Author: Peter Berger
Date: 13:19:37 08/29/05
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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
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