As we all know, it can be very difficult to pair round 4 of a 4 player section. I’m curious as to thoughts as to how to pair the next (final) round.
# Name/Rtng/ID St/Tm Cb/Sx Rd 1 Rd 2 Rd 3
1 TF B 3 W 2 W 4
1427 1 1 2
2 AK ---- B 1 W 3
1423 X 1 2 2½
3 CB W 1 B 4 B 2
1352 0 ½ 1
4 JM bye W 3 B 1
1275 1 1½ 1½
Alex Relyea
2 options
pair 2 vs 4 since they are the only pair that hasn’t yet played
4-2 and 3-1 (equalizes colors for 3 and 1 even though it doesn’t reverse the earlier colors they played)
pair 2-1 and 3-4 to pair the top scores
First let’s make this crosstable readable:
[size=150][code]
score colors
1 TF 1427 2.0 bww W3 L2 W4
2 AK 1423 2.5 -bw X W1 D3
3 CB 1352 1.0 wbb L1 D4 D2
4 JM 1275 1.5 -wb B D3 L1
[/code][/size]
(With a little help from my friends, including myself, in the Displaying crosstables in the forums sticky at the top.)
Hmm. Why would AK get a forfeit win and JM a full-point bye? e.g. if it turned out JM had requested a half-point bye, and this fact was overlooked during the pairings, shouldn’t JM have been given a half-point instead of a full point?
Anyway –
In this sort of “round robin plus one”, I might consider fudging the colors in round 3 to make the probable colors after round 4 come out even, and to give each player one white and one black in the repeat pairing.
If, for example, after round 2 it is already obvious (or highly likely) who the leaders will be after round 3, these two (potential) leaders can be assigned opposite colors in round 3, so that after the leaders are paired again in round 4, all colors will come out even (both overall colors and rematch colors).
But in this case, with the byes and all, the objective in round 4 might be different. Instead of pairing the leaders, a better objective would probably be to insure that each player has played each other player at least once. This means pairing 2 vs 4, and thus 1 vs 3.
This gets wild in this case, because to alternate colors in the repeat pairing, you’d have to switch all the colors in round 3, giving one player two whites and no blacks, and another two blacks and no whites. Still, you already know what the round 4 pairings will be, so why not? Everybody ends up with two whites and two blacks, or two of one and one of the other, and the colors come out balanced in the repeat pairing. It’s worth it, but try explaining that to the players after they see the round 3 pairings and colors!
Bill Smythe
I figured AK got a forfeit win against a fifth player who was a round one no-show.
Actually, AK drew a player who was disqualified for the section.
Alex Relyea
I see you went with option two, and that the seven games had two draws, four wins by black and one win by white.