Here's a nice solution from my friend Bryan Clair at Saint Louis University. It doesn't use integrals, and I think the calculations are easier to understand than in Gareth's solution. ---------- Forwarded message ---------- Let's say W wins, beating R in the first round. W beats X in the championship. X beats Y in the semis Y beats Z in the 2nd round. Z beats U in round 1. We want to know if R is better than Z. So: There are 16 players. Pick W. Now there are 15. There are 6 players in W's bracket besides R. They don't matter. That leaves 9 players, and we want to know if R is better than the 3rd/4th finisher from the other 8. Well, 1/9 of the time, R is the best of them. Otherwise (8/9 of the time), X and the three players on X's side of the bracket don't matter. Then we have 5 players left, and we want to know if R is better than the 2nd place finisher from the remaining 4. 1/5 of the time, R is the best of them. Otherwise (4/5 of the time), Y and the player Y beats in round 1 don't matter. Then we're left with R, Z, and U. Then from these 3 players, is a random player better than the winner between the other two? Yes, 1/3 of the time. So: 1/9 + (8/9)*(1/5 + (4/5)*(1/3)) = 71/135 Pretty clear you could do some kind of induction and handle more than 16 players, or treat some (maybe all?) other comparisons. Bryan