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Topic: Bug: tournament tie-breaks
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There is a bug in the tournament tie-break system as applied to round-robin tournaments. This was pointed out by Beco in the `Off-Topic forum'; I'm reposting here to draw attention to it and suggest ways of fixing the problem.
The first tie-break is the average score of the opponents. However, this can never separate two tied players in a round-robin!
Suppose we have four players, A, B, C and D, with A and B tied for first place. The average score of A's opponents is (B+C+D)/3 and the average score of B's opponents is (A+C+D)/3. But A=B (the players are tied, so they must have the same score), which means that (B+C+D)/3 = (A+C+D)/3 -- the two players have the same tie-break score.
A correct way of doing this is to use the Sonneborn-Berger system. This assigns the following score to each player: total score of the defeated opponents plus half the total score of the drawn opponents. (If you want, you can then divide by the total number of opponents but there's little point since everyone has played the same number of games.) This system rewards players who score well against people at the top of the table.
As a secondary tie-break, it's probably best to use the score in the games between the tied players, which gives a good indication of who's better.
Using highest rating as a tie-break seems dubious to me. Suppose player A, rated 1500 ties with player B, rated 2000. A has done much better than his rating suggests so shouldn't he be rewarded for that by being declared winner? Player B has done worse than his rating suggests he should have -- shouldn't he come second?
Note, also, that any tie-break system is likely to perform quite badly in the case of two players tied in a four-person round-robin.
Dave.
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First of all, I do agree that the tie-break system could be improved, and I will at some point.
But your A,B,C,D example assumes that the tied players always played on the same group and have the same opponents. That is in fact not true for most cases, if they played in different groups in the rounds before the final then the opponents for each one will be different.
Miguel
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Hi Miguel,
I think there could be some true in his words, but I am not sure if the problem allways happens.
I think, as pointed by Andreavb in other topic ( http://www.queenalice.com/topic.php?id=9540&page=2 ) , the problem only appear when the winners make 12 points.
For example:
http://queenalice.com/tournament.php?id=1057&round=1
Group 1: 4 players / 6 games each
The winner Andreavb make 12 points letting 0 points to her opponents.
The other 3 players will play 4 games against each other, that means 12 points.
It does not matter if they will score like
xmas 2
J_Ricardo 2
ravague 8
or
xmas 0
J_Ricardo 4
ravague 8
or whatever...
This 3 players will always sum 12 points.
Then, if in the other group, the winner also made 12 points (likely to happen because we are talking about the two winners of the tournment), his opponent will also score 12 points.
So, lets say S1 is the opponent sum of group 1, and S2 for group 2. Lets say W1 is the winner of group 1, and W2 is the winner of group 2. And the opponents are A1, B1, C1 and so on.
Total point in a group is T, always 24.
T1 = A1 + B1 + C1 + W1 = 24
T2 = A2 + B2 + C2 + W2 = 24
S1 = A1 + B1 + C1 = 24 - W1
S2 = A2 + B2 + C2 = 24 - W2
If W1=W2 then S1=S2;
Lets say W1=W2=12
Then S1=S2=12
But if W1=10, then S1=14
and...
Keeping W2=12, then S2=12
This means that in case of tie in second round, scores of the W1 opponents (S1) is higher then S2, and the weird thing is that it is better to go to the next round with 10 points than with 12!
Got the idea?
Cheers,
Beco.
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| your A,B,C,D example assumes that the tied players always played on the same group and have the same opponents. That is in fact not true for most cases, if they played in different groups in the rounds before the final then the opponents for each one will be different. |
The tie-break uses data from earlier rounds?!? I have a number of problems with this.
1) Choosing the player whose opponents had the highest average score is exactly the same as choosing the player with the lowest total score through the tournament. That doesn't seem to be right. Having a lower score doesn't necessarily mean that the opposition was stronger -- it might just be indicative of carelessness or...
1.1) In particular, it means that the best strategy to win tournaments is to win the early rounds with the lowest possible score. That means deliberately losing all one's games once one has enough points to guarantee qualification for the next round.
1.2) If the tied players also scored the same total number of points in the earlier rounds (not unlikely, since most round winners seem to have ten or twelve points), the tie-break doesn't break the tie.
2) Why is information from earlier rounds only used when the score is tied? If the information is relevant, surely it should be incorporated into the score for the final, whether players are tied or not?
3) It would be better to just use rating, which includes more information than performance in earlier rounds of the given tournament. (I.e., it includes the results of games outside the tournament.)
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