Problem 17 Solution: "Seeds of Victory"

No. The problem is that there are 12 ways of choosing a Final Four with seeds 1,1,2,3, and only one way of choosing a Final Four with seeds 1,1,1,1. If I choose the Final Four with seeds 1,1,1,1, I will have a 1 in 38.02 chance of success. However. if I choose a Final Four with seeds 1,1,2,3, then I will only have the correct Final Four if both (a) the final four has seeds 1,1,2,3 (this has probability 1 in 14.05), and (b) I correctly predicted which of the twelve (1,1,2,3) combinations would happen (this has probability 1 in 12), for a total probability of 1 in 168.6.

To give a simpler analogy, let's say I flip two biased coins that each have probability 0.6 of coming up heads. It's easy to see that the probability of two heads is 0.36 while the probability of one tail and one head is 0.48. But that's because the "one tail and one head" is really two outcomes, (head, tail) and (tail, head). If I was forced to predict the results of each of the two coin tosses separately, a prediction of (heads, heads) would be right with probability 0.36 while a prediction of (heads, tails) would be right with probability 0.24. And in March Madness you predict each conference separately, not the Final Four seed distribution as a whole. The situation above is the same, where "1,1,2,3" is really 12 different outcomes. ((1,1,2,3),(1,1,3,2),(1,2,1,3)

, etc.)

(Yes, I did send this analysis to Jacobson. I just did it a few minutes ago, so I haven't heard back from him yet, though.)