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Now the first ELITE tornament is finist with win by my icelandic friend IAMC. He was tied with SaintBazoka but IAMC is the winner by his higher ELO as it seems. not justice to me as it is more achivement for the lower rated player..perhaps this is like in the real world were the rich man is always higher than the poor one. I saw some days ago that some player was crazy because of this situation were he was placed second... 
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| not justice to me as it is more achivement for the lower rated player.. |
Interesting... I tend to agree. A lower rated player that tied with a higher rated one definitly made a better performance than the expected.
For now, congrats to IAMC... He is the 1st Auto-Elite winner!
But Miguel should take a look in this suggestion and change the rule.
Cheers,
Beco.
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Hi, this is from www.queenalice.com/about.php :
Tie-Breaks
When two or more players obtain the same score in the final round of a tournament, the following tie-break methods will be applied in order:
Averaged opponent score: This value is computed by adding the scores obtained by all the opponents in all rounds (ignoring the games played against the player) and dividing the total by the number of games played. The player with the highest value will be the winner, presuming that he/she had a tougher competition.
Rating: The player with the highest rating will be the winner.
Random: A random player will be the winner.
iamc win because of method 1.
I suppose Methods 2 is not used in a lots of cases. It is improbable that the averaged opponent scores are equals. But yes, it seems to be unjust.
I think Method 2 could be erased.
Sorry for my spanglish. Cheers, Francisco.
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Hi Francisco,
I think iamc won because method 2... Look:
http://www.queenalice.com/tournament.php?id=198&round=4
Round 3:
Tie-break - Averaged opponent score (x1000)
iamc 1021
saint_bazoka 1021
Round 4:
Tie-break - Rating
iamc 2110
saint_bazoka 1951
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Funny thing is that it is suposed to finish more games in tie-break method 1 (by average opponent), but here we have another case:
http://www.queenalice.com/tournament.php?id=778&round=4
Round 3:
Tie-break - Averaged opponent score (x1000)
dstrutz 1079
sakhan 1079
Round 4:
Tie-break - Rating
dstrutz 1887
sakhan 1789
It is a coincidence, I think...
Bye,
Beco.
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Yes Beco. I was wrong.
I looked at "round=4"
and I thought it was the first tie-break.
Thanks.
Francisco.
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Since ratings are one of the key elements in tournaments, primarily Swiss tournaments for pairings they must be considered in tie breaks as well. Typical Swiss, you pair the top half against the bottom half (ratings).
You might argue that the lower rated player has had the better performance and should be declared the winner however; the correct decision is to award the tie break to the player with the higher rating.
This argument has been around for a long time, the number of points in a tournament is not the only factor to consider, depending on the tournament you can consider, colors, performance, ratings and draws versus wins to make a decision.
To make a long story short, it was decided years ago that the most consistent method of declaring a tie break (if no play off games) was to award it to the higher rated player because this player has already proven his/her strength (with the higher rating / performance) where as the lower rated player has only proven he /she has the potential with his/her performance in this tournament.
To put it in another way, the same principle is used in the World championship match, at the end of the match (games) if there is a tie; the champion retains his crown, as it is in all forms of competition.
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