Perfect Bracket Odds

I am sitting at my local sportsbar in Covington, KY (Willie’s). They are running an NCAA March Madness bracket contest. I was asked by a patron what the odds are of picking a perfect bracket. I thought it was an interesting question, and so I did a little research.

Ignoring the play-in game Tuesday evening, if you just flipped a coin for each of the 63 games, the odds are (0.5 ^ 63) = 9,223,372,036,854,780,000 (9.2 quintillion) to 1 of getting a perfect bracket. If you assume that the #1 and #2 seeds will each win their 1st and 2nd round games, and that the #3 seeds will win their 1st round games, that leaves 43 games to get right. The odds of flipping a coin and getting a perfect bracket in this scenario is (0.5 ^ 43) = 8,796,093,022,208 (8.8 trillion) to 1.

By way of comparison, the odds of winning the Powerball is about 200 million to 1, or about 44,000 times more likely than picking a perfect bracket under the second scenario listed above. You’re about 50,000 times more likely to win Mega Millions with its odds of about 175 million to 1; you’re about 12.5 million times more likely to be struck by lightning (estimated to be 700,000 to 1).

So, let’s say you think you have a line on 43 of the 63 games, and have to guess on the remaining 20 games. Assuming that you are correct on the 43 that you picked, the odds getting the remaining 20 games correct are still slightly more than 1,000,000 to 1. (This hypothetical is somewhat of an over-simplification, but you get the idea).