As a club TD, I have a question about pairings. The event is done with, but I ask this so that I know for the future.
We had 6 people in the top section for round 5. This is one of those 1-game a week for 5 weeks tournaments, so frequent 0-point byes. Here was the situation going into last night:
B = Black, W = White, NP = No Play, Sorted by score then rating
1 - B W NP W - 3 Points - Rating = 2087
2 - W B B W - 3 Points - Rating = 2019
3 - B B W W - 2 Points - Rating = 1825
4 - W B W B - 1.5 Points - Rating = 2175
5 - NP NP NP NP - 0 Points - Rating = 1936
6 - W NP B NP - 0 Points - Rating = 1874
As for who has already played who:
1 has already played 2, 4, and 6
2 has already played 1, 3, and 4
3 has already played 2
4 has already played 1 and 2
6 has already played 1
So which of the following pairings is right given who has already played, and color issues (like 1 and 3 both in dire need of Black, but due to face based on score/rating/history)?
3 - 1, 5 - 2, 6 - 4
Or
5 - 1, 6 - 2, 4 - 3
We went with the latter to avoid the triple-white for 3, but which is correct?
You should have paired 1 and 3. The requirement that players are paired with others in the same score group, or in the closest possible score group, outweighs avoiding three consecutive instances of the same color. Also, the correct pairing is 1 (white) vs. 3 (black), not 3 (white) vs. 1 (black). Both players are due black to avoid three consecutive whites, so you need to look backward in the color history for the most recent round in which the players had different colors. In the third round, player 3 had white and player 1 did not have a color. Thus, player 3 gets black in round 5. The pairings 5 (white) vs. 2 (black) and 6 (white) vs. 3 (black) are correct.
The first. Correct color is way down on the list of desiderata for pairings. Getting the score groups as correct as possible takes precedence. The second floats both the 3.0’s down to play 0’s when you have a way to pair one of them with the 2.0. Of course, given that it’s a rather free form tournament, the “best” pairing would probably be to have 1 play 2 again, but if that’s out, then have 1 play 3.
This I know for a fact is not “1 gets White”. 3 Gets White because equalization comes before alternation.
1 has had 1 Black and 2 Whites. 3 has had 2 Blacks and 2 Whites. Therefore, if these 2 faced, 1 would definitely get Black. The problem was we weren’t sure whether they should have faced because of the triple-White for 3. I guess we know now, and thanks for that clarification, but unless there’s a rule change I’m not aware of, equalization takes priority.
Why did you have 6-4 instead of 4-6? They are both equalized. Black was the last color each played. In the fourth round 4 had black and 6 had no color, so alternating from that would give 4 white.
Oops … sorry, yes, 4-6 is absolutely correct. I also made a mistake in my original post in that I had 6 paired with 3 (instead of 4). again …
I believe it is incorrect to argue “equalization takes priority over alternation” in assigning colors when pairing players 1 and 3. In both cases, the players have had white in the last two (played) rounds. Avoiding three consecutive instances of the same color takes top priority over both alternation and equalization. (Note that the rule governing three of the same color in a row [I don’t have my rule book with me, so I can not give a reference to a rule number, but it’s in 29E, probably 29E5] says that no player receives the same color in three consecutive rounds unless there is no other way to pair the score group. This trumps considerations of either equalization or alternation. Therefore, both 1 and 3 are “equally due” black.)
The rules is “29E5f. Colors in a series. No player shall be assigned the same color three times in a row, unless there is no other reasonable way to pair the score group or unless necessary to equalize colors.”
So color equalization DOES take precedence over not getting three in a row.
After the comma you ncluded the old logic (4th edition and earlier) to erroneously imply it was necessary to consider to get the correct result. Score and rating dueness is only looked at if the full color history is the same, so your statement was correct if you had ended the sentence where the comma is.
With the fifth edition, if both players are equally due then you first look at the most recent round where the colors differed and use that round (in this case giving player 6 some color and player 4 white).
If 6 had skipped round three and had black in round four then round three would have been the last round with a difference, resulting in player 4 alternating from that and getting black while player 6 thus gets white.
If both players had gone WBWB then the score (and rating for equal scores) would have been used to determine which player received the color due that player.
You’re right, Jack. The second section of Rule 29E4, under the heading Pairing players due the same color, explains how to assign colors for two players who are due the same color. The first and highest priority rule is “If one player has had an equal number of whites and blacks, while the other has had equal colors, the player who has had unequal colors gets due color. Example: WBW gets black over BxW. Note: x denotes any unplayed game - full-point bye, half-point bye, forfeit win, forfeit loss, etc.”
The rule that Ken is citing is the 4th priority: looking at the most recent round where the players had different colors.
I understand your reasoning, but I disagree with your interpretation. I think this is an ambiguity in the wording of the rule. I believe that the phrase “unless necessary to equalize colors” should be interpreted as “unless necessary to equalize colors for the player receiving the same color three times in a row”, not as “unless necessary to equalize colors for other players”. For instance, suppose at a U.S. Open (or other event of eight or more rounds) a player had the (rather improbable) color history WWBWWBB. I interpret the phrase “unless necessary to equalize colors” to mean that, in this case, it is permissible to assign the player black in order to equalize that player’s color allocation.
It would be interesting to see what the rest of the rules committee thinks.
As an interesting historical note, the prohibition against giving a player three of the same color in a row used to be absolute, meaning that one would actually break up a score group if there were no way to avoid doing so. Again, I’m doing this from memory, but I believe there is a variation (29E5f1?) that makes the prohibition absolute in all but the final round.
The change from being an absolute prohibition to being subordinate to not breaking up a scoregroup happened in the fourth edition of the rule book.
What is the source that leads you to this interpretation, that the actual intent of the rule was to limit the equalization priority to the player receiving three colors in a row? I presume it is something more than just that you think this is the way the rule should be. I have certainly never heard of this, from any source.
You may well be right but I would like to see some documentation, some indication that this issue was discussed by those drafting the rule.
If there are only two 2000+ rated players in the highest score group who are 3-0, and have the colors WBB, then it seems inappropriate to break up the score group and pair them against lower rated players in the next score group. They have to play. Rather than assign colors, I would have them flip for color. We used to do that at one time before computer pairing programs. I would skip the rule of the higher ranked gets color in that case. No one complains because they had a fifty-fifty shot at White in the last round. In small tournaments this sort of thing happens frequently enough that alternation, equalization rules have to be flexible. You are trying to get a clear winner. Now if the scenario happens in lower score groups where the players are likely to be closer in rating, then breaking the score group and pairing down to the next score group is possible.
That’s not how I interpreted it. In pairing the score group, the 200 point limit doesn’t apply when trying to avoid pairing someone who would get 3 in a row, but if you can’t avoid either pairing two people with (say) WWBB or one with WWBB with another with BBxW then you assign the colors on that board based upon the standard equalization and alternation rules.
I also made a mistake in my original post in that I had 6 paired with 3 (instead of 4).