## Calculating March Madness

I did join a little group of people competing to try calling the various NCAA basketball tournament brackets. It’s a silly pastime and way to commiserate with other people about how badly we’re doing forecasting the outcome of the 63 games in the match. We’re competing just for points and the glory of doing a little better than our friends, but there’s some actual betting pools out there, and some contests that offer, for perfect brackets, a billion dollars (Warren Buffet, if I have that right), or maybe even a new car (WLNS-TV, channel 6, Lansing).

Working out what the odds are of *getting* all 63 games right is more interesting than it might seem at first. The natural (it seems to me) first guess at working out the odds is to say, well, there are 63 games, and whatever team you pick has a 50 percent chance of winning that game, so the chance of getting all 63 games right is , or one chance in 9,223,372,036,854,775,808.

But it’s not quite so, and the reason is buried in the assumption that every team has a 50 percent chance of winning any given game. And that’s just not so: it’s plausible (as of this writing) to think that the final game will be Michigan State playing the University of Michigan. It’s just ridiculous to think that the final game will be SUNY/Albany (16th seeded) playing Wofford (15th).

The thing is that not all the matches are equally likely to be won by either team. The contest starts out with the number one seed playing the number 16, the number two seed playing the number 15, and so on. The seeding order roughly approximates the order of how good the teams are. It doesn’t take any great stretch to imagine the number ten seed beating the number nine seed; but, has a number 16 seed *ever* beaten the number one?

To really work out the probability of getting all the brackets right turns into a fairly involved problem. We can probably assume that the chance of, say, number-one seed Virginia beating number-16 seed Coastal Carolina is close to how frequently number-one seeds have beaten number-16 seeds in the past, and similarly that number-four seed Michigan State’s chances over number-13 Delaware is close to *that* historical average. But there are some 9,223,372,036,854,775,808 possible ways that the tournament could, in principle, go, and they’ve all got different probabilities of happening.

So there isn’t a unique answer to what is the chance that you’ve picked a perfect bracket set. It’s higher if you’ve picked a lot of higher-ranking seeds, certainly, at least assuming that this year’s tournament is much like previous years’, and that seeds do somewhat well reflect how likely teams are to win. At some point it starts to be easier to accept “one chance in 9,223,372,036,854,775,808” as close enough. Me, I’ll be gloating for the whole tournament thanks to my guess that Ohio State would lose to Dayton.

*[Edit: first paragraph originally read “games in the match”, which doesn’t quite parse.]*

## nebusresearch | The Math Blog Statistics, March 2014 12:09 am

onTuesday, 1 April, 2014 Permalink |[…] Calculating March Madness, and the tricky problem of figuring out the chance of getting a perfect bracket. […]

LikeLike