Bracketology

The NCAA basketball tournament offers some fun exercises in combinatorics and probabilities. Here’s some ideas to think through.

Suppose Iowa has even odds against any team they play i.e. Iowa beats any team in the tournament with probability 1/2. What’s the probability that Iowa wins the tournament? A possible answer is 1/64 – if each of the 64 teams have the same chance of winning, then each team wins with probability 1/64.
However, what if we assume nothing about the relative strengths of other teams except that Iowa has even odds against each of them? For example, suppose Indiana beats Illinois with a 90% probability. Iowa’s chance of winning the tournament is still 1/64! Iowa wins the tournament if it wins 6 games in a row, which has a probability of $\left( \frac{1}{2} \right)^6 \equiv \frac{1}{64}$ and this does not depend at all on how other teams fare against each other!
How many games are played in the tournament in all? The direct way to calculate this is to just add up the numbers of games in the first round + second round etc i.e. $32+16+\dots 1 = 63$ .
There is however, a more fun way to calculate this number. Note that every game results in exactly one team being eliminated. At the end of the tournament, 63 out of 64 teams must be eliminated..
More generally, if we can predict each game with probability p, how large must p be to give us even odds of getting a perfect bracket? Clearly, we need to solve for $p^{63} = \frac{1}{2}$ or $p=\frac{1}{\sqrt[63]{2}}$ . This number has fewer nines in it than one may expect..