suppose 16 tennis players enter an unseeded single-elimination tournament, and that the better player wins every match. after 4 stages and 15 matches, the best player (who we call W) wins.
on average, which of the two following players is better?
a) the player who loses to W in the first round,
or
b) the player who won the first match, but then was beaten by the player who was beaten by the player who lost to W in the final round?
the answer, obtained by a 16-fold integral, surprised me.
Is the answer the same for a transitive "better" relation, as it is for an intransitive "better" relation?
i was thinking transitive. all bets are off for intransitive versions.
Clearly player A is a random selection from {everyone but W}. So is the "non-W half" in which B lies, whose "average average" therefore equals A's average.
Within this group, B is "the player who won, then lost to the person who lost to the winner" and we want to know whether B is above or below average in this half.
The non-winning half of this group is a random selection of four people from {everyone but the best in the group}. Call them P<Q<R<S. The non-winning half of *this* group is a random selection of two from P,Q,R. Call them X<Y. And then B equals Y.
OK. So, B's rank within {X,Y} is 1. So her average rank within {P,Q,R} is 4/3. So her average rank within {P,Q,R,S} is 7/3. So her average tank within the 7 non-best members of her half of the original tournament is (if I've got my head round this correctly) 7/3 . 8/5 = 56/15. So her average rank within the whole of her half of the tournament is 71/15. That's bigger than 9/2, so on average B is worse than halfway down her half of the tournament, which means on average B is worse than A.
No 16-fold integrals required, but this is the kind of thing I find very easy to get wrong. Did I?
your answer agrees with mine. if we think of players' abilities as uniform random numbers from 0 to 1, then after the tournament, E(A)=1080/2295 and E(B)=1024/2295. erich