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. erich
On Monday 05 September 2011 22:00:10 Erich Friedman wrote:
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.
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? -- g
Is the answer the same for a transitive "better" relation, as it is for an intransitive "better" relation? On Mon, Sep 5, 2011 at 2:00 PM, Erich Friedman <efriedma@stetson.edu> wrote:
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.
erich
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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
On 9/6/11, Erich Friedman <efriedma@stetson.edu> wrote:
... 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
To some extent, the algorithms used to rank players in chess, tennis etc. reflect this phenomenon. At one time, a highly-rated chess player would refuse a match with one much lower-rated, on the grounds that regardless of who won, the weak player's rating would improve at the expense of the strong. I once refereed a somewhat poorly attended tournament, at which the stronger players were finding themselves under-occupied after quickly winning their games; so in between times, I challenged them to friendly matches, which unknown to me subsequently somehow or other found their way onto the official returns. When the annual league ranking list was published, scanning down columns revealed one idle but fortunate individual who had managed to land halfway up the table on the strength of only 3 games. Scanning across rather lengthy rows eventually identified this ingenious operator as none other than --- Fred Lunnon
participants (4)
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Erich Friedman -
Fred lunnon -
Gareth McCaughan -
Tom Rokicki