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Topic: Tournament winners tied
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Suppose we have a round-robin tournament with players A, B, C and D, and that players A and B are tied for first place. It will always, as Beco points out, be the case that the average scores of A and B's opponents are the same. Why? Because the average of A's opponents' score is (B+C+D)/3
and the average of B's opponents' score is (A+C+D)/3. But A=B (they're tied for first place so must have the same score!) and, so, (A+C+D)/3 = (B+C+D)/3.
The usual way of constructing a tie break is the Sonneborn-Berger system. Here, a player's tie-break score is the sum of the scores of the people that player beat, plus half the sum of the scores of the people s/he drew with. (There's no difference between taking the total and the average.)
Note also that taking the opponents' average rating as a tie-break gives exactly the same result as choosing the lowest-rated of the tied players as the winner. Why? Well, letting A, B, C and D stand for our four players' ratings, now, with A and B still tied in first place, we have
A's average: (B+C+D)/3
B's average: (A+C+D)/3
But (B+C+D)/3 > (A+C+D)/3 exactly when B>A. That is, A wins the tie break exactly when his rating is lower than B's.
I'm going to post this as a bug report to the help and support forum, too.
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