This week the question is: Can You Play the Favorite?

March Madness—the NCAA basketball tournament—is here!

The single-elimination tournament consists of 64 teams spread across four regions, each with teams seeded 1 through 16. (In recent years, additional teams beyond the 64 have been added, but you needn’t worry about these teams for this week’s puzzle.)

Suppose in any matchup between teams with seeds M and N, the M-seed wins with probability N/(M+N), while the N-seed wins with probability M/(M+N). For example, if a 3-seed plays a 5-seed, then the 3-seed wins with probability 5/8, while the 5-seed wins with probability 3/8.

In one of the brackets, the top four seeds remain (i.e., the 1-seed, the 2-seed, the 3-seed, and the 4-seed). If case you’re not familiar with how such brackets work, at this point the 1-seed and 4-seed face off, as do the 2-seed and 3-seed. The winners then play each other.

What is the probability that the 1-seed will emerge victorious from this region?

And for extra credit:

As before, the probability that an M-seed defeats an N-seed is N/(M+N). But instead of 16 teams in a region, now suppose there are 2^k teams, where k is a very large whole number.

The teams are seeded 1 through 2^k, and play in a traditional seeded tournament format. That is, in the first round, the sum of opponents’ seeds is 2^k+1. If the stronger team always advances, then the sum of opponents’ seeds in the second round is 2^k−1+1, and so on. Of course, stronger teams may not always advance, but this convention tells you which seeds can play which other seeds in each round.

For any such region with 2^k teams, what is the probability that the 1-seed emerges victorious from the region?

Highlight to reveal (possibly incorrect) solution:

Program .

In order for the 1-seed to win, they have to beat the 4-seed and the winner of the 2-seed/3-seed match. The first win has probability 4/5. The winner of the other match is the 2-seed with 3/5 probability and the 3-seed with 2/5 probability. The 1-seed is victorious in this 2nd match with probability 3/5 * 2/3 + 2/5 * 3/4 = 7/10. Combined with the 1st win, the overall probability for a winner 1-seed is 4/5 * 7/10 = 14/25 = 56%.

My program confirms this.

And for extra credit:

There’s probably a smart way to combine all the above fractions into a sum, that then turns into, I don’t know, sin(2^k). 😁 I will however stick to my program, fiddled (!) a little to let k grow and see what happens. The probability quickly converges to 51.6%.