Odds of a perfect NCAA March Madness bracket

March 22, 2013  |  Statistics

Math professor Jeff Bergen explains the odds of picking a perfect bracket.

The first probability is based on a 50/50 split of correct picks, which is like using fair coin flips to pick winners. Bergen doesn't really go into how he calculated the second probability, but that smaller number comes up by bumping up the probability of picking the right team for each game. I think he's using an average probability of slightly less than 70% (based on simulation results from this old Wall Street Journal column).

That's why businesses can offer up million dollar prizes. In all likelihood, no one is going to win, which turns out to be a great business model for insurance companies who back these contests:

If millions of people enter a particular contest, it might seem like the chance of someone winning is suddenly in the realm of possibility. But there's a catch: This scenario assumes everyone maximized their chances by picking mostly favorites, so those with the best shot at winning are likely to have identical entries. These contests generally protect themselves from big losses by stating they'll divvy up the loot if there are multiple perfect brackets.

These favorable conditions make insuring these prize offers a good business, as the Dallas company SCA Promotions has discovered. SCA, founded by 11-time world bridge champion Robert D. Hamman, has taken on the insurance risk for roughly 50 perfect-bracket prizes -- including a Sporting News offer of $1 million in 2001, according to vice president Chris Hamman, the founder's son. In the 12 years it has been doing so, SCA has never had to pay out a claim.

4 Comments

  • Having trouble w/ the link to the WSJ article on those insurance companies.

  • Probability of picking a bracket that earns 0 points: 1 in 4.3 billion.

  • Say,what?

    1) If going with the coin toss (p = ½), the probability is infinitesimal. There are 32+16+8+4+2+1 = 63 games. 2 ^ 63 (2 to the power of 63) = 9.223372037 * (10^18) = 1 in 0.18 zeroes, then 92233…

    2) If going with the higher odds for some favorite teams, there are still upsets. The #15 seed Florida Gulf Coast upsetting #2 Georgetown would have a probability of 1/100. A #15 going to Sweet 16 would have had a chance very close to 0 (it had never happened before!)

    3) For all intents and purposes, the probability to have a perfect bracket is nil, nihil, nada… Even if we have ALL favorites win all the time with an unreal probability of 90% (9/10) is very low. It would be (9 ^ 63) / (10 ^ 63) = 1.31 *10^60 / 1 * 10^63 = .0013 = 1.3%. Who came with that outrageous figure, 70%.

    Reality is you’ll never have 90% favorites who will win all the time. It’s called March Madness because there are upsets every year…

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