Color assignment for rematches

Running Chess Tournaments
1

Last weekend I directed a four-round Swiss with four players. Not intentionally - we just didn’t get as many players as we’d hoped for. One of the players suggested that the format could be changed to a three-round round robin, i.e. a quad, but the organizer was insistent that all four rounds should be played. Results were:

Round 1:

2302 vs. 2405 0-1
2360 vs. 2200 1-0

Round 2:

2405 vs. 2360 ½-½
2200 vs. 2302 ½-½

Round 3:

2200 vs. 2405 0-1
2360 vs. 2302 1-0

Round 4:

2405 vs. 2360 1-0
2302 vs. 2200 0-1

Each player got two whites and two blacks, with one rematch. It was unfortunate that the 2360 had to play black vs. the 2405 in both games. Is there a way I could have avoided that while staying within the rulebook?

A tournament that I directed a few times in the 1990s, the Monadnock Marathon, was a 12-round Swiss, and the rule there, announced in advance as a variation, was that players could be paired against each other for the second time starting in round 9, and the colors in a rematch would be the opposite of what they were in the first game between those players. Since no such variation had been announced at the tournament last weekend I felt that I had to go by the rulebook.

When I paired round 3 I thought about reversing the colors on both boards to avoid the problem on the top board in round four:

2405 vs. 2200
2302 vs. 2360

Assuming the 2405 and the 2360 played in the last round, the round 4 pairings would have been:

2360 vs. 2405
2200 vs. 2302

Each player would have had two whites and two blacks, and the 2360 and the 2405 would have had the opposite colors from what they had in their round 2 game. The 2200 and the 2302 would be playing with the same colors that they had in round 2, but arguably it’s more important for the colors to be right in the top score group than in the bottom score group.

Would this color reversal in round 3 have been legitimate under the rules? Can anyone suggest another way of improving the round 4 colors?

2

Sounds like this organizer put you in a pretty tough spot. :slight_smile: I’m not sure the pairings you made can be improved. If you’re going to pair entirely by the rules, all the way up to the last round, I think I’d have paired it the same way you did. Any significant change to your pairings would require some, ah, creativity.

With that in mind, I would like to offer a possible alternative suggestion. It seems clear that you will have some rematches no matter what you do. So I might have considered pairing the third round without regard to rematches, focusing only on score and color. I believe this would have produced the following matchups (player score in parentheses):

2360 (1.5) - 2405 (1.5)
2302 (0.5) - 2200 (0.5)

The upside to this is that the 2360 and 2405 are guaranteed to get two games against each other with equal color distribution. The downside is that a draw on board 2 could make for some ugly pairings in the last round. In that scenario, the natural round 4 pairings will give you the same opponents for the third consecutive round.

Another idea: You could have paired the first three rounds as a quad, and then used Swiss pairing rules for round 4 without regard to rematches. This way, you can’t have any three-peat matches, and the top players in the event will get to play in the last round. Color distribution is the potential issue here: since the last round of a quad is a coin flip for color, someone could end up with three straight Blacks (here is where Harold Stenzel tells us about Rule 29E5f :smiley: ).

All in all, I think your pairings make the most sense, and are most defensible from a rulebook point of view.

3

I think the neatest way would be to have the repeat matches in rounds 1 and 2. Essentially you would have one double round followed by two single rounds. For example:

round 1:
3-1
4-2

round 2:
1-3
2-4

Then be careful in round 3, so that nobody will end up with a third white or a third black in round 4. The following, for example, would work:

round 3:
1-4
3-2

round 4:
2-1
4-3

The pairings for all four rounds could be posted at the start of round 1. Since the players’ scores would not be known when the pairings are posted, nobody could squawk about the event not being “Swiss” enough.

Two players would receive WBWB or BWBW, the other two WBBW or BWWB. And, colors in both repeat matches are reversed. That’s as good as it gets. It is mathematically impossible, in this situation, to alternate all players in all rounds.

When the tournament is outside the box, think outside the box. Don’t be a slave to “rules” that were designed for other situations.

Bill Smythe

4

The problem with that is that other players might have entered the tournament at the start of round 2 or even round 3 with byes for the missed games. Rumor had it that a fifth player had said he was going to play in the tournament, although as it happened he never showed up. I felt that the first opportunity for doing something “out of the box” was at the start of round 3, when it was clear that no one else was going to enter the tournament.

5

Geez, Bob. I was going to post that. :stuck_out_tongue:

Also, I’d prefer that the players’ scores decide the pairings for at least the last round. It’s not a “normal” Swiss, but if it isn’t a round robin, then completely pre-determined pairings are, IMO, less than desirable. I’d rather give the top scorers a chance to settle the event with a head-to-head matchup in the last round, even if colors are FUBAR.

6

I think in the situation described, what happened was the most correct outcome. Particularly since the peceived iniquity is identical colors on an unavoidable rematch.

Playing with Black is a part of chess, and the rules determining who plays that part when generally ought not be disturbed.

7

Thanks for your comments, everyone. I’ve had this situation before with four players in a four round Swiss but in previous tournaments it was generally with lower-rated players who didn’t complain when I followed the standard Swiss pairing rules. In a tournament with four masters, assigning the correct colors becomes more important. Another problem is that the players want to be able to prepare openings against each other. One reason I didn’t reverse the colors in round 3 was that I didn’t think of that until Sunday morning before the round. If I were going to do that I should have told the players about it on Saturday.

Here’s a possible improvement over Bill’s proposal. Assuming there are no upsets it has the advantage that the rematches are in the last round, which makes it easier to revert to normal Swiss pairings if a fifth player shows up, and the rematches are between the top two players and the bottom two instead of 1 vs. 3 and 2 vs. 4.

Round 1:

3 vs. 1
4 vs. 2

Round 2:

1 vs. 2
4 vs. 3

Round 3:

1 vs. 4
2 vs. 3

Round 4:

2 vs. 1
3 vs. 4

The disadvantage is that if there are upsets I either have to change the pairings to account for the actual score groups instead the anticipated score groups, which would mess up the color assignments, or stick with the above pairings, which would mean the tournament leaders wouldn’t be playing in the last round.

On the whole, maybe it was best to stick to the rulebook instead of trying to anticipate the likely last round pairings and trying to improve them.

8

Or, announce/publicize that if opponents face each other a second time you will reverse the colors if possible (you might like that “possible” to give yourself some wiggle room if/when the rare occurrence of a huge color problem develops due to the announcement of the color reversal). Simple announcements to solve knotty problems are good things.

9

I think what I’ll do in the future when I have four players in a four-player Swiss is to announce at the start of the first round that I’ll be using modified Swiss System pairings. I’ll try to make sure that players are given reversed colors in a rematch but I won’t guarantee it. As I see it, it’s more important to equalize colors than to reverse colors in a rematch. If I can reverse colors on only one of the two boards I’ll give preference to reversing colors for the highest score group.

The perfect result would be that each player gets two whites and two blacks, the last round has the two highest-ranked players playing against each other and the two lowest-ranked players playing against each other, and colors are reversed on both boards from what they were the first time the players faced each other. The problem is that at the start of the tournament I don’t know which players will have the highest score in the last round. Since the most likely matchup is 1 vs. 2, I’ll choose pairings so that if those do end up being the two highest-ranked players the pairings will be perfect, but otherwise I’ll do the best I can.

Here’s a slight change from the pairings I gave earlier which I think will be easier to remember. In the first round I’ll toss for color and give the top two highest-rated players the SAME color. In round two I will AVOID pairing the two highest-ranked players. If one or more additional players enter the tournament I’ll revert to normal Swiss System pairings.

If the highest-rated player gets black in round one and there are no upsets the pairings will be:

Round 1:

3 - 1
4 - 2

Round 2:

1 - 4
2 - 3

Round 3:

2 - 1
3 - 4

Round 4:

1 - 2
4 - 3

Comments?

10

I don’t understand this. Aren’t the two boards independent in terms of colors, all else being equal. That is to say, I have decided that Annie is going to play Boris. How do their colors change what happens on the other boards?

Alex Relyea

11

As an example of what I meant, in my first post in this thread I said I was thinking of reversing the colors on both boards in round 3. That way the colors would be reversed in the rematch on board 1 instead of board 2 in round 4.

12

The arithmetic of color balance in a quad-plus-one tournament (four players, four rounds) is tricky indeed, and some of the results seem counter-intuitive.

There are essentially eight possible pairing schemes, represented below by the numbers 101-104 and 201-204. Without loss of generality we may assume that the first-round pairings are A-B and C-D (first-named player has white) in all the schemes. We are not assuming (not yet, anyway) that A is the highest-rated, nor D the lowest-rated, etc.

In schemes 101-104, each player is paired in round 2 against an opponent whose round 1 color was the opposite of his own. Scheme 101 balances colors for all players after round 2. Schemes 102-104 each leave at least two players unbalanced after two rounds.

Schemes 201-204 pair each player against an opponent who played the same color in round 1. These schemes each leave two players with unbalanced colors after two rounds.


-------------------------------------------------------

scheme:      101  102  103  104     201  202  203  204

-------------------------------------------------------

rd1:         A-B  A-B  A-B  A-B     A-B  A-B  A-B  A-B
             C-D  C-D  C-D  C-D     C-D  C-D  C-D  C-D

rd2:         D-A  D-A  A-D  A-D     A-C  A-C  C-A  C-A
             B-C  C-B  B-C  C-B     B-D  D-B  B-D  D-B

-------------------------------------------------------

rd3a:        ***  A-C  C-A  ***     D-A  ***  ***  A-D
             ***  B-D  D-B  ***     C-B  ***  ***  B-C

rd4a:         repeats rd1 with       repeats rd1 with
              colors reversed        colors reversed

-------------------------------------------------------

rd3b:        ***  ***  ***  ***     D-A  ***  ***  D-A
             ***  ***  ***  ***     B-C  ***  ***  B-C

rd4b:         repeats rd2 with       repeats rd2 with
              colors reversed        colors reversed

-------------------------------------------------------

rd3c:        A-C  ***  ***  ***     ***  ***  ***  ***
             B-D  ***  ***  ***     ***  ***  ***  ***

rd4c:         repeats rd3 with       repeats rd3 with
              colors reversed        colors reversed

-------------------------------------------------------

Of course, in any four-round four-player tournament, one of the rounds (customarily round 4) must be a repeat of an earlier round, preferably with colors reversed. In the table above, rounds 3a and 4a represent attempts to repeat round 1, while 3b and 4b try to repeat round 2, and 3c and 4c try to repeat round 3. Impossible combinations (those involving same-color rematches or 3-1 overall color imbalances) are indicated by asterisks. For example, scheme 101 does not allow round 1 to be repeated without color issues.

Since all eight schemes use the same first-round pairings (you knew there was a reason for that, didn’t you?), the TD may postpone the decision as to which scheme to use until the round 1 results are in. For that matter, he may also postpone the assignment of letters (A,B,C,D) to players (top-rated, 2nd, 3rd, bottom-rated) as long as the colors and opponents just played are consistent with the table. (For example, letters A and C must be assigned to the players who just played the white pieces, and A and B must be assigned to two players who just played each other.)

After round 1 is over, it is up to the TD to guess (based on results so far, ratings, etc) which of the first 3 rounds is likely to become the best candidate for repetition (with colors reversed) in round 4. Then the TD chooses a scheme (101-104 or 201-204) compatible with this candidate.

For example, if players B and D both win in round 1, and are also the highest-rated players, the TD may decide that it would be a Good Thing for these two players to meet twice. He could then choose scheme 101, where B and D would meet in rounds 3 and 4, or one of the schemes 201,204, where these players would face each other in rounds 2 and 4.

Of these three options, scheme 101 appears to be the least flexible. With scheme 201 or 204, the TD would still have the option, after round 2, of yanking the B-D rematch by repeating round 1 rather than round 2. So much for making the colors work too well in the early rounds of a small tournament.

In any case, look at all the asterisks in the table. If you simply wing it, odds are you’ll run into a color problem, either a same-color rematch or a couple of 3-1 overall imbalances. A quad-plus-one must be planned carefully.

Bill Smythe

13

Thanks for your analysis, Bill. Very interesting.

The line in your table for repeating round 2 doesn’t look quite right. As far as I can see, 201 to 204 should all be:

D-A
B-C

This makes sense, because if you’re repeating round 2 all four players should have the opposite colors in round 3 from what they had in round 1, and the above matchups achieve this.

It looks like I got it half right: in round 1, give the same color to the top two highest rated players. Otherwise, if you pair the players in round 2 you’re in columns 101 to 104, which can’t repeat round 2, and if you pair them in round 3 you’re in columns 201 to 204, which can’t repeat round 3. I suppose you could pair them in round 1 and choose appropriate colors in round 2, but I’d rather give them the same color in round 1 and delay the 1 vs. 2 matchup. According your table, especially with my revision, it looks like the top two players SHOULD play in round 2 if they both win (or at least don’t lose), which will most likely be repeated in round 4 with reversed colors, but there’s still the option of repeating round 1.

14

I know this isn’t a normal Swiss event, but I just do not like the idea of predetermining all four rounds. I guess this is a difference of philosophy.

Why not just pair the first three rounds as a quad, and let the chips fall where they may for round 4 pairings? The worst thing that can happen is that someone gets 3 of a color in 4 rounds - but at least this way, the top players by score are guaranteed to decide the tournament among themselves. This seems to be more in spirit with the Swiss format. I also imagine players would be much more content with that, especially the player(s) chasing the leader(s).

15

To determine where the chips should fall for round 4 pairings, I think they should do Rock, Paper, Scissors, Lizard, Spock to determine the first person to get White, and then Rock, Paper, Scissors to determine the second person to get White, and then flip a coin to see who gets to play the first player.

16

Boyd, as I understand it Bill’s latest proposal, unlike his earlier one, wouldn’t predetermine all four rounds. The first three rounds would be paired as a quad, but the order in which players met each other and the color assignments would be chosen to maximize the chance of making the color assignments perfect, i.e. that each player gets two whites and two blacks, the top two players play each other again in the last round, and the colors in the rematch are the opposite of what they were the first two times those opponents played each other. If I’d done that in the tournament I directed two weeks ago the 2360 would have been more content because he would have had white in his second game against the 2405 instead of being black in both games.
But if there were upsets so that the 2405 and the 2360 weren’t the two highest-ranked players going into the last round I would have made a different pairing in that round, even if meant the color assignments weren’t perfect.

17

I didn’t get that sense from reading the latest proposal.

This seems to me like the TD must pre-determine round 4 pairings based on the provided tables at the end of round 1.

Am I mis-reading this? If so, I apologize to Bill for my faulty interpretation of his most recent proposal, and to the readers for the irrelevant discussion I undertook on same.

18

It’s possible that Bill did intend that the pairings would be predetermined after round 1, in which case I disagree with him. To me it’s more important to pair the top two players in the last round than it is to make the color assignments perfect.

19

Yes, that’s correct. Although the round 1 pairings look the same in all eight schemes, keep in mind that at this stage the TD has not yet decided which player is A, which is B, etc. A-B-C-D aren’t necessarily 1-2-3-4 (by rating) respectively. How to match the letters with the numbers, and which of the eight schemes to use, can be decided after the round 1 results are in, and should be chosen in such a way as to make things as nice as possible in the later rounds:

And then, after round 2, if things didn’t go as expected, the TD would have the option of either (a) sticking to the chosen scheme anyway, so that colors would come out right, or (b) abandoning ship, pairing by score, and tolerating the bad colors. And if option (b) is chosen, the TD might still want to reverse one or both round 3 colors if it seems likely that round 4 will come out better as a result.

Incidentally:

Probably so. But the asterisks indicate that, no matter how you maneuver the colors in round 3, you’ll end up with less than perfect results after round 4.

I’m too lazy to look it up. I know how Rock, Paper, Scissors works, but what are the rules for Rock, Paper, Scissors, Lizard, Spock?

Bill Smythe

20

If the pairings go (column 202):

Round 1: A-B, C-D

Round 2: A-C, D-B

Round 3: D-A, B-C

Round 4: C-A, B-D

wouldn’t the pairings be perfect? Each player has two whites and two blacks, and colors are reversed in the rematch.