color allocation

Running Chess Tournaments
1

Player # 9 is rated 1672, has one point after three rounds, has had BWW and is paired with player #6 who is rated 1811, has 1/2 point, has had WxW. Who gets white in round four? And where in the “Official Rules of Chess” can I find the answer? Seems to me that #9 should get Black, which equalizes his colors at 2 B and 2 W. Also giving #9 white would give him 3 whites in a row, which is prohibited. Yet, SwissSys says #9 gets white! Help, please.

Elberto

2

Have you taken a look at the Pairing Logic in Swiss-Sys? When in the pairing / results window click on tools on the drop down menu to find Pairing Logic. It may or may not help but that is the first thing I would look at.

3

SwissSys paired it correctly. The ratings and points did not matter. The color history of both players was all that mattered and was properly taken into account.

4

Color allocation given the pairings is completely deterministic. #9 has +1 W and #6 has +2 W, so since #6 has a stronger need for B, he gets it. The 3 in a row should come into play in deciding whether to pair these two, given that the colors are really, really wrong. But once they’re paired, the greater imbalance in colors controls.

5

Can you clarify? Clearly the relevant question is how the unplayed game for player 6 is to be handled. Since this is OP’s first post here he may not be well versed in these matters. For that matter, any reader who isn’t a TD may not be well versed in these matters.

To my uninformed way of thinking, there are two plausible approaches:

  1. It may be treated as either a W or a B, whichever is more convenient. If you treated it as a White, the pattern WWW would obviously force him to get a black in round 4. If you treated it as a Black, then the pattern WBW would get the white pieces in round 4.
  2. It should be treated as if it never happened. In that case, WW should get priority for the black pieces over BWW.

I always thought approach 1 was the correct approach, but I’m not a TD. Clarification requested!

6

It’s sort of like #2, but not quite—the position of the x can matter, it just doesn’t count as a color for determining imbalance. If xWW plays WxW, then xWW gets Black. Both are plus 2 W, so both have the same “claim” to B. Thus you work back from round three until their colors differ, which is in round 2 where xWW has W and WxW has nothing. So xWW is “more” due for Black than WxW. You only go to the higher ranked tiebreaker (higher ranked gets due color) if the color histories are absolutely identical.

7

I am a TD, and questions like this explain why I don’t direct much these days…but really:

My first reaction is that this pairing should be avoided at almost any cost. If it’s the last round, you wind up with a player who has three straight of the same color to end with 3-1 colors. If it’s not the last round, then that player absolutely has to have Black the next round no matter what. (And really should have another Black for round 6 is there is one.)

If it’s a smallish event—which is the only way this pairing could seem forced in the first place—with little on the line but rating points, I would not hesitate to get creative. Can Player 9 get dropped farther down by a spot or two?

Here is where the FIDE approach of no 200-point limit for color equalization and no +3 colors—meant for big-time events with Masters—helps pair small events with class players. Go figure.

8

I’m afraid I disagree with Mr. Doan’s analysis, but I do agree with his conclusion. Perhaps I’m confused by the statement that “both players have +2 Whites.”

Player 9 has color history BWW, so he has had one more white than black. Player 6 has color history WxW, so he has had two whites and zero black. Giving either player white would result in that player receiving the same color three times in a row, so there’s no avoiding that problem.

As player 9’s color history is more imbalanced, he has a stronger claim on his due color for equalization. It is not even necessary to look at color history, and in fact using color history to determine color allocation leads to an incorrect result (giving player 9 black as he had white in the second round).

I do agree with Mr. Mark that this pairing should be avoided if at all possible. Rule 29E5f requires the TD to do everything possible short of breaking up the score group to avoid giving a player the same color three times in a row. (Interestingly, in the FIDE Swiss pairing rules, the ban on the same color three times in a row is absolute, and the arbiter is to break up the score group if it is not possible to avoid such a pairing within the score group.)

9

Sorry. I changed the situation from the OP to create a situation where the position of the x mattered.

10

I apologize to Mr. Doan. After more careful reading, I realize that in his first reply, he did identify player 9 as having one more white than black, while player 6 has two more whites than black.

11

Check rule 29E4, in the paragraph Pairing players due the same color, on page 151 of the 6th edition (or page 144 in the 5th edition):

2. If both players have had an unequal number of whites and blacks, the player with the greater total color imbalance gets due color. Example: WWBW gets black over xWBW.

In the situation laid out by the OP, player #9 is out of balance by 1 (one more white than black), while player #6 is out of balance by 2 (two more whites than black), so player #6 gets due color.

Bill Smythe