New Rule Suggestion re: Color Allocation

Add a new Variation 29E8 (or make it earlier and renumber some of the other 29Es):

In pairing the last round of an event, after correctly pairing all boards with regards to color equalization, all other boards shall be assigned colors using the announced event tie-break procedure where the player with the highest tie-breaks receives White. It is recommended that, when this variation is used, written notice be posted before the start of the tournament, and it must be used in all such cases without exception.

This would mostly take effect in last rounds of odd-numbered events. This seems a lot fairer to me than due color pairings where Swiss randomness can royally hurt people sometimes. Here’s one example by round from a recent real life event:

2076 player: 1-B1650, 2-W1797, 3-B1975, 4-W2104 (avg. 1881.50)
2074 player: 1-W1521, 2-B1788, 3-W1696, 4-B1918 (avg. 1730.75)

In round 5, entering final round with perfect scores, they play each other with the 2074 player receiving White despite the 2076 player having to play an average of 150 points higher rated opponent per round. Why should the 2076 player get punished yet again with Black by the vagaries of the Swiss system? Computers should be able to handle the tie-break element without blinking.

  1. You don’t indicate which type of tie-break to be used and your example indicates you want the opponents’ average rating used as the tie-break.

  2. The last two rounds are often the most difficult. Your suggestion would make it more likely to have people with either two whites or two blacks to finish the tournament. Since white has the first-move edge that may be seen as giving such players an unnecessarily advantageous edge. If you don’t feel that white has a first-move edge then there is no reason to play with the color allocation rules.
    Just wait until one 4-0 player went BWBW and is higher rated than another 4-0 that went WWBB. Your change would result in BWBWW vs WWBBB with the lower-rated player wondering why the player was stuck with black during all of the last three rounds when facing the toughest opponents.

Why should the rule say what tie-breaks are being used? Obviously, the tie-breaks being used at the event. This has nothing to do with opponent averages - although having a higher average rating of opposition is generally a corollary of higher tiebreakers.

If someone has two blacks to end the tournament as a result of this rule, then oh well, he should be grateful he had an easier road to get to the score in the first place. That’s not ideal but it more fairly balances the breaks than the alternative. As far as three blacks is concerned, that would be incredibly rare, and it’s a sticking point for most people, so I think you’re right to suggest a small amendment, “Except for 29E5f…” would be a reasonable addition to the above rule.

As I read the proposal, if the lower rated player had the better tiebreaks because he played the tougher schedule, he would get White, so it would be WWBBW according to my reading of Jeff’s example.

Correct. In my example, if the opponents of the 2074 player had great results, it sucks for the 2076 player, but at least the rule would give him black for an arguably fair reason instead of randomly give him black.

I’ll simply disagree with your suggestion (which allows a player to finish with BWBWW vs WBWBB), disagree with your statement that due colors are random, and disagree with the apparent premise that tie-breaks themselves are not particularly random (I remember one 4-round tournament where two 4-0 players each played four different 3-1 players, with some of them getting 3-1 by scoring upset wins).

I agree with jwiewel’s statement. On tie-breakers, sometimes they are of no value, but serve practical purposes.

Two players finishing 4-0 could not have done any better. It was the pairing method that determined who they played, and they both met the challenge completely. Playoffs, speed games if you wish, test the players against each other. Time contraints at a site might make that impossible, but I think it is desireable if any prizes would otherwise be determined by tie-breakers.

Two players at 3 1/2-1/2, could have done better. Using tie-breakers could indicate whether one player had tougher opposition. Then again, if they had played each other, they both met the challenge presented by the pairing method completely.

Of course, all of this should be announced before the first round.

All the best, Joe Lux

I think it was the late Ira Lee Riddle who once said “Tiebreaks are a mathematical process designed to irritate the maximum number of players.”

This sounds like a great way to increase that number.

Swiss system idiosyncracies annoy me more. Oh well - agree to disagree.

There’s no reason why you can’t run a tournament using this rule, as long as you announce it in all pre-tournament publicity. Let us know how it works out.

Alex Relyea

Note that this might be the only color allocation rule that assumes that having White is an advantage. Is there another?

Do you always know the answers to questions before you ask them? For what it’s worth, this is a non-point, as you can always change the “White” concept to the player with the better tiebreaks gets to choose his color.

Ick. Some of the players with better tiebreaks will be assigned white or black because of color equalization. Some will be assigned white because of better tiebreaks. The pairings wouldn’t necessarily state whether the white was because of tie-breaks or equalization.
Now you have three options.

  1. defer printing the final pairings until after you’ve asked every player with an option which color they want (just hope that none of them went out for a bite to eat and planned to make it back maybe 10-15 minutes after the start of the round).
  2. manually go through the pairings and mark each such game as one where the player listed as white really gets to choose their color.
  3. simply give white to the player winning the tiebreak and let any that are interested come up to you and ask if they are allowed to choose black.

Round 5 (final round) Pairing Sheet

BoardWhiteBlack
1…Alekhine…Fischer
2…Morphy…Lopez
3…Euwe/Anand…Anand/Euwe — (Euwe chooses color)
4…Karpov…Kasparov
5…Botvinnik/Kramnik…Kramnik/Botvinnik — (Botvinnik chooses color)

Not that hard to do, and there might be easier methods. Probably best though is the original, just give the player with the better breaks White.

Why?

Ah, the eternal question of the young child.

I’ve played in quads where the player winning the toss took Black. I lost both times. The player taking Black was a young kid. LOL

On the one hand, I like bbentrup’s suggestion, because I personally have a better record with black than with white. On the other hand, it does offend against the presumption underlying other official USCF rules (for instance, prize allocation) that decisions should be made by the tournament director, not by player choice.

There is a little bit of a catch-22 here that will require the first such events to be fairly small.
To do this in a timely matter at a large event, the pairing programs would have to have the ability added. Before spending the resources to add that ability, the people who sell the pairing programs would require a shown demand for the ability. That shown demand would come from events where this was already done.

A gotcha to worry about is properly recording the results of the games when the color is not pre-determined. The pairing programs are set to enter the results of white, so that can’t be done until white has been determined for all games. If neither player circles who played white/black (that will happen in some of the tournaments) then would the default be that the first player listed had white? Inputting the results of the final round would be a bit delayed as the various games have their colors assigned.

I’ve seen pairing programs that have alpha pairing lists. In those, the player is listed as either B (black) or W (white). It sounds like C (choice) and D (depends on opponent’s choice) would need to be added as possibilities.

An additional rule would need to be added if a player’s opponent has the choice and is late. Currently the player would start the opponent’s clock (if black) or first start their own clock, move, and then start the opponent’s clock (if white). Since the player wouldn’t know the colors, they’ll need to be given the instructions (possibly to simply start the opponent’s clock, but that could cause some confusion in multiple time-control events if the opponent opts to play black).

The thrust of the rule variant is to help players of higher ratings where having an extra Black is a huge disadvantage. Sometimes that can’t be helped, of course, but when their last round opponent not only has the advantage of White from the randomness of Swiss pairings (usually by having been granted White in round 1 by luck of the coin flip) but also had the advantage of easier pairings along the way, it would be nice to balance things out. By reversing colors in this situation you mitigate against the bad luck involved. This rule is not really aimed for class-B/C and below players where having an extra Black is not a huge deal. If anyone can come up with a better wording based on this idea, I would appreciate it.