you haven’t been paying attention all semester and haven’t a clue which of [25%, 50%, 60%] is correct. Your buddy knows this, and bets you 4:1 that you’ll get it wrong (thinking those are even odds).

Do you take the bet?

~~~~~~~~

First off, and because you’ve paid so little attention, the available choices [25%, 50%, 60%] might as well be [Apples, Pears, Mangoes]. So we’ll go with that:

“25%” = Apple

“50%” = Pear

“60%” = Mango

You check the quiz and see the professor has listed Apple (ie – 25%) twice. While this seems odd, nothing says s/he can’t mark ~both~ (a) and (d) correct. So you’re faced with:

a) Apple

b) Pear

c) Mango

d) Apple

A glimmer of high school statistics comes to mind:

— if the correct answer is [A]pple and you choose (a) then you win

— if the correct answer is [A]pple and you choose (b) then you lose

…

— if the correct answer is [P]ear and you choose (a) then you lose

— if the correct answer is [P]ear and you choose (b) then you win

etc.

Brute forcing it, you write this out in table form:

A a 1

A b 0

A c 0

A d 1

P a 0

P b 1

P c 0

P d 0

M a 0

M b 0

M c 1

M d 0

total outcomes: 12

successful outcomes: 4

your chances of winning: 4/12 = 1/3 = 33.33%

Now that you’re feeling pretty good about yourself, a bit more HS stats comes back. You recognize the problem can be modeled as two independent random events — two rolls of the dice — the first roll being (your ignorant view of) where the professor hid the answer (Apple, Pear, Mango) and the second roll being your random selection (a, b, c, d):

(odds of winning if Apple)(odds of Apple)

+ (odds of winning if Pear)(odds of Pear)

+ (odds of winning if Mango)(odds of Mango)

= (2/4)(1/3) + (1/4)(1/3) + (1/4)(1/3)

= (1/6) + (1/12) + (1/12)

= 1/3 = 33%

So you bet your buddy $20, write “33.33%” on your index card, turn it in…and hope for a good outcome.

but you forget that the so called ‘pear'(b) in your question is also possible IF a) and or d) are correct since the probability of getting 25% is 50%. there really is no other way than without numbers

I don’t follow: the values “25%” and “50%” cannot both be correct/true — only one answer is allowed. We don’t know what the answer to the question is, just that it is one of [“25%”, “50%”, “60%”] — and only one — and we don’t know what the “25%” / “50%” /” “60%” actually ~mean~ only that one of them is ‘correct’, at the whim of the professor (maybe s/he rolled a dice too).

…the percentages:

a) 25%

b) 50%

c) 60%

d) 25%

are just numbers…and a choice you have to make (a/b/c/d => apple/pear/mango) — they have nothing to do with the ~probability~ of being your being correct. Because the actual question “25% of what?”, “50% of what” is undefined; it’s simply “25%” right or wrong.

…change the question to:

if you choose [a, b, c, d] below at random, what is the chance you picked the correct fruit? (as decided by me before class)

a) Apple

b) Pear

c) Mango

d) Apple

Actually – I made an assumption that the professor was forcing the random choice over [a, b, c, d], but the question is ambiguous (just says “choose an answer to this question at random” which could also mean over the values [“25%”, “50%, “”60%”] and where (d) becomes moot and odds are also 33.33% — way quicker :-)

…and factoring my original equation:

(2/4)(1/3) + (1/4)(1/3) + (1/4)(1/3)

becomes

(1/3)(2/4 + 1/4 + 1/4) = (1/3)(1) = 1/3 = 33.33%

which means [a, b, c, d] don’t matter. duh. I need more sleep. :)

So…I’m going the “Random” doesn’t mean “Stupid” route…and here’s my at least consistent solution to a brain-teaser that might have a logical answer and not just a paradox.

First…I think the question implies that it’s a multiple-choice question…in fact, the specific one given. Now, since I’m asked to be “random”…I’ll be random…but like I said, let’s not choose “stupid” first.

My main meta-assumption to this problem is that if it’s a “good” multiple-choice, it has only one right answer, therefor, A and D can be ruled out. Therefor…I have only two-answers to choose from…and if I choose “randomly” as required. I get 50% as my answer. (being right 50% of the time…which just happens to be one of the possible answers).

So – I’m going with B after ruling out A and D as being the same.

Joe

I believe you are changing the parameters of the question by ruling out “bad answers”… you have 4 to choose from at random. 25% chance you would get it right, but two of them have 25% so that’s 50%, answer B… but then you can only get B 25% of the time at random… so the answer flips back to A or D which is 50%. There’s no answer… it’s a circular loop if you stick with the parameters of the question.

I don’t own the answer…but I truly know Maurits Escher would be fuckin’ proud of this…

Probability is A (or D) with an answer of 20%

Statistically the answer is B with an answer of 50%

This problem demonstrates the difference between probability and statistics. Statistics would take into account the data contained within the answers. Probability would not. It would concern itself only with the ramdom chance of 1 of 4 answers.

Oops, typo “Probability is A (or D) with an answer of 25%”

There is no answer, because there is no question

Not true…I think :) Pretty sure the question (rephrased) is “if you randomly pick (a) (b) (c) or (d), what is the probability the person who wrote this on the board will say ‘correct!’ ” — then there are two assumptions: (1) the correct value is one of the choices given: [“25%”, “50%”, “60%”], and (2) s/he has already resolved the correct value (ie – it doesn’t matter how, just that it is no longer random/undecided…even though ~we~ don’t know what it is yet (like Let’s Make a Deal: the goat (or Lexus) is already behind ~one~ of the doors ;-) )

I was wrong, you are correct.

1. Probability that you would randomly pick 25% as the answer, i.e. select A or D = 50%

2. Therefore, correct answer is B (50%)

3. Probability that you would choose B, which is the correct answer, is 25% which is in agreement with statement 1

you are geneous.

The only winning move is not to play

100/4=25%

The answer is 50%

; this question implies you’ve already chose an answer so the chances of it being correct is 50% -it’s either right or wrong they aren’t asking what’s the answer to the non existent question

hmmm… assuming there’s been no exchange of information yet (no tells), just because I’ve decided to buy a car doesn’t mean the odds of me getting it are 50/50 — for example, there may only be a 10% chance the dealer will sell it to me.

Since there is technically no question, I get to make up my own question. I would only make up a question to which I already knew the answer. Therefore the probability that I would select the correst answer would be 100%

I’ll bet you a drink that you won’t buy me a drink :-)

As I was going up the stairs,

I met a man who wasn’t there,

He wasn’t there again today,

Oh how I wish he’d go away.

Hi, guys, your professor made a mistake. This problem, when was first posted two years ago, has (c) being 0%. Now it becomes 60%. Should that change affect your answer?

http://answers.yahoo.com/question/index?qid=20100220073507AAUNplo

Not mine – just another arbitrary value. ;-)

It is just beautiful.

It may be (high chances) useful to read the following link, but may be (high chances) not….

http://en.wikipedia.org/wiki/Law_of_excluded_middle

:D

Or may be this ones either…

http://en.wikipedia.org/wiki/Fallacy_of_the_excluded_middle

The probability one of these options (A, B, C, D) to be correct is = 0.

The fact that the answers have percentages as an answer means nothing to the question. it doesnt matter which one you choose. The percentage will always be 25%.

The chance to answer correctly if choosing at random is 25% (for answers of the type {A, B, C, D}).

The flaw people make from there is in noticing that selecting 25% as an answer still has a 50% chance of being wrong, but that’s not the question they asked. They asked about random selection, and selecting at random from four possible choices {A, B, C, D} will always be 25%. That you have only a 50% chance of deliberately choosing the right 25% answer from {A, D} does not change what you would select if choosing at random.

Further notes: the puzzle does have a correct answer and is not a paradox. Proof as follows.

If A is correct, the chance of randomly choosing it is 25%, which is what A says, so A indeed may be correct.

If B or C is correct, this does not hold. Both would create a paradox if correct, so they are not correct.

D follows A, and it may indeed be correct. The chance of choosing it randomly is likewise 25%.

A and D cannot both be correct. Again, that’s a paradox, as you would choose A or D 50% of the time if picking at random, and thus they could not be correct unless only one of them is. This also shows the answer must be either A or D, and not “25%” in the abstract.

So there’s some chance (P) that A is correct, and some other chance that D is correct (which is 1-P). If there’s a 25% chance to choose either, that’s

=> 25% * P + 25% * (1-P)

=> 25% * (P+1-P)

=> 25%

You can add the 50% chance of not being correct (by choosing B or C at random) as well, but 50% * 0 is just 0.

So there is no paradox, and we have a complete answer. While we cannot know which one of A or D is the correct answer, it is one of them, and the chance of choosing it is 25%, no matter how likely one or the other is.

Has anyone counted the number of answers in this thread that conclude one of the 4 answers a,b,c or d. It may give you the solution to the question.

The answer is that you have a 25% chance of being correct. The correct answer is either A, B, C, or D; onnly one of those four is correct. The answer is not 25%, or 60%, or anything like that; just A, B, C or D.

If you choose A thinking that 25% is correct, you may be wrong in that D is the real answer. Same goes with choosing D.

You can choose only 1 of the four choices and a random choice gives you a 25% chance of being correct.

Dido

What are the chances of a meteor striking the earth on land?

Parallel to dividing by 0, the answer is undefined. This is a beautiful self-referential question and as such has no answer. I do like the idea of adding .0 as another option, just to further confuse the matter. I urge everyone to read Godel, Escher, Bach by Douglas Hofsteder for a lay-person-appropriate exploration of self referential problems.

But ‘right’ is not defined; it is an abstract idea… it is about truth. truth, according to correspondence theory implies that a proposition matches reality. an answer can not be right in its self. it is a philosophical joke. There is no reality to match it to. unless it is an ontological truth they are referring to and that also is a truth that can not be answered. just because a question can be written it does not mean it can be answered… and don’t go saying that the answer is 0 then…..

The answer depends on how many of the students the teacher finds sexy, because those are the only ones getting it right.

its B….coz dere are four options…….p(for every option to b correct)=1/4,

p(for option to b false)=3/4…….as 25% occured twice, so to choose a random answer 4C1 /2! and so on so b will b d correct answer……

B is the correct Answer as 25% is repeated twice. So chance of selecting 25% is 50% i.e 2/4*100%

Matt J is right if he says “You have a 50% chance of selecting A or D which is the probability of randomly selecting the correct selection B”.

The point is, however that as soon as you randomly select A or D, the random answer is not correct, since it should be B. so if B is the right answer, the probability to choose the right answer is 25%, which is not the value of B. conclusion: B can not be the right answer.

you can do the same reasoning with all the other answer.

Once this is digested, we can make the problem more fun by changing C to C = 0%.

I wonder what you think about it…

It’s not a paradox. The answer to the question is not one of the 4 given choices: there is no correct answer.

It seems like a paradox because we are trained to expect that one one the 4 choices is correct.

if you chose c = 0% it becomes a paradox, since if one chooses at random C, he can ether be correct or wrong, if he is correct, the chance of randomly choosing C is 25%, which is not the value in C, so he is wrong. BUT if he is wrong (and the other answers are no solutions – which is the case) then the probability to be wright is 0%, which means that C is the right answer, BUT if C is the right answer, the chance of randomly choosing C is 25%, which is not the value in C, so C is wrong. BUT if C is wrong (and the other answers are no solutions – which is the case) then the probability to be wright is 0%, which means that C is the right answer, BUT… and i could copy the text once again, but I think you got it…

C is at the time right and wrong -> this is paradoxal

so i think the problem was originally with C = 0%

the way it is written here is not really mind blowing…

I like the discussion. I think it boils down to semantics. If you assume as the question implies, that the option selected must match the probability of selecting that option, then it is a paradox and there is no correct solution. In this case changing 60% to 0% would make it more interesting.

If you assume that there is no connection, then the actual percentage answers are irrelevant, and so using any 4 options(i.e. blue,yellow,green,blue) of which 2 are the same, would result in 12 possible outcomes. (4 options for you abcd, 3 options for a correct response as chosen by the creator of the question) Of these 12 outcomes , 4 result in a match of you randomly selecting the option that matches the creator’s selected response. So that would be a probability of 1/3.

Thanks for making my head spin.

60%

Working out the answer to this question in the points above reminds me of the Princess Bride and watching Vizzini figuring out which glass the poison was in…

The irony is that this is not a good question because actually none of the answers are correct. 25% would be the correct answer only if there were one correct answer. 50% would be correct only if there were two correct answers–but there aren’t. For the question to work, 50% must be listed twice, 25% once and 60% once.

Actually that doesn’t work either, but the point remains that none of the answers are correct.

25% can’t be correct because it appears twice out of 4 options so you have a 50% chance of picking it. 50% can’t be correct because it appears once out of 4 options so you have a 25% chance of choosing it. 60% can’t be correct because 60%=3/5 and there are 4 options so the answer is either 0% or “The question is not logical.”

BEST ANSWER, but still not the final answer because human nature won’t allow it. This question is paradoxical and there is no right answer. If you try to solve it mathematically, you realize you are at a dead end and must revert to logic, but again you second guess your reasoning when you reread the question and feel the need to choose an answer from the list. This is classical circular logic and if you’ve commented, you’ve completed half the circle. THERE IS NO RIGHT ANSWER AND YOU DON’T HAVE TO CHOOSE ONE RANDOMLY. Look at all the comments, hilarious.

Try it with C = 0%, in this case C can’t be right but it can’t be wrong ether.

Answer: 1 in 4 chance your answer is correct.

Reason: Under assumption, “one” of the answers is correct, now ignore the actual values. If you choose at random, there are 4 options, each having the same likelihood of being chosen, none more likely than another, therefore if you randomly choose, the answer is 1 in 4 or 25% that it is correct. (The actual values are irrelevant, they are a distraction.)

Don’t think to hard.

If 25% is correct then the chances of you picking it at random is 2 out of 4 which is 50% so it cannot be correct.

Think a little bit harder

The whole point is the randomness, you are thinking to much about a simple problem.

Again: assume 1 answer is correct, randomly choose 1 out of 4, what are the chances it is correct, 1 in 4…. simple, don’t be blindsided by the numbers, they are irrelevant.

If you read closely, it didn’t ask you to choose the correct answer. It merely asks what the chances of picking it are. That is where 90% of the people who have commented here have gotten lost.

So once, again, don’t think to hard my friend.

Its not actually asking a question that has a % as an answer……so the right answer is ‘the question makes no sense and doesn’t have an answer) or in math terms NA, NaN, etc

It would have an answer if we’d asked “what is the chance of correctly picking the correct answer from the following question”,

Question:

“What is 20%+ 5%

A) 25%

B) 50%

C) 60%

D) 25%”

Chance of correctly picking the right answer

A) 25%

B) 59%

C) 60%

D) 25%

This question is not a paradox question, it’s just a “SIMPLE WRONG QUESTION” where your teacher/professor mistype the abcd answers.

the question is:

“If you choose an answer to this question at random, what is the chance you will be correct?

base assumption:

– only one correct answer

– “random” mean not see the value of the answer

since there is 4 possibility and only 1 correct answer, then change to be correct is 25%

the answer is 25% (if it’s not a multiple question then just write down 25%)

when you try to choose your answer by not “random” (find answer with 25% value) you will confuse since there is 2 answer with 25% value.

Just stop, and and told your teacher/professor to revise the question/answer.

It will become a paradox if you back again to the question, and your will never find “convergent” answer because the answer will changes every time you back to the question.

it’s what we call Non Linear Question (~non linear equation) :D

y=f(x)

where;

x=f(z)

z=f(y)

wkwkwkw…

The real problem is that the question is a circular reference, which, for good reason, always generates an error in spreadsheets.

I started a long-winded comment, then realized that Jon stated it very well.

Actual, the question isn’t self-referential/circular:

In saying “If you [do xyz], what is the chance you will be correct?”, the question does NOT require the value chosen by randomly selecting from the available answers to ALSO equal the odds of having done so.

In other words: it is perfectly legal for the odds to be 33.33% that you will choose the right value from (25%, 50%, 60%) by randomly selecting from (a,b,c,d) …which it is, and has been shown several times:

P = (1/3)(2/4) + (1/3)(1/4) + (1/3)(1/4)

P = 1/3 = 33.33%

The answer is clearly 0 (well, modulo the question being a little vague). And this isn’t a paradox, unless every badly constructed multiple choice where none of the options are correct is a paradox.

The really interesting thing about the question is that, while there is no logical paradox, the answer kind of makes no sense at all regardless….. So it’s really a shame about the multiple choice.

Clearly it isn’t clearly anything as shown by the length of this discussion. :)

33%. With two same answers in four, your chance becomes 1 in 3… I like Scharnout answer too, though: 50%, either right or wrong

It might be a matter of precision (?Perhaps?).

If first of those 25% was 25% and the other was actually 25.000000000001% )(confidently rounded by the chalker) then choosing the 25% would be a correct answer, so the answer to the probability question would be 25% – the chance of choosing that 25%.

And whats a part per trillion between friends?

But I guess if I believe my last para, then choosing the 25.0000whatever% would be 99.9999% correct – Mmmmh!

Bob J.

25%

Actually the answer is E) 20%

Correct me if the answer is wrong

OK:

P = (1/3)(2/4) + (1/3)(1/4) + (1/3)(1/4)

P = 1/3 = 33.33%

Correct!

The answer is easy, just pay attention to the statement, especially the comma, the chance of being correct by choosing any option is 0% (zero)

I don’t see how you arrive at 0%.

The question is:

“If you [do something], what is the chance you are correct?”

where “[do something]” is to “choose an answer to this question at random” which in effect is to use a 4-sided die to pick one of (a, b, c, d).

Since if you roll the die enough times you’ll get all of a, b, c, d, the only way the chances would be 0 is if none of (a=d=25%, b=50%, c=60%) is correct.

Wnat is the chance you will be correct?****

if it were a dice, certainly there would always be a side, but the chance to know which side would be in this situation and the side chosen containing the real probability is zero.

Again, so long as one of the available answers (a,b,c,d) points to a ‘correct’ value (25%, 50%, 60%), the odds cannot be zero.

For this question, ‘correct’ simply means the person grading the test will mark it so. It does not mean for example the odds are 25% that you will pick either a=”25%” or b=”25%” (the odds of that is 50%).

oops – I meant: “…a=’25%’ or d=’25%’…”

you are looking for the situation not for the question.

the chance of you choose any alternative is 25% (100/4), but the chance of be correct with your chosen as zero.

look for the question man:

“What is the chance you will be correct?”

I think a lot of people are misreading the question: the odds of being ‘correct’ are not tied to the numbers in the list: a,b,c,d.

Imagine that I put into a hat 6 slips of paper marked with (“25%”, “50%”, “60%”, “15%”, “65%”, “10%”). Then I decide the ‘correct’ answer to my question will be whatever slip of paper I pull out of the hat (at random) before class. Later I pull out the slip “65%”, walk into my class class of 1000 students and write on the blackboard:

“If you choose an answer to this question at random, what is the chance you will be correct?

(a) 25%

(b) 50%

(c) 60%

(d) 15%

(e) 65%

(f) 65%

(g) 65%

(h) 65%

(i) 65%

(j) 10% ”

Later on when I grade the papers, I expect to mark ~500 ‘correct’ (not 10%)

Oh god!!! you are trying imagine one question to justify your answer, god…

but we have 4 alternatives and we know all of them, if you change others terms can find other answer, but for this problem the answer requires some interpretation only.

read the question carefully, analyze each answer and tell me how much chance has to be correct with any choice

BUT LOOK FOR THE PROBLEM, DON’T TRY IMAGINE OTHERS…

But that is because I know the ‘correct’ answer is “65%”.

But the students do NOT which answer will be marked correct: it could be ANY of (“25%”, “50%”, “60%”, “15%”, “65%”, “10%”). So for them, the odds are:

P = (1/6)*(1/10) + (1/6)*(1/10) + (1/6)*(1/10) + (1/6)*(1/10) + (1/6)*(5/10) + (1/6)*(1/10)

P = 1/6 = 16.67% (…also not 10%)

Your math is good, only need to pay attention to the problem ^ ^

good night ^ ^

No – also I misread the question at first. Then realized I was reading something into the question that wasn’t there: the question asks you to choose (at random) an answer from a list of possible answers (“a=25%”,”b=50%”,”c=60%”,”d=25%”) no more or less than that. Then it asks you what are the odds that you picked the ‘correct’ – without telling you what is ‘correct’ so any of the 3 values in the list (25%, 50%, 60%) might be the correct one. I therefore changed the equation to match the question (already given above).

g’nite.

It’s actually 50%, either you are right or you are wrong :D

Don’t go to Vegas…where you either win or lose. ;)

The correct answer, in the words of the one and only Obelix – “These Romans are crazy!”

0%

The question don’t says that one of these answers (A, B, C, D) is the correct.

Exactly, the question only ask “what is the chance you will be correct?” chosing one of the answers…

60%

The answer is simple: the Cat is both alive and dead.

if we have the answers

a 25%

b 50%

c 60%

d 25%

then the cat is simply dead

but if you take the answers

a 25%

b 50%

c 0%

d 25%

then the cat (or C) is both alive and dead

but c) 0% isn’t in the question. The answer is 0% and because it isn’t an option this makes the question illogical.

multiple choice questions with no right answers are not illogical, just unexpected. There is nothing that would violate any logical rule if dont have a correct answer!

try to think it trough with C) 0%

I’m pretty sure the original problem was that way.

Maybe the original problem was different to the one above. But we aren’t here to investigate or speculate on what the original problem was. If c) 0% was an option we would have a 25% chance of picking it at random so therefore 0% would be incorrect.

ah you are getting there! go on thinking…

if C is incorrect, the probability to chose at random the right answer is 0% which is the answer C, this means that C is correct, but if C is correct the probability to chose C is 25% which is not C, so C is incorrect, but if C is incorrect… and so on and so on…

Some people are saying we should ignore the values given and say 25% is the answer. Imagine it were a different question and that one of the multiple choices was correct. Would you still advocate ignoring the values given? No. Then why are you changing the question or ignoring the choices given in order to ‘shoe horn’ your answer in? If the question was, “when given four choices what would the chance of picking the correct answer at random be?” Then the answer would be 25%. But that IS NOT the question. The question is a multiple choice and it gives four (which is actually three because 25% appears twice) choices. The question does NOT ask for a value, it asks which of the options a) b) c) or d) is correct. nowhere does it say ‘for example’ or ‘ignore the four choices below’. It says which of these choices is the correct answer.

Doesn’t matter what is the question, the answer of one of the A, B, C, D, is one quarter of 100. The answer is 25% If your answer is one of these A, B, C, D, means your chance is 1/4 of 100 to get to correct answer, which is 25%

How can it not matter what the question is? Where does it say “read this question then change the question to suit your answer?” Where does it say “ignore the values given and the number of times 25% is given as an option?”

this is the interesting problem, if you chose C) 0%

in this case C is both right and wrong (exactly like Schrödingers Cat is both dead and alive)

The simplest way to solve this is to wait and see what the smart kid (Kevin Taylor in my class) puts and copy his answer. I’m going to write ‘5318008’ on my calculator and scratch my initials into the desk while I’m waiting.

The answer to the question is: the probability of hitting random choosing is 0

If there is a correct answer, and assuming equality in probability, the probability of one right in four options is 25%. But we have two options of 25%, so the correct answer is B) 50%. There is only one option of 50% out of four. So, reading the question “If you choose an answer to this question at random, what is the chance you will be correct?” If I choose at random an answer where only one out of four is correct, my probability of being correct is 25%, which is not the answer, but only the probability of being right which is what is being asked.

if you randomly picked a answer and one of them is right then you have a 25%chance of randomly picking the right answer. same as guessing. now if the correct answer is available twice then you have a 50% chance of picking the right answer. its like if you put the letters A B C and A in a bag, whats the probability you will pick A out of the bag. so for the example given the answer is you have a 50% chance of being right.

but then if the answer is 50 then that brings you back to 25%. wow nice one

the trick her is that this is not a multiple choice question. the answers are set up in multiple choice but your answer to this question should not be a b c or d, your answer should be 25%. not A or C, BUT 25%

Exactly, my friend! Math is amazing!

Hypothesis 1: the probability of guessing correctly is 25%.

Analysis: if the probability of guessing correctly is 25% (our hypothesis) then the probability of guessing correctly is 50% (two chances in four). This is a contradiction. Therefore the probability of guessing correctly is not 25%.

We can eliminate the other choices in a similar manner. Therefore, none of the choices are correct.

telemoxie this is about logic. none of the choices are the answer. the answer is 25%. its note A. or B or C or D. its simply 25% why because 1 out of 4 is 25%, if you can pick 25% twice then the odds increase to 50%, in witch case we come back to 25% because 50% can be picked once. listen to what is being asked.

a, b, c, and d are art of the question. the answer is 33%

lol the answers are examples for the question being asked. how do you get a third when you have four chioces is beyond me. the instructions dont say eliminate one answer because it appears twice. you have to use all the information to develop your answer

If you choose an answer at random, you have a 25% chance of choosing (A), 25% for (B), etc. So, you have a 25% chance of choosing the answer “25%” (correct), another 25% chance of choosing the answer “50%” (wrong), 25% chance of choosing “60%” (wrong), and a final 25% chance of choosing “25%” (correct). So overall you have a 50% chance of choosing the correct answer.

so whats the odds of picking 50%? the 50% takes you back to 25% because it only appears once in a choice of 4. thats the loop trick to this question.

Sid, the correct answer is only one and it is 50%. The chances you pick the right answer at random are one out of four, so the answer is 25%. Do not relate to what is written in the options. The question is only what is the probability of being correct. So the answer to that question is a number and not an option available on the text. Forget the text. The answer is “25%”, not option A, B, C or D. Just 25%.

i wrote that 9 comments ago

I can’t believe some of the arguments given here. There is no correct answer! It’s also not a paradox; it’s just that the question and answers are written poorly.

why do you say it is not a paradox?

Yes, and I wrote it 13 comments ago. But now I started reading some of the more than 400 comments and some people had already wrote the same idea. No one can read 400 comments.

actually… Not proud of it but I’m afraid I got addicted to it. :)

Since “this question” is not well defined the answer is asymptotically equal to zero since the universe of possible answers is so large. If it is assumed that the answer must be a whole percentage, then the answer is 3/101. There are three answers allowed but there are 101 possible whole percentages. The question does not assume the correct answer is among the multiple choices.

If c is the correct answer wich of these answers is correct.

a) b

b) d

c) a

d) c

The statistics are just there to distract you from the paridoxical nature of the question.

What is more likely, that this is some kind of deep, self-referential, paradoxical riddle or that someone at Flowingdata f’d up and shouldn’t have posted this turkey?

I will still give my answer: by saying B you are sure to be right. Explanation: by choosing 25%, you necessarily have two answers (A & D) out of four right. 2/4 = 50%.

By choosing 50%, you only chose between 25 and 60%. As it is either one or the other, it’s necessarily one chance out of two: 50%.

However, the question still disturbs me a little bit, because by picking an answer randomly as the question asks, I would pick one out of four, and even without looking at the answer, I should say 25%.

Assuming at least one of the four lettered choices is correct, which must follow for the question to have any meaninng the chance of success is 1:3 since two of the choices are the same.

ONLY ONE ME LOOKING FOR MY LIFE SHARE WITH WORLD TEACH.

This thread has devolved into nonsense. It’s obvious no one is reading other people’s posts.

B

Many people here have been assuming too much in their “calculation” of the “answer.” The question asks if you randomly choose an answer to a question, what is the chance your choice will be correct. It doesn’t ask “which of the following choices is correct.”

The probability of a correct answer can only be calculated based on the distribution of correct to incorrect answers and a sample size of responses.

Since this isn’t stated anywhere, we will never know. We can’t assume there is only one correct choice or that the probability is simply 50% (right or wrong).

For example:

Say you have 6 apples and 4 oranges. What is the probability you pick an apple at random from these 10 objects? This is simple to calculate because you know the distribution of objects.

Now if I gave you a bag of 10 apples and oranges, without knowing the distribution, it is impossible to tell the probability of picking an apple at random, but it is somewhere between 10 and 100%, I promise.

Similarly if I gave you a bag of 6 apples and an unknown number of oranges, you wouldn’t be able to calculate a probability because you don’t know the total number of objects in the bag.

Now if I give you a bag of apples and oranges and you don’t know how many total objects are in the bag or how many of each, it is just like asking the original question. We don’t know the distribution of the answers, so we can’t calculate a probability.

There is insufficient information to answer this question as written. It doesn’t matter what A,B,C,D represent.

Now if you make a bunch of assumptions, you may be able to narrow down to a choice of the given solutions.

The chances of a correct answer are normally 50% (right or wrong). But the question puts you in an universe of four possible answers (25%). The right answer would be indeed 25% if they were all different from each other. But considering we have two times “25%”, your chances grow to 50% (B). ;)

Mandei beeeeemmmm!!!

(melhor até que o autor da pergunta) %P

The question is not asking “which of the following four choices IS the correct answer.” The question is asking what the probability (chance) of randomly picking the correct answer from four choice. The answer is 1 out of 4, or 25%.

I’m with you! All this people and their assumptions! If they just stick to answering whats asked…

Exactly! And two times 25% equals 50%, which is the final answer. ;)

and if 25 is there twice then you have a 50% chance of being right. then if the correct answer is 50% and 50% only appears once then the probability goes back to 25%. once it goes back to 25% it goes back to50%. once it goes there, it goes right back to 25%. and it keeps going that way. an unending equation . that’s the paradox.

25%

A=25%

Guys the answer is B – 50%.. my reason is as follows:

The question asks what are the chances of chosing at random, the correct answer to this question. Now, there are only two possibilities:

1. The answer you chose is correct

2. The answer you chose is incorrect

Please note, nowhere is the question giving you options of what the answer is – the options are only giving you the probability – so either you are right or wrong. Hence the probability is 50:50..!

B:

only one answer is cerroct out of four, so if we choose randonly answer is %25 (not A or D) , and as A & D are both %25 so we have 2 chances out of 4 to choose the correct answer, so our chance will be 50% and B is correct,

This answer is 50%, but NOT B. This is an open question on a multiple choice question, not a multiple choice question itself.

Read the question again and forget about the answers. its 25% since there are 4 answers. If there was

a) red

b) green

c) yellow

d) blue

this question would be so much easier. Yet people seem to fail to see the answers given, are part of the question.

(50%*1)/3+(25%*1)/3+(25%*1)/3=33%

@Sven: the answers shoudl then still be:

a) red

b) green

c) yellow

d) red

So if the actual answer was green or yellow it would be 25%, if the actual answer was red it would be 50%…

The only winning move is not to play.

It is 25%, because we do not know the question, just imagine when each answer is written by another person and the question is ” Who wrote answer B?”

The question is only a formal (grammatical) question, without possible answer.

I don’t understand the joke.

If it is a joke.

I think that most of you misunderstood the question.

The question is “What is the chance to be correct?” and not “Wich of the 4 alternatives is correct?”

By choosing one of the alternatives given above, the chance to be correct is 0% (regardless how the random choice is distributed). So the answer for the question is 0%. Actually the question does not say PICk the right answer from the 4 given alternatives, but just askes you for giving the chance to be correct IF you picked one of the given answers.

Oh come on, it’s simple. The answer is 0%.

I believe that by generalising one can deduce when this question will have correct answers. Let X be any percentage. The question will have a correct answer if the proportion of answers equal to value X equal value X.

For instance, if X = 25% and one out of four answers is 25% then this is the correct answer.

There is a problem with this hypothesis. What if 3 answers are 75% and the 4th is 25%? Then we have fulfilled the requirements twice. We can fix this by adjusting the rule to say, ” Let X be any percentage. The question will have a correct answer if the proportion of answers equal to value X equal value X. This is conditional upon only one distinct answer fulfilling this requirement.”

Btw, I realise this could be written more precisely but hopefully I have gotten my point across. Finally, In the given question none of the values fulfill the requirements but what about if X = 0%? 0% of the answers are 0% so does this mean the answers have this value? The only problem with this is that it is assumed that a multiple choice question should list a correct answer.

Thoughts?

Obviously a and d are incorrect as they contradict themselves, so I’d pick randomly out of the other two and get the correct answer of 50% half of the time.

who’s to say the %ages are the answer, it may be the letters :-)

“If you choose an answer to this question at random, what is the chance you will be correct?”

Excuse me, what is “this question”? Has it been asked?

Sometimes you have to back away from the tree to see the forest. This is less a math question than a psychological conundrum.

“IF you choose an answer to this question at random” then you do not actually weigh the answers, but act blindly to their validity (or invalidity). So, you have a 25% chance of being “correct”, assuming there is a “right” answer, because there are four choices (disregarding the value associated with those choices). The question’s REAL paradox is in where you record your answer; If you methodically choose one of the given answers, you are not choosing “at random”.

For a real answer to this question, see http://en.wikipedia.org/wiki/Russell's_paradox

Additionally, for conversation on this, see http://www.reddit.com/r/math/comments/c2p7u/multiple_choice_if_you_choose_an_answer_to_this/

i totally agree with gaurav marwaha. If you randomly choose an answer, the probability is 50:50 (the selected answer is right or wrong). It’s as simple as that. CMIIW

ok, i’d try it a completely different way and forget (mostly) about maths:

let’s try to forget first about the question and the answers. the question is “xxxxxx xxx xxxxxxxx xxxx?”, and we also don’t understand the answers. in very many multiple choice questions (i start with three answer options A, B and C), the correct answer is the middle one. if you don’t understand the question and/or the answers, or have simply no idea of the correct answer, you should choose B. chances are quite good that it’s correct. also, people have the tendency of choosing the middle answers, they usually don’t pick the extremes.

add the fourth option, now we have A, B, C and D. according to what i just wrote, there is a tendency that the correct answer is one of the middle ones. at least most people would choose either B or C if they don’t have a clue (=randomly). so we can forget about A and D.

that leaves us with two possible (random) answers: B and C. back to the start of the question: “what are the chances to pick the right answer?”

B: 50 %

C: 60 %

as now we only have two possible answers, i take a look at the percentages and say B is correct, 50%.

There is a logical problem on this, but it’s no paradox, it’s a human error:

This logical dilemma has the logical problem of just one premise, not two as it needs:

Premise 1: By what chance will you be right?

Premise 2 (missing): Number of (different) answers.

______

Conclusion: With four (different) optional answers, the chance is 25%, with three it is 33% and with two it is 50%

Taking the given answers as a hint, you can either think of three options (then no answer is correct) or four.

Logics say, there are three answers given, which are:

A: 25&

B: 50%

C: 60%

A: 25% (D = A, therefore D is ruled out)

Then no answer is correct, for it is 33%.

Naive assumption shows four options: A, B, C, D. But in this case, the given answers can’t be the potential answers, therefore the correct answer would be “25%, whatever letter that option would be”.

Sheesh.

A and D are not the same! Based on the serif of the so-called 2 in D, and the circles on the solidus in it’s percent sign, it’s obvious he meant to write 35%.

Anyway, unfortunately, the picture missed options E) none of the above and F) All of the above.

Nonetheless, the correct answer if G) 42, which isn’t even there, but the choice of true randomness.

If we think of this as all one question, we inevitably get a contradiction as dedoco described in the beginning. Therefor, it must be two questions and it does not specify what “this question” is. In other words, suppose instead it read:

“If you choose an answer to “What is 1+1?” at random, what is the chance you will be correct?”

and you take random to mean any choosing real number, then there is one correct answer out of infinity. 1/infinity is 0. I therefor agree with those who have suggested that option C should be 0%. This also works if there is more than 1 possible answer. The only flaw to this logic would be if any possible answer is correct in which case it would be infinity/infinity, but it told us not to think too hard :P

“If you choose an answer to this question at random, what is the chance you will be correct.”

a) 25%

b) 50%

c) 60%

d) 25%

Can a) be correct? “The chance of choosing the right answer randomly is 25%”. Well if this was the case, d) was also correct and therefore the chance would no longer be 25% so both answers become false.

Can b) be correct? “The chance of choosing the right answer randomly is 50%”. The probability of choosing b) out of a pool of 4 answers is 25% regardless of what the answers are even if they were all the same! But we already stated 25% as a false answer.

Can c) be correct? “The chance of choosing the right answer randomly is 60%”. Same thing as obove, you’ll end up with 25% chance.

Can d) be correct? See a)

This is called a “self reference problem”. It’s a paradox in the end.

Raphi, for B) could say that or you could also say the probability of choosing 50% out of a pool of four answers if indeed 50% (and not 25%) because you are actually taking it out of a pool of two answers instead (50% and 60%). From 4 answers, two of them (50% correct) are 25%, and the other two (50% wrong) are 50% and 60%. Considering that, the alternative B) could represent the 50% chance for the correct answer (the two times 25%).

By the way: If c) was 0% that wouldn’t change a thing:

“The chance of choosing the right answer randomly is 0%”. This might be right for the answers given above but in the very moment 0% becomes a possible answer it becomes a false answer, as there is now a 25% chance of getting “0%”. Self reference guys, self reference. :P

How can you choose an answer if there is no question asked?

P(0.25) = 0.5, P(0.5) = 0.25, P(0.6) = 0.25. In order for a correct answer to exist, shouldn’t it need to satisfy P(X) = X?

Fun question, but no solution.

Jesus, how do I opt out of receiving updates on this thread from hell?

%33 bence

0%

If you were to limit yourself to the four choices listed, the answer is 0%.

But neither limiting your answer to those choices nor even a numbered percentage is specified in the question. Therefore, the number of answers, as well correct answers could be limitless. For example: if your random answer was “dog” or “banana”, that would be incorrect. If your random answer was “slight” or “slim”, that could be correct.

People, I already gave the right answer. It is 50%! You may stop arguing now. The answer is been given! :)

I am reminded of the “left leg is both the same” (Answer to “Why is a duck?)

“this question” is not defined.

So the answer is Undefined, Indeterminate, Unknowable, Undecidable, … any answer is as valid as any other answer because they are all equally Meaningless.

Since “this question” is not defined there is no way to understand the statistical distribution associated with the “question” because there is no defined “question”.

If the question was “can you guess the height of a person correctly within 1 cm?, what horse will win the Melbourne Cup in 2012?, will a coin toss result in heads or tails?, … ” this would be a Defined question.

So the reference to “randomness” or non-randomness (deterministic) is completely irrelevant because “this question” is not defined, very fishy, smells bad, a red herring so to speak.

This looks like a grammar question, not a statistics question. And because nothing is being asked, there is nothing to be answered.

Great thinking. But I think you folks need more constructive projects to work on.

What part of “try not to think too hard” did you guys not understand? :D :D

If God can do anything, can God make a rock so big that he cannot lift it?

I think it is 50%. There is a 50% probability of guessing one of the 25% answers, which is the probability of guessing any one of four answers.

My version of this paradox:

“If You choose an answer to this question at random, what is the chance You will be correct?

a) one of three,

b) two of three,

c) one of three.”

You’ve got only to think this:

The chance of choosing “25%” is –> 50% (2 answers of 4)

The chance of choosing “50%” is –> 25% (1 of 4)

The chance of choosing “60%” (or whatever) is –> 25% (1 of 4)

So, if the answer and the chance of choosing it cannot be equal,

THERE IS NO RIGHT ANSWER!!!!!

So, the right answer should be 0%. And that isn’t an existing answer in the options.

Truly, the correct answer is C since I am right most of the time.

The key is not to think too hard. None of the four answers is correct. The question does not insist that you pick any of the four offered answers. Your can make up your own answer. Therefore, the answer is there is a zero probability that any of the four answers is correct.

The question is perfectly sound.

Because the question asks you to evaluate the chance that if you pick a random answer from the list you will find the right answer to the question. The right answer is obviously 50%.

Why? Because the chance of picking that chance (50%) is one in four -> 25%. This answer appears twice on the board, so the chance of picking 25% is 50%.

The question is very very clear, let’s break it down: “If you choose an answer…” supposes that you DO make a choice. If not, end the exercise right here. But since you have chosen to answer (“at random” mind you): “What is the chance you are right?…” (the question seeks a probability as the answer, that’s what “chance” means”). Not the chance of “being right” but the chance that “your single choice is right”

Heisenberg anyone? Pin the 4 probabilities to the board and throw a dart at it. As soon as the dart leaves your hand it is flying towards an answer that it will invalidate by landing on it. But at the moment the dart leaves your hand each of the answers is neither right nor wrong because you haven’t made a choice yet. “If you choose” is the landing of the dart. “you will be correct” is the uncertainty at the time of guessing.

0% is the empirical, thoughtful, wordly answer after you are done guessing… hindsight is? a) 20 b) 20 c) 20 d) 20

B.

You have a 25% chance of correctly choosing 50% at random, which is the correct answer, since you had a 50% chance of choosing 25% at random when there was a 25% chance of picking the right answer all along in the first place.

Oops, that B. should be and A.

Oops again, should be A and D.

Wait, now I’m confused …

You guys are complicating things too much..

what is the probability of answering this question right , when the answer is chosen at random.

Now assuming a b c d have equal probability , the prob of guessing right is 25%… Since two options A and D have this answer. if we choose among a b c d , we ‘re right when we choose a and d..

SO that brings us to 2 right choices out of 4 , or 1/2*100 = 50 % !!! Pls dont talk about paradoxes in irrelevant places , just because u’ve heard the term !! and there’s that fool called PAOLO who’s given a shitty answer !!

Die richtige Antwort “E) 20 %” wurde beim Fotografieren leider abgeschnitten, um das Bild interessanter zu gestalten.

I’d say the correct answer is zero (or 0% if you prefer). It’s not telling you you have to choose one of those answers, it’s saying that if you were too, what would be the chance of getting it right. And the probability of randomly picking a correct answer from an answer bank that contains no correct answers is zero.

Alternatively: What is the probability a square has unequal sides?

a) 50%

b) 25%

The answer is zero, because regardless of what’s in the answer bank, all a squares sides are equal.

The logic on why there’s no correct answers:

If 25% is correct, you have 50% chance of getting the answer -> no good.

If 50% is correct, you have a 25% chance of getting the answer -> no good.

60% is just stupid.

While it’s possible to construct a self-referential question in the same vein that has a correct answer, this one is not written that way.

Answers corresponding to A,B,C,D are irrelevant. The (actual) question remains what are the chances of answering correctly a multiple choice question.Given that there IS a correct answer, the answer is 25% (1 out of 4 answers).

Taking your solution in account, Since A&D are same, so it is 50%(B) (2 out of 4 answers)

Well, yes. Actually it is. Its just that the “25%” answer should not be drawn from the A,B,C,D set of answers. Thats the tricky part of the question..This is a probability question about a question (which is not actually given to us), who’s answer is not actually included in the given set of answers..

This is a typical “nerd sniping” problem. See http://xkcd.com/356/

Is there an answer (to the XKCD resistance problem)?

It depends on how you define “correct” in picking an answer.

The answer is 25%. There are 4 answers to choose from 1/4 is 25%. Done. Not hard.

Stephen, two answers are 25%, so that doubles the chance of choosing a correct answer to 50%.

If only one answer was 25%, that would be the correct answer. However, it’s 50%, as two of the answers are both 25%, thereby doubling your chance of choosing the correct answer.

OK it’s easy …. B) 50%

When you chose you can be right or wrong. So you have 50/50 :)

If I have 4 choices, and one of those answers is correct, there is a 1 chance in 4 that I’d be correct. Or 25%.

Since 25% appears twice, I have a 2 out of 4 chance of choosing 25% as the right answer. Therefore, I have a 50% chance of picking the right answer.

B is the correct answer.

This may be the “best statistics question ever” because you can interpret the data any way you like and then justify your own answer by math/logic/grammar/disdain.

Since answers A. & D. are the same (25%) and all answers are equally posible to be chosen, the chance to be correct is 33.3%.

No ?

As stated, it’s not a paradox — it has an answer: 0%, because each of a,b,c, and d are wrong. But if you replace c) 60% with 0%, you get a kind of Gödel statement: within the context of the question, it’s a recursive contradiction, which can only be evaluated outside the question.

Here is the answer. I will give it to you if just PLEASE STOP going on…

The question is the first set and the answers are the second set. The question albeit poorly worded is self-referential. It becomes Russell’s Paradox.

From wikipedia…

According to naive set theory, any definable collection is a set. Let R be the set of all sets that are not members of themselves. If R qualifies as a member of itself, it would contradict its own definition as a set containing all sets that are not members of themselves. On the other hand, if such a set is not a member of itself, it would qualify as a member of itself by the same definition. This contradiction is Russell’s paradox.

(http://en.wikipedia.org/wiki/Russell's_paradox)

The concept is a broad examination into many math concepts. This picture making the rounds is a poorly written attempt to get one to think about Russell’s paradox. Now, please stop trying to solve a paradox.

Although there is a paradox among the choices a,b,c, and d, the question itself is not a paradox. It has a simple answer: 0%. No matter which choice, there is no possibility of its being right, therefore the answer to the question is 0%. That’s why somebody way back in the comments suggested replacing 60% with 0%, because the question then turns into a kind of liar paradox. As originally stated, there’s an answer which happens not to be among the multiple choices. With the replacement, it has no numerical answer.

1/3

There are three possible correct answers: 25%, 60% or 50%. We don’t know which one is the right one, so we have to add the three distinct probabilities that one answer is correct AND we choose it. There are three possible answer choices, so each has a 1/3 probability of being the correct answer. Now we can calculate the individual probabilities and add them.

Now, let’s assume that 25% is the correct choice. If this is true, then we have a 1/2 (50%) chance of choosing the right answer at random because it is both answer choice A and D. In order to calculate the distinct probability that 25% is the right answer AND we choose it, we have to multiply. Therefore 1/3 x 1/2 = 1/6.

We can perform the same operation for answer choices 60% and 50%. Since each of these has only a 1/4 chance of being selected, the distinct probability that 60% (or 50%) is the correct answer AND is chosen is 1/3 x 1/4 = 1/12.

Finally, we have to add the three distinct probabilities: 1/6 + 1/12 + 1/12 = 1/3

Any answer invalidates your assumptions, so just pick the answer that feels the most logical.

The probability for guessing correctly on a multiple choice with 4 answers is 25%, but…… since there are two 25% possibilities, then there is a 50/50 chance of getting the right answer. B 50% is the answer.

The answer is 0%. That is the only stable answer. If you randomly guess one of the four choices you will be correct 0% of the time..

That’s exactly right. Although the choices are recursively inconclusive, the question itself is not a paradox. It has a direct and simple answer, which happens not to be among those 4 choices. But if you replace c) 60% with 0%, as someone suggested way back, then you get a liar paradox that has no numerical answer.

We have to first be informed as what “this question” is before we can choose any of the alternatives. Why does it not phrase it to refer to “any question”?

my assumption: out of the 4 options 1 has to be true…and that i am allowed to tick only 1 option…

considering the above either 25% is correct or 50% is correct or 60% is correct….

ticking 25% as my option will occur with a probability of 1/2

ticking 50% as my option will happen at a probability of 1/4

ticking 60% as my option will happen at a probability of 1/4

so in effect my probability should be between 1/4 and 1/2….that is between 25% & 50%…but i have no option which is between 25-50%….and to start with in my assumption I have considered that atleast 1 option is correct…but my solution now shows that is not the case…

hence, the question which is being asked cannot be answered at all…it has no mathematical significance at all!!

Its a paradox. When you arrive at one answer for the question the probability changes thus changing the answer. 1st it changes from 25% to 50% when you discover that there are two answers that say 25%. But now this makes the answer 50%. However this makes the answer 25% again because there is only 1 solution that says 50%.

You guys are all making it too hard. Read the instructions carefully. “Don’t think too hard”. Does it say answer out of the list presented? No, those listed possibilities are just a red herring because when you see a list of options people naturally assume you have to give an answer out of the list presented.

If you think outside the box, there is an infinite list of possible answers. Write in your answer. It is effectively 0% or 1/infinity. Don’t circle any of the options because they are all wrong.

The answer is 0-100%. All the letters and percentages are part of the question since they are written but they don’t have to be part of the answer. It doesn’t say anywere that the answer is multiple choice

the answer is 50%… you are either right or you are wrong…

I saw this linked at http://wmbriggs.com/blog/?p=4608

Just thought I’d cross-post my two cents over here.

Everyone is assuming all the figures are in base ten. I make no such assumption. I assume all the answers are unique. Therefore one of the 25% is a different number base. You have only a 1 chance in 4 of randomly picking the answer containing the base ten number 25%.

I think we may use fixed point theorem to understand this problem, and the answer is: there is no answer!

If you say it’s 25% you’re wrong since the actual chance of picking A or D randomly is 50%.

If you say it’s 50% you’re wrong since the actual chance of picking B randomly is 25%.

I’ll pick C then, it’s certainly wrong but I like that certainty over the circular logic nightmare.

The fact that the choices are 25%, 50%, 60% and 25% are irrelevant apart from the fact that two of them are the same (the choices could be apple, orange, pear, apple). Basically you don’t know what this question is and you don’t know what the answer is (for all you know the question could be “What percentage of 24 is 6”, the answer would be 25%, and if you’d randomly chosen A or D then you’d be correct). The chance of you answering this question correctly IS 33% because there’s only three choices that are correct.

You know when I think about this there are two answers, if the correct answer is B or C then you’ve got a 25% chance of getting it right as a random choice could be one of 4. If the correct answer is A or D then it’s 33%.

The implicit assumption is that the percents are part of the problem. But the question could be read as follows: One of the letters (A, B, C, or D) will be randomly selected as “the correct answer”. If you choose randomly also, what is the chance you will select the correct letter? The answer is 25%. This is the only interpretation of the question that has an answer listed.

i’m not going to wade through the 500+ responses to see who said 100%, but that is the correct answer. why? http://en.wikipedia.org/wiki/The_Treachery_of_Images

There’s this other question that’s been floating around on the ‘net that granted, is asked in a misleading manner, but essentially uses the same “trick”: http://edudemic.com/2011/10/can-you-spot-the-mistake/

The answer is B. I didn’t consider the given choices, I just based my answer on simple binary logic. When I answer at randon, either I’m correct or I’m wrong, so my answer is that I have a 50% chance of being right, ergo B.

Like this one…

:-)

There is no question.

The probability of choosing one of the answers is 1/3.

1/3 is not a choice.

The question says “what is the chance you will be correct”.

The chance is therefore 0% – you will *never* be right if you choose one of those answers.

100%.

If I choose the answer then I am going to be right.

This is a pretty good diagram explaining why none of the given answers are logical: http://tiny.cc/luminousProbabilityCurve

Nice one !

It has a simple answer, and the question itself is no paradox. At the risk of being empirical, just evaluate each random choice. (a) is wrong and it doesn’t imply that (b) is right. It only implies (b) is right if you assume that (a) is right. But if (a) were right, it’d be wrong: contradiction. It is necessarily wrong and doesn’t imply anything else. If (b) is chosen, it’s just plain wrong because its likelihood is only 25%. So wrong, no contradiction. It doesn’t imply that (a) is right — it just implies that if (b) were right, then (a) would be right, and that would be a double contradiction. But that’s to assume that (b) is right. You don’t have to assume that with (b): it is simply wrong, and (a) does not imply it. (c) is obviously wrong. (d) same as (a). All are wrong therefore the likelihood of being correct is 0%. That’s the simple answer.

If you replace c) 60% with 0%, you get an interesting example of the distinction between internal and external contradiction: x contradicting y versus x contradicting itself or x contradicting y contradicting itself. If I say “I am a liar” then I’m contradicting myself, and it is both true and false. If I say “you are a liar” and you are not saying something recursively contradictory, then I’m just contradicting you. But if you say “x is a liar” and what x said was “I, x, am a liar” then you’re in as much of a self-reflective contradiction as x.

This question is the simplest of the three –the answwer to the question is “the choices are all wrong, therefore the likelihood is 0%”. a,b, and d imply each other if you assume each is the right answer, but that’s not the question. The question just asks the likelihood of a right answer, not whether if there is an answer, what’s the likelihood of that answer.

If you replace c) 60% with 0%, then you get a true liar paradox.

The answer is a) or d). My reasoning is that a) and d) refer to the same value, but are different answers–if you fill them in on a scantron sheet they will be treated by the computer as different responses.

yes i like ur response…. it can be that way. becoz if we are treating it as random selection so the answer would be as u said and if we use question logic rather than just than probabliity it would be that A and D cant be answer becoz they are same and one question cant have same answer……… so B or C can be answer. Which means……. 1 out of 2…….which is 50 percept………… But 50% cant be right becoz of Random selection. so 25 percent is the answer.

I appriciate ur answer

Maybe it is said before but the answer is very simple as long as you think of the definition of probability.

”The probability of an event is the ratio of the number of cases favorable to it, to the number of all cases possible when nothing leads us to expect that any one of these cases should occur more than any other, which renders them, for us, equally possible”.

According to that definition we have (a+d)/(a+b+c+d) = 2/4 = 1/2 = 50%

Maybe it was stated before but i was too lazy to read all the comments.

and also lazy to carefully read the question :-)

who said a) and d) are favorable cases? if you choose b) as an answer then b) is the only favorable case , isn’t it?

and that would lead to 25% probability (according to a definition)

:-)))

This is not a statistics question! It’s a probability question!

Here I am summarizing a lot of what has been said, except I explain it as a joke or a cartoon type statement, rather than a real probability question.

This question is making fun of multiple choice questions, like a cartoon makes fun of a politician for instance. The humor is that students will sometimes select answers randomly from a multiple choice selection. Out of 4 different answers, there would be 1/4 chance to get the correct answer. But if the same answer is offered twice out of 4 choices, the student is really left with only 3 choices, will probably not answer blindly, and will therefore have 1/3 chance of getting it right. If the student answers blindly, s/he has 50% chance of selecting one answer, and 25% chance of selecting one of the remaining 2.

In the case of this question, there are two references to an answer and it therefore includes a play on words. The first reference to an answer is “If you choose an answer to …” The possible answers offered are: 25%, 50%, and 60%. The second reference to an answer is in answering the question “what is the chance you will be correct?” If one of the answers is correct, the chance one is correct is 33% (provided one selects after looking at the choices). And if the answer is not one of the choices offered, the chance of being correct is nil.

Then reading the question again, one notices that the answer needs to be for “this” question. But neither of the answers for “this” question (either 33% or 0%) are among the options. Therefore the chance of being correct in answering randomly is nil. And that is the punch line to the joke: Answering randomly always leads you to the wrong answer and the student who thinks s/he has to guess, will be fooled!

Sounds good to me! I did not read all of the comments, but this one strikes as simple and clear.

Thanks :-)

The probabilities are a delightful obfuscation. Underneath is almost a semantic paradox (“The sentence is false”). Remove the probability obfuscation and you have:

How many of the answers listed below are correct answers to this question?

A) 1

B) 2

C) 2.4

D) 1

You can make it a full-fledged semantic paradox by following an earlier suggestion and adding 0 as one of the answers. (You might use it instead of the C) answer that’s given, or add it as another possibility, adjusting the probabilities accordingly.)

I went where you went momentarily John…. but that is a different question!

hey, this is my first attempt at anything on LInkedin, and being a math guy ( but actually weak in statistics and logic ) see similarity to a probablity exercise where you are led to think that any answer might have a probability of being WRONG 75%. Then you multiply .75 X .75 X .75 X .75.

i am not going to read ALL 579 comments before me, but……i see the ”question” as simply “….choose an answer….” thus making ALL answers as ‘appropriate’ ( satisfying the ‘question’ of choosing an answer ) and

so what looks like 4 multiple choice answers, are REALLY part of the question, which then has to be answered as a fill in the blank, and that answer is 100%.

did anyone else come up with that ‘logic’ ? And to Raymond Johnson, i ask, am i correct ?

Hey i think i have got the answer.

look If we look as per out question……….

A and D are same. But for a right answer there cant be two options. which means there are only two answer either 50% or 60%. so if 1 is out of 2 is answer so its 50% which is the answer.

But who told u that there can not be 2 same answers in this multiple choice question? You have to choose 1 but that does not say that there can’t be 2 equal answers

So this is why Wikipedia is not always right. There is a crowd gathered here and many of the contributors don’t know what they are talking about and aren’t listening to those who are trying to help (even if some of the ‘help’ does not make any sense to the rest of us).

There are no paradoxes in this question. The question is self referencing, which is why there is an answer (rather than the other way around). There are two questions on top of each other, but you have to read and answer both to get the answer. The answer requires very little math and just a few seconds of logic.

The answer is B.

If you cannot work out why, there is no short or pithy response that is going to help you here. The facts have been stated above, but there is always someone looking to ignore the obvious.

For all those choosing “None of the Above” and “if only I could insert ‘zero’ at option C then the question would be complete”, you are only reading one of the two questions written on the board.

It’s an example of the Russell / Zermelo paradox just as Darin stated above. If the problem offered a definitive solution surely it would have motivated less debate. The many arguments for 1:3 arrive at this answer by relaxing the problem request to ‘choose’ a correct answer. If instead, the respondent were to merely ‘state’ an answer, then the multiple choice options would form no part of the problem and the answer from probability would have no basis (since there would no longer be a set with three members we were required to pick from). Either the option choices factor in or they do not. If they do we arrive at a paradox. If they do not, then either the question is not well-formed or else there is an infinitesimally small chance (i.e., 0%) of selecting a correct answer to a probability question with unstated (i.e. infinite) outcomes. This last interpretation is surely not so interesting..

The correct answer is not offered :-)

It’s a feedback loop. Doesn’t work.

Correct answer can be either of the three options – 25%, 50% or 60%

if the right answer is 25%, then the conditional probability of choosing 25%, given 25% is the right answer = 1/3 * 1/2 = 1/6

if the right answer is 50%, then the conditional probability of choosing 50%, given 50% is the right answer = 1/3 * 1/4 = 1/12

if the right answer is 60%, then the conditional probability of choosing 60%, given 60% is the right answer = 1/3 * 1/4 = 1/12

Since all the three options are mutually exclusive, probability of choosing right answer would be 1/6 + 1/12 + 1/12 = 1/3

Given that the number of possible answers to this question is infinite, then the probability distribution corresponding to random selection is a uniform distribution over the interval [0,1]. The probability density at any given point on this interval, including the point for the correct answer, is zero = 0%.

To those who suppose that there is a finite number of answers: What is that number? Let us call it M = number of answers to the question, M is a fixed and given, positive integer. Taking any two answers within this set, a third answer can be created by linking the answers with “AND”, creating element M+1 of the set. Thus, the number of answers, while it might well be an integer, is not countable. In the mathematics of probability, this would lead once again to the uniform distribution over [0,1], and the same answer as above: 0%.

The answer must be ‘B’ – 50%

I choose one out of the 4 choices so thats a 25% probability. but then there are two choices of 25% – A & D.

Now, The probability of choosing one of A and D is 50%

then the answer is D that means…the chance of answering D is 25% then you have two answers…then the chance is 50%…25%..50%….

please, somebody Ask Marilyn.

I get 50% as correct. Alternatives a.25% and d.25% must be eliminated as being correct since a mc question can have only one correct answer or else the question is invalid and they are the same. That leaves b.50% and c.60% as the only possibly correct choices . The questn does not stipulate among which solutions Imust choose just to choose randomly. I randomly choose among b.50% and c.60% and would automatically be right 50% of the time since b. 50% is the only answer that fits. Thank you

if you rephrase this question just a bit, it would become “if you choose a number from 1 to 4 at random, what is the chance you will be correct?” it’s meaningless.

Like they teach us in school, multiple questions might be solved by elimination.

However here, you can eliminate all answers (this has been explained by many commenters already).

So the “answer” would be just like any other multiple-answer question with no correct answer: Calling the teacher and saying “Um… I think there’s a mistake in the test, none of the answers are correct.”

No paradox.

What if you changed c. to 0%? (as some suggested)

The same logic follows. this answer can also be eliminated.

But what if you changed c. to “None of these answers are correct.”

Now you’ve bitten the bullet…

You can still eliminate all the answers like before (c is incorrect because if it was a+d are and c isn’t),

but then if you call the teacher with the same story she’d just say: “Well, if that’s what you think, mark c!”

So you can’t conclude no answer, because that just makes c correct. But if c is correct…

This is Paradox.

if we say that “a” is correct this means that “d” also is correct and we know that correct ansver should be only one so “a” and “d” are not correct. in this case “b” also is not correct because any choice is either thue or falce and chances of this is the same(50%) for one each so if we say that correct is “b” this means that our answer is true for 100% and this will be against our answer(50%). so we have only one answer and this is “c”. to my mind “c” is correct answer

but if “c” is corect this means that this is correct for 100% so in this case we cannot to make choice because this is paradox and I agree with razdaz

If you choose an answer to this question at random, what is the chance that you will be correct?

A) 25%

B) 50%

C) 60%

D) 25%

It’s 0% (rounding up). The “question in question,” as I see it, is in the eyes of the beholder. It’s not a paradox, so much as it’s philosophical.

It can’t be 33%, by the way- because the question clearly states “at random,” while offering 4 choices, two of which being 25%, giving 25% a distinct advantage over the other options.

If the options were:

A.) 25%

B.) 50%

C.) 60%

D.) 10% (for example)

It would be logical to gravitate toward 25% (side note: thinking about the provided answers eliminates the randomness of the riddle, by the way).

In simpler terms, the whole thing is a crapshoot. It’s 0%- and my rationale is quite simple. The question asked, isn’t “the question.” Short of having telepathy, I can give you an answerable question, but who knows what your odds are of randomly getting it right are? Furthermore, what are your odds that any of the choices available are viable answers to my question?

I think this is a nice way of the person who created this puzzle to say “lighten up.”

It’s not math… but it is.

Aaron says, “It’s 0%- and my rationale is quite simple. The question asked, isn’t “the question.” Short of having telepathy, I can give you an answerable question, but who knows what your odds are of randomly getting it right are?” Paraphrasing, the odds of getting the right answer at random is zero, because there are “too many” possible answers.

Quite easy:

Probability of answer being correct multiplied with probability of that answer being chosen…

In sum that becomes 1/3.

P(25 correct) is 1/3

P(25 chosen) is 1/2

P(50 correct) is 1/3

P(50 chosen) is 1/4

Same for 60

Gives 1/6 + 1/12 + 1/12 = 1/3

You’re making this too complicated. There are only three choices (25, 50, 60). The chance that a random choice yields the correct result is 1/3, or 33%,

The correct answer is to not answer the question.

Don’t let your own assumptions trick you! If you read the question carefully and literally, you conclude that “this question” has not actually been defined. It is some question the specifics of which we know nothing about. It could be a totally random question, in which case you would probably have little chance of getting it right, or by the same virtue, it could be a question regarding one of your fields of interest, in which case you would have a large chance of getting the correct answer. But these, again, are just assumptions about the nature of a question which hasn’t even been asked. Since the author has left “this question” very general, we too must follow suit – in order to give an unbiased and objective answer –and provide a general answer, of which there is only one: generally, one has a 50% chance of being correct when answering a general question – either they are correct or they aren’t. Therefore the only fitting answer to such a vague question is B) 50%.

Hi, Drake, I agree with your analysis of the problem. Mathematically, the probability that pertains to this kind of situation is 0%, rather than 50%. You might think of the situation as being one in which “too many” possible answers are correct, and your chance of getting the one correct answer is the inverse of the number of possible answers. See my post above, and also the comment by Aaron. Aaron essentially says the same thing that you and I are both saying. -) Again, nice analysis, Drake.

Hi Wesley, thanks for the discussion. I also agree with 0% being the correct global answer in the sense that out of all the possible questions that can be formulated, you’d know the answer to such a small subset of these that the probability of you getting a question for which you know the answer is essentially 0%. The fact that you are choosing an answer at random adds an additional degree of complexity – now you have to account for the fact that you could correctly answer a question by pure chance and, conversely, you could randomly select an incorrect answer to a question to which you knew the correct answer. These probabilities are small and probably mostly cancel each other out. The real problem is that we are all trying to find a logical answer to a question which precludes finding such an answer because it is poorly formulated. In any case, operating under the ASSUMPTION that while the question is nonsensical the answer must be one of the 4 choices offered, in arriving at the answer of 50%, I employed a zeroth order approach in which, in the absence of any additional information, I strictly considered the local proability of an individual answering a question correctly at random. This answer is just as stupid as the question so it seemed fitting, LOL.

The obvious answer (after reading through this discussion) is “not very good”.

Actually the correct answer is 0%, so you have 0% chance to find it on a blackboard.

When you choose something “at random” there must be some probability distribution. Most people here are assuming a uniform distribution, but the problem doesn’t specify one. For any of the answers given you can choose a distribution that gives you that answer for the corresponding percentage of times. For example, if you randomly pick A %20 of the time, B %10 of the time, C 60% of the time, and D 10% of the time, then C is the correct answer.

Lane, it sounds as though you are looking for a probability distribution for the answers themselves. That would be correct if the problem involved some process that randomly generated an answer, and the issue was “how likely is that answer to be correct?” Continuing on with your analysis, in the example where C came out 60% of the time, the “60%” would be a reasonable (statistical) estimator for the mode of the distribution.

However, the problem on the blackboard is a different one. It presumes that you or I choose an answer, and that our choice be random. In the colloquialism of statistics, that means that we make our choice without having any particular reason to favor one choice over another. Thus, the choice itself is the random variable, and the distribution of that choice is the one relevant to this problem. Given that we have no reason to favor one choice over the other, the uniform distribution applies.

You make another point in some of your posts, namely that the correct answer is one of the four choices on the blackboard. That is debatable. However, even in that case, choosing one answer at random involves assigning the same probability to each of the four choices. This is a discrete uniform distribution, one that applies to coin tossing, dice, cards, and so on.

Again: we choose at random. That means that we could make any choice available to us with equal probability. It is not that the answer is presented to us as the outcome of a random process, and that we then have to estimate how likely it is to be the correct one.

Choosing an answer at random and “some process that randomly generated an answer” are the same thing. “Random” does not imply a uniform distribution, it only implies that there is some distribution. If statisticians are assuming anything, it’s that a random variable has a normal distribution, not a uniform one, but careful statisticians know that is an assumption that must be checked.

Indeed I am assuming that this question follows the rules of a multiple choice question in that the correct answer must be listed as one of the choices. I think that is a safer assumption than assuming that we choose the answer randomly according to a uniform distribution, for which there is no evidence.

A distribution that favors one answer over another is still random, it’s just not uniform. We can easily define distributions that give discrete answers with unequal probabilities. Consider a spinner that labels 3/5 of the disc as ‘C’; this would pick C 60% of the time. It’s discrete and it’s random.

If we are speaking of a single guy, randomingly choosing an answer, there are not correct answers, and so the probability is 0%.

But, what about a great number of groups (let’ say 100.000 groups) of 20 people each?

Among all those groups it will be someone where exactly 10 people randomingly have choosed “B”, someone where exactly 12 people have choosed “C”, ans some other where exactly 5 people have choosed “A” or “D”; that means that each of the answer has a (little) probability.

Waiting to find the way to do an analytical calculation, i have tried a simulation and the result is about 2.4%

The question is not fully specified. Any of the answers can be correct if the right distribution is chosen. We don’t know what the right answer is, but we know it’s not 0% or 2.4% because they are not among the choices.

The answer to that question is zero percent. First, figure out what the correct answer is. You will realize the correct answer is not listed. Therefore, the probability to pick it up is zero.

But 0% is not listed, therefore 0% is not the correct answer. Any of the given answers can be correct, depending on what probability distribution is used to randomly choose one. The problem does not specify a distribution, so we need more information. One thing that is certain is that the distribution is not uniform.

Lane, “choosing at random” is a colloquial expression in statistics. It is a valid situation for you to have a bias to choose “C” 60% of the time, so that your probability distribution would be some kind of modal distribution around event “C”. But “choosing at random” is not meant in that sense. It is meant in the sense of making a choice without favoring any one choice over the other, about having no information that would tend to make one choice more probable than another.

As far as your suggesting that a spinner would pick “C” 60% of the time if 3/5 possible outcomes were labelled “C”: That, too, is a uniform distribution in which three outcomes have the same value. You make a valid point for those who are looking at the A, B, C, D of the problem, and see that two outcomes have the same value, “25%”.

Lane, although I personally disagree, I won’t question your statement (in later post that the answer must be one of the four on the blackboard, because your arguments in other posts are clearly predicated upon that assumption.

Finally, Lane, thank you for pointing out that “choosing an answer at random” is mathematically equivalent to having “an answer generated by some process” (such as a lottery). You are correct, and my supposed distinction was not valid.

At first when I looked at it I was going to say 25%, but since there are two 25 percents there are two interpretations.

First assume its a multiple choice question if so:

Since there are two 25% then 25% cannot be the right answer. So, we are left with two choices and a probability of 1/2 of getting the right answer. So based on this scenario the answer would be B 50%.

Second assume its a theoretical question if so:

Then in this case the answer can be any probability and we are not just limited by the 4 options. You could justify 33% because the chance of 25% being right is 1/2 * 1/3 (25%, 50%, 60%) = 1/6, chance of 50% being right is 1/4 * 1/3(25%, 50%, 60%) = 1/12, and chance of 60% being right is 1/4 * 1/3 (25%, 50%, 60%) = 1/12. We can find the expected value by summing the probabilities (1/6) + (1/12) + (1/12) = 33%.

or maybe said another way:

(also my way of an apology to Keil :) )

________

If you take the position (some said constructed ;-) ) that there IS AN ANSWER: one and only one of [25%, 50%, 60%] and the question is unknown/irrelevant (for example: Instructor tells Grader before class: “The answer to today’s quiz is 50%…don’t ask why; not important.”) , then folks have shown a bunch of times why 33.33% is the answer.

________

If you take the position that there IS A QUESTION: “this question” (ie – self-referential) and the percentages listed are your odds [right now] of choosing a ‘correct answer’, then a bunch of other folks have said (although maybe not in this way):

Since the odds of choosing an answer presented once in the list is 25%, twice: 50%, etc.:

1x => 25%

2x => 50%

3x => 75%

4x => 100%

Then independent of any ‘correct answer’, your only choices are to go with one of:

“25%” (via A or D): 2x => odds are 50% (FAIL)

“50%” (via C): 1x => odds are 25% (FAIL)

“60%” (via D): 1x => odds are 25% (FAIL)

and since structurally all choices result in ‘FAIL’ then there can be NO ‘correct answer’.

and btw – substituting “0%” / “none of above” / “none possible” onto the list doesn’t change things since once added they’re subject to the same table:

“0%” / “none of above” / “none possible”: 1x => 25% (FAIL)

Greg, I think it helps to make an assumption about whether the answers are, in fact, limited to the ones on the blackboard. If they are, then Pranay’s analysis is valid. If not, then there is a strong case for choosing “0%”, as stated in my earlier posts.

Reviewing Pranay’s reasoning, while I personally think it is debatable whether the problem on the blackboard states that the correct answer is given by A, B, C or D, suppose, for argument’s sake, that it is. You have to select A, B, C or D at random, with probability ¼ for each letter.

The question then becomes “What is the probability that you have chosen the letter that is associated with the correct answer?” In general, “How likely is any one answer to be correct?” Assuming that you have no knowledge about this likelihood, and that no one answer is more likely to be correct than any other, then each of the three answers, “25%”, “50%” and “60%” has a probability of 1/3 of being correct.

Now suppose you make 600 random picks of A, B, C or D. On average, you will choose each letter 150 times. Since A and D have the same value, “25%”, you will choose “25%” 300 times. On average, the value “25%” will be correct 100 times. Choosing B means choosing “50%”. Out of 150 selections, “50%” will be correct about 50 times, on average. Similarly, D, leading to “60%” will tend to be correct around 50 times.

Out of roughly 600 picks, you will choose the correct answer, on average, 100 + 50 + 50 = 200 times: you will tend to be correct 200/600 times, or 1/3, as many posters, and Pranay above have suggested.

Other posters have suggested that the correct answer might not be listed on the blackboard. I agree with those posters, but that leads to a different answer than 1/3.

No one has shown that the answer is in { 25%, 50%, 60% } and that the answer is also 33.33% because that is impossible.

33.33% is not in { 25%, 50%, 60% }.

Wesley – thanks — and then with more time for this to rattle around in my head, I’ve come around to earlier points that:

“If you choose an answer to this question at random, what is the chance you will be correct?”

first part only specifies selection criteria (basically: pick randomly from the list below), and the second part asks only “what is the chance you will be correct?” Correct at what: picking randomly below?

There is nothing in the question tying the random selection (1 in 4 or 1 in 3, depending on your religion) to values/answers that can be good/bad outcomes with which to calculate probabilities. One can construct an argument (and I have) that “25%” / “50%” / “60%” are meaningless for our purposes (the unknown question argument), combined with the assumption (as I did) that one of them (and only one) must be correct (else our calculations don’t work).

But then those are some big leaps that the question also doesn’t speak to (or even imply) . So ultimately we don’t know the question, and we don’t know anything about the answer (is it in the list? does it exist? are there multiple or range of? or maybe it is to quack like a duck! (as CB said) ) . Ultimately: there is no question (other than the elusary “this question”) and therefore there can be no correct answer.

(my strong belief — weakly held. :-) )

http://xkcd.com/953/

This is not a paradox and there is no correct answer. Its just a simple case of the incorrect use of language.▲

The problem with the exercise is that it refers to a question that hasn’t been asked. The grammatical formulation: “if you choose an answer to this question, what is the chance you will be correct?” Asks the chance of correctness for a question that hasn’t been formulated. Which question am I to be correct or incorrect in answering? Or else it is self-referential and the question is that is being referred to as “this question” is the question “what is the chance you will be correct?” In which case the question is meaningless because this particular type of question refers to another question and cannot furnish a criteria in itself for correctness of incorrectness. The question “what is the chance you will be correct” is a question that implies a previous question that has criteria for being right or wrong. The question requires a previous question which there is a separate set of answers which could be broken down by percentages. It is an unhappy utterance in the game of asking questions, and the answer to the question might as well be “purple monkey dishwasher.”

Or else the question could be asked: if you are given a set of four numbers, 25, 25, 50 and 60, what is the chance that you will pick a particular number in the set, and if so is it the same as a number in that set? In which case, the number 25 can’t be in the same set twice. One could imagine someone saying “who said anything about sets?” But, then, I would challenge them to show me another way of actually formulating the question that leads to the problem. “Given four options of which only one is correct, what is the percentage chance that a person will randomly pick the right answer?” The answer is 25. Even if that answer shows up again, it is still 25 and we simply have a case of having two right answers. This is not impossible by any means for the purely mechanical process of evaluating a multiple choice test. There is no way to reformulate this question in such a manner as to reproduce the original problem, and the way it is formulated is an unhappy utterance that is, strictly speaking, meanigless. Case closed.

We might be thinking along the same lines where, after a bit more head scratching, I hit on:

If you choose an answer to this question at random, what is the chance you will be CORRECT?

a) 50%

b) 25%

c) 75%

d) 50%

Which took me waaaay back to Oct/28, this combination, and the question: what is ‘correct’?:

Abe — October 28, 2011 at 3:05 pm

“This is an incomplete question. The first part of the question says “if you choose an answer to “THIS” question at random.” What is the “THIS” question? Is it the second part of the question, which is, “what is the chance you will be correct?” If so, first, we need to know the correct answer to original “THIS” question. Then only we can determine whether the correct answer is present in the answer choices.”

alex — October 28, 2011 at 3:05 pm

“An interesting variant of this problem would have the answers choices of:

a) 50%

b) 25%

c) 75%

d) 50% ”

since both “25%” and “50%” match their odds of being selected (at random), then [a,b,d] are good choices and therefore odds are (3 of 4) or 75%, yeah? Which that means is also good, yeah? but does this make it (1 of 4) or (4 of 4)? …and if (4 of 4), are chances 100% or 0% since “100%” isn’t on the list?

It seems we (most of us anyway) have been assuming ‘correct’ == ‘matches what I calculated’. But I can calculate (and mean) different things…many in fact, and occasionally more than one at the same time. :)

(also pretty sure this isn’t a paradox in the logical sense since the problem is logically incomplete — rather than logically conflicting)

..and agreed: Case closed. :)

This is clearly a joke. The chance of selecting a correct answer at random from a 4 item multiple choice question is 25%. However because two items are 25% there is actually a 50% chance of selecting this answer if it was chosen at random. But if the correct answer was 50% then there would only be a 25% chance of getting it at random as it is only one of the four choices, making that false as well.

I disagree

It may be a joke, and the author may even be assuming a uniform distribution. The problem does not specify a uniform distribution, however. “Choosing at random” means “choosing at random”; it does not mean that you have to choose answers with equal likelihood. The mistake that most people here are making is to assume that one chooses the answers with equal probability, and then they try to compute the correct answer from this, but it only leads to a contradiction. There are an infinite number of distributions you can choose from for which don’t lead to a contradiction, but assigning 25% probability to each answer is not one of them.

I don’t come from a maths background, but this works well for me as a joke about students randomly picking answers in multiple choice exams, by posing a question about random selection in multiple choice questions that does not have a correct answer. I don’t really understand how picking one out of a group of four at random does not give each item a 25% chance of being picked, it doesn’t really relate to how I percieve it but I’m interested in the theory if you could explain it.

This “uniform distribution” argument is a red herring. There is no statistical distribution of the correct answer. It’s either a, b, c, or d. What difference does it matter if the answer is A or B if I am going to pick an answer at random? The real issue is that there are multiple questions being asked.

When I was in school teachers would tell us that if we wanted to pass multiple test without reading anything then choose C all the way down. Years ago I read somewhere that C was changed to B and now that’s what should be chosen.

Unbelievable (well — not really) that a teacher would tell kids this. But at the end of the day all that matters is that the students ‘pass’ and the teacher keeps getting paid.

I really hope to meet you all at the poker table someday.

Dear God. I have never seen so many utterly idiotic answers to a perfectly facile question. Almost everyone here is HORRIBLY over-thinking, probably to make themselves feel smarter. The only clever thing about this question is that it appears to be designed to demonstrate one’s propensity to heavily over-think a problem if it’s phrased in a self-reflexive way.

There is nothing difficult, complicated or paradoxical about the question. It is phrased in such a way as to make you think too hard and disregard the answer because it seems too simple.

The answer is B. That’s that. There’s no more to be said about it. It’s just B. 50%. You can all go home now.

LOL 600+ comments and that’s your takeaway?

I figured it was probably some sort of trick question where the actual answer is some philosophical crap. But I went ahead and took it at face value and chose B 50%.

The question doesn’t ask you what letter the answer it is. It asks you what the chances are. 50 percent is right. B (of which there is a 25 percent chance) is sort of wrong.

When you randomly answer any question, your either have the correct answer …or you dont…. so the answer is 50%

/thread

I disagree.

Following this logic, I would be rich by now. That would be the same as winning the lottery, either I win or I dont, 50%.

ding ding ding

Yes but the question does not ask how likely am i to win the lotto, it states “if you answer randomly to this question how likey are you to be “correct”.

Hence only two out comes can be atained correct or incorrect from answering the question randomly…

50/50 chance

its all about the question and it dosnt ask you to pick 6 ramdom numbers with a chance to win $$$…

um:

I have picked a number between 1 and 10. Guess what it is. What are the chances you are correct?

o shit i can not see your moves

Likelyhood of choosing “25 %” = 50 % != 25 % => (A) and (D) are wrong

Likelyhood of choosing “50 %” = 25 % != 50 % => (B) is wrong

Likelyhood of choosing “60 %” = 25 % != 60 % => (C) is wrong

Likelyhood of choosing “0 %” is 0 % => 0 % is correct

And this one (via alex)?

(which caused me to rethink things)

(same question)

a) 50%

b) 25%

c) 75%

d) 50%

That one is more confusing. The way my mind tries to figure this out, for that one I’d choose B – 25% if I were in a pessimistic mood and C – 75% if I were feeling lucky.

Same (0), since the two first leads to a contradiction (that two different answers are correct)

I sort of agree…but not 0%:

“If you choose an answer to this question at random, what is the chance you will be correct?”

The answer to the question is 25%? But that’s the selection criteria “choose [abcd] at random” — then what? I think its a leap to you then look to see if the thing you just calculated happens to be the value you just selected. So again as others have pointed out, the 1 in 4 is a condition not an outcome.

[and anyway – on a multiple choice test yes there’s is the one kind of ‘answer’ (the A/B/C/D), as in “mark your answer with a #2 pencil” but you’d never say that you got the answer A/B/C/D right or wrong, you’d say you got 1/3 or 3.14 (or whatever the value is) right or wrong. So to me the 25% as a value is a total red herring.]

I choose B 50% because A and D marked each other out , so all that’s left is B and C ..

the key is choosing at random, and therefore you have a 1 in 4 chance, thats 25%

The answer is very simple…. 50%

the reason for it is that if u read the question then the answer has only 2 possibilities, “yes” or “no”, which 50% probability of each occurrence.

50%

Funny, the question is part of the question, sort of recursive, thus undefined.

You can read it as:

what is the chance you will be correct: A(25),B(50),C(60),D(25) if you choose an answer to “what is the chance you will be correct: A(25),B(50),C(60),D(25) if you choose an answer to …”.

You can’t answer it, but it’s funny because it seems that it can actually have a meaning.

the way I see it..we can not answer this question..if the choices were all different it would have been 25%…but as two of the answers are the same…it will be 33%..then the answer is not given..then the probability is 0%..

The answer is B – 50% This goes into my theory of absolute relativity, in which you have a 50% chance of picking the correct answer and a 50% chance of picking the wrong answer.

50/50 Chance of being right or wrong

I have picked a number between 1 and 10. Guess what it is. What are the chances you are correct?

I guess 6, and there is a 50% chance that I am correct and a 50% chance that I am incorrect.

Think of it this way:

a) Dog

b) Cat

c) Giraffe

d) Dog

What was the question? Seriously, is there a question?

I think you’re on the right track (and by the way, this has been pointed out several times…which I ignored). And after much certainty, then doubt, then frustration, this variation nudged me along:

If you choose an answer to this question at random, what is the chance you will be correct?

A) 50%

B) 25%

C) 75%

D) 50%

Which you pick depends what you believe to be correct. The question is mute (it tells you how to selct (‘at random’) but not what is success/failure)

said another way (also earlier and better than me in this same thread):

Isn’t the question:

“If you choose [*] at random, what is the chance you will be correct?”

The first part tells you how to choose, but nothing in the question about right/wrong, success/fail. We assume because we picked correctly then that calculation must be, well, ‘correct’.

(“Look, I did seven riffle shuffles and cut the deck between 1/3 ~ 2/3 of the way down just like you asked, what do you mean [6 of clubs] isn’t a winner?”)

Thats pretty simple. Four answers, on correct (assumption). 1/4 = 25%

Yes, it can be argued that there are TWO 25% in the question. This means that either A OR D is correct. In that case, the probability of picking the right one is 50%. So the final probability is 0.5*0.25 which is 0.125. This is obviously not in the answers, which means you wasted you precious time reading this for nothing.

I would be correct 100% of the time. Read the instructions again, then follow them.

The only way to win is not play the game .the question implies choice when there is none .first off if you choose,secondly at random,the answer is there until you read the question because once you read the question and look at the answers you change the reality of the answer by looking at the answers you are force to think about it which in turn takes the random aspect out of the equation and any answer after that is wrong.unless you are blindly picking an answer without knowing the choices you will be wrong .just like the reality engine it works till you try and observe it forcing it into your reality.

remember the question if you choose which implies the 5th choice not to choose.

The answer cannot be A or D, because in either case, the probability would be 25% (A or D) and, at the same time, 50% (two chances out of four). By the same token, the correct answer cannot be B or D, because in either case, the probability of getting a correct answer would be 50% or 60% (B or D) and, at the same time, 25% (a chance out of four). Hence, the correct value is different from all possible answers and, since there is no chance of a correct answer, it must be 0.

Many people are balking at the seeming contradiction here, but I don’t see it. The selectors A B C D are actually different from the answers 25 50 60 25. If I rephrase the question to “There are four envelopes containing 25, 50, 60, and 25 dollars respectively. What is the likelihood that you will choose an envelope with 25 dollars in it?” Is not the answer 50%?

Yes, but the key question is that the answers represent likelikehoods, not dollars. You can have 25 dollars with probabilty 50%, but you cannot have 25% probability with 50% probability. That’s the reason why the problem is tricky: self-referencing, so to speak.

Correction: I should have said “25% probability and 50% probability, at the same time” instead of “25% probability with 50% probability”

The correct answer is 25% as there are 4 possible answers – A | B | C | D.

So, assuming one must be correct you have a 25% chance of selecting the right answer.

The paradox appears to exist since there are two answers that align with 25% A & D. This falsely give the impression that you have a 50% chance of selecting one of the two 25% percent answers and hence the right answer – yet that assumes that A. 25% and D.25% are not unique answers due to the combination of letter+percentage…..

The key to the riddle is that you must choose at random — so analysis is forbidden — and there for you have a 1 in 4 chance.

I think that, to make sense for the duplicate 25% in A and D, one must suppose that the purpose of the proposer was to create a question that mimicks those which are usually found in multiple-answers tests, where you mark in a form what you think is the correct answer. If such is the case, A-25% and D-25% are in fact two different answers, with a likelihood of 25% each.

Following your process, and while these odds are 25%:

A) 25%

B) 50%

C) 60%

D) 70%

These are not:

A) 25%

B) 25%

C) 50%

D) 60%

(odds of picking answer “25%” is 50% (independent of success/failure)

25%

If the question were “What are the odds of

randomly choosing the correct answer from

4 multiple choices, 2 of which, A and D, are

the same (and assuming the correct answer

is included), the answer would be 37.5%.

Chance of an answer being correct x

chance of picking that answer:

A 50%x25%=12.5%

B 25%x25%=6.25%

C 25%x25%=6.25%

D 50%x25%=12.5%

Total=37.5%

(It’s not 33.3% because even though there

are only three distinct answers there are

still four answers in total.)

If the question implies that the correct answer

is limited to the four choices listed then it is a

clever and/or annoying paradox: if 25% is the “correct”

answer then the “correct” answer becomes 50%, and

vice versa, so that the answer bounces back and forth

between 25% and 50%, the average of which,

coincidentally enough, is also 37.5%.

B=50%

Following your process, actually it’s 33.33%:

(1/3)*(2/4) + (1/3)*(1/4) + (1/3)*(1/4) = 1/3 = 33.33%

sorry, what I mean: on a multiple choice test, I don’t see how you decouple the 4 choices A/B/C/D from the 3 possible outcomes “25%”/”50%”/”60%”

(which is/are ‘correct’ is a whole different question)

Let us stipulate that there must be a correct answer and that we must accept the simplest set of assumptions sufficient for there to be a correct answer.

One can dismiss the assumption that there is only one correct answer. One can dismiss the assumption that the randomness is uniform.

I therefore postulate the following:

The simplest set of assumptions that provides for a correct answer is: the random function converges to one of the answers, and the random function used is one that deviates least from a uniform random function.

With the first component alone, any answer could be right and therefor we do not have sufficient assumptions to have a correct answer. With the second component convergence will be biased towards the answer most closely approximating equal probability. I assert that therefor the answer is “A or D” and leave as an exercise for the reader to prove that a the combined probability of two of four elements being 25% deviates less from a uniform random function than one element of four having a probability of 50% or 60%.

The difficutly is: are the choices (a) (b) (c) and (d)? or the contents of each answer. If you prefer not to think too hard, the correct answer is a or d. Since you cant pick both, either one is correct. Not just statistics here but a little game theory, eh?

Let’s make some equations (P(right) = probability for choosing the right answer)

and sorry that my english isn’t as good, i’m not a native speaker.

I: P(right) = ( rightAnswers ) / ( possibleAnswers)

II: rightAnswers = P(right) * possibleAnswers

I+II: P(right) = ( P(right) * possibleAnswers ) / (possibleAnswers)

P(right) = P(right)

1 = 1

Therefore no statement can be made about P(right), bringing me to the conclusion that the question is impossible to answer, if we don’t know the number of right answers.

And if we assume it to be 1, 2, 3 or 4 a paradox occurs, because the choices we have contradict this assumptions.

Math error:

(while I and II are consistent, they are also the same)

I: P(right) = ( rightAnswers ) / ( possibleAnswers)

#swap sides

( rightAnswers ) / (possibleAnswers) = P(right)

#multiply both sides by possibleAnswers

possibleAnswers * (rightAnswers)/(possibleAnswers) = possibleAnswers * P(right)

rightAnswers = P(right) * possibleAnswers

II: rightAnswers = P(right) * possibleAnswers

The answer is 50 percent.

The question asks what the odds are of picking a random multiple choice when there are four options. We all know the odds are 25 percent (assuming there is a correct answer to the question). Since two of the four answers in this particular multiple choice are 25 percent, we have 50 percent random chance of picking the right answer.

The question doesn’t say we have to pick the right answer from the choices given, but it twists the knife by offering the correct answer as one of the multiple choices, making it seem that the odds of picking the correct answer to the question (50 percent) is 25 percent.

But the question doesn’t ask us to pick the correct answer by chance or choice. It asks us what the odds are of picking the correct answer to a four option multiple choice question are if the correct answer is repeated once.

In this case, answers B and C could be anything. The right answer is 50 percent. Not B.

Not exactly, if all of a,b,c, and d are incorrect. You crucially assume that there is a correct choice among a,b,c or d, but there is no indication in the question that one of those is, except the assumption that multiple choices usually include a correct question. It’s a natural assumption, but not entailed by the question.

Moreover, why stop at that assumption and not try to find the correct answer, since you assume that one of them is correct in a normal multiple choice question? Once you try to test your assumption, you find it false. But one doesn’t have to go that far: a multiple choice question with two identical answers is already non normal.

The chances of getting a correct answer from among a,b,c, and d at random is the question. If none is correct, the chance is 0%. Since any choice of a,b,c,and d will be wrong, 0% is the answer to the question of the chances.

Apologies if someone has already written this but isn’t it a question about probabilities of probabilities? The chance of picking any one answer at random is 1/4. But two of the answers are 25 per cent so shouldn’t I add together the chances of picking 25pc with the chances of not picking 25pc? 1/2+2/3=3/5, or 60pc.

So, the answer is c, and my chances of picking c remain 25pc!