Author: Robert Hyatt
Date: 19:26:01 08/29/05
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On August 29, 2005 at 16:19:37, Peter Berger wrote: >On August 29, 2005 at 14:51:21, Robert Hyatt wrote: > >>Just noticed something strange. Crafty and Junior finished with scores of 6.5, >>which would (should) mean we tied for places 5/6. But somehow the final >>crosstable has crafty 5th and Junior 6th. The "SB" tiebreak somehow favors >>Crafty. I'm not sure how this is computed, but in a RR, it would seem that >>tiebreaks are pointless unless it somehow factors in that we had 6 blacks, or >>that we had black against stronger opponents and white against weaker opponents, >>or whatever. >> >>Anyone care to enlighten me on how that particular tiebreak works? I notice >>that Sjeng and Shredder are listed as tied for 3/4, with the same SB scores... >> >>Would seem more reasonable that Crafty/Junior are listed as tied for 5/6... > >Something agreed on in the players' meeting. Sonneborn Berger as explained by >Gábor was used as tiebreaker in case of equal points. > >It's the most common system in human RR tournaments. Wins or draws against >opponents who scored better count more. In this case there is *some* epical >justice in it as Crafty won against Junior which is the major reason for the >difference in SB. Not that a tiebreak for the 5th place is of great importance. > >To add some more irrelevant stuff: last year Buchholz was the general tiebreaker >in Ramat-Gan (most common in Swiss) , which meant that Fritz and Crafty finished >in a dead (and somehow futile) race for 4th rank. Yet all tables on webpages had >Fritz ahead because CB uses SB automatically creating the tables ;) . > >Peter I suppose anything goes to promote one's product. :)
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