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Ahhca
Hi friends,
I understand that Bucholz system (where sum of all opponent's scores are added up to get a player's score) is one of the most effective and widely used Tie break system in Swiss Chess Tournaments.
My doubt is, say in a 8 round Swiss Tournament, what happens if 2 players I met in rounds 2 and 3 withdraw from the rest of the tourney from round 4?
How will this be taken into account while calculating my Bucholz score? Because it is obvious that such a player will lose out if the sum of all players are considered because 2 of the opponents wouldnt have scored for 5 rounds!
Is there any rule or something in FIDE/USCF to factor this?
Any suggestions/opinions will be very helpful.
Thanks and Regards,
Arun
rooperi
I've never thought about that. It seems like a good question.
rigamagician
I think that if they have played less than half their games, their results are stricken from the record. Perhaps you get a point for a bye, but a low tiebreak score. Fischer withdrawing from Sousse Interzonal in 1967 is a famous example of a player who withdrew, although that was not a Swiss.
I found the relevant passage in the FIDE Handbook:
"Handling of unplayed games for calculation of Buchholz (Congress 2009) There are two points of view:
Another correction from the FIDE handbook:
"If a player withdraws from a Swiss-system tournament the points scored by him and by his opponents shall remain in the cross-table for ranking purposes. Only games that are actually played are rated."
echecs06
That is a TD's nightmare, been there, done that! It is true that , in case of withdrawing players, their points and ratings remain in the final computation of the tie-breaker. People think it's unfair, but experience shows it is the most fair system to use.
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