Author: John Merlino
Date: 18:47:01 01/02/02
Go up one level in this thread
On January 02, 2002 at 21:39:54, Jason Williamson wrote: >The logic as far as I understand it is that with an even number of rounds its >possible to have for example 4 blacks and 2 whites due to alternation and >equalization rules (in a 6 round swiss). With an odd number of rounds the >chance is reduced and you are more likely to get 3-2 rather then 4-1 (I don't >think I EVER had to pair someone with colors THAT bad in a 5 rd swiss but in 6rd >swisses I have had to give 4-2 many times.) I'm sure that this is a valid argument that doesn't come across in the raw math. From what I recall in the directions to tournament directors in the USCF rules, they say that, if it can at all be avoided, no player should EVER at any time during the tournament have played as Black two times more than he/she has played as White (with a possible exception being after the first two rounds). Of course, I know that most of the other rules for pairings take precedence over that "suggestion", meaning that (for example) 4 blacks and 2 whites can occur frequently. But, for example, if you had 30 players in a tournament, and didn't have enough time for 7 rounds, would you really choose to have 5 rounds (the mathematically optimal solution) instead of 6 (the "add one round" solution) just because you prefer an odd number of rounds? jm
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.