Going to be 4 rd swiss. 6 players for Rd 1.
Rd 1. (Player 1 / 1831 is white ) (Player 2 / 1788 Black) (Player 3/ 1509 White) (Player 4/ 1474p Black) (Player 5/ 938 White) (Player 6 / Unr Black) Pairings no problems.
1 vs 4 5 vs 2 3 vs 6. (Players 1; 2; 3 winners of rd 1)
Rd 2 2 players joined. (Player 1 / 1831 is white ) (Player 2 / 1788 Black) (Player 3/ 1509 White) (Player 4/ 1474p Black) (Player 5/ 938 White) (Player 6 / Unr Black) (Player 7 / 1741) Player 8 / Unr) Parings no problems. 2 vs 1 7 vs 3 4 vs 8 6 vs 5 (Players 1; 6; 7; 8 winners of rd 2)
Rd 3 WinTD pairings call for (1 vs 7 6 vs 2 8 vs 3 5 vs 4 ) QUESTION IS: why not the due color pairing for rd 3 be ( 1 vs 7 3 vs 2 8 vs 6 5 vs 4) any one explain WinTD pairings? So I will understand the idea behind WinTD.
The “natural” pairings (we’ll call them Alpha pairings) for round 3 are, in board order (white listed first):
1 (1831) vs. 7 (1741)
8 (UNR) vs. 2 (1788)
3 (1509) vs. 6 (UNR)
5 (938) vs. 4 (1474)
Of course, board 3 can’t happen, because that would be a rematch of a round 1 game. The problem is that 8 has to play someone in the 1-point score group if at all possible. So, since he can’t play 2, let’s try dropping him to play 3. This presents a bit of a color issue, but since 8 is in the higher score group, he gets his due color. So now, we have Beta pairings:
1 (1831) vs. 7 (1741)
8 (UNR) vs. 3 (1509)
6 (UNR) vs. 2 (1788)
5 (938) vs. 4 (1474)
Well, these are reasonable pairings. But we don’t like the color issue on board 2. So maybe we can fix the color problem if we do this (Charlie pairings):
1 (1831) vs. 7 (1741)
8 (UNR) vs. 6 (UNR)
3 (1509) vs. 2 (1788)
5 (938) vs. 4 (1474)
So now, everyone has their “due” color.
(Note here that Beta pairings are what WinTD came up with, and Charlie pairings are your proposed alternative.)
The problem with Charlie pairings is that 8 gets a much weaker opponent (at least, by rating) than he does in Beta pairings. USCF rules say that the difference between 3 and 6 (8’s prospective opponents in Beta pairings and Charlie pairings) is too big to switch them when the only reason you’re doing it is to improve colors. Ergo, Beta pairings seem to be correct.
(Caveat: I did all this by sight, and without a rulebook in hand. So I may have made a clerical error somewhere that would change some or most of the above. If I did, please accept my apologies.)
Boyd is correct, but one of his reasons needs a minor correction.
The 8-3 game does not give white to 8 because he is in the higher score group (higher ranks when players are equally due colors). Both players are in the one point group. 8, however has an unplayed game and has played zero white and one black, needing white to equalize. 5 had played one white and one black and needs white to alternate. Equalization has a higher priority than alternation and thus 8 gets white.
Once 7 is pulled from the one-point score group to play 1, there are four players left in the score group. 2 and 3 are in the top half and 6 and 8 are in the bottom half. Correcting colors can be done by doing an interchange to put 3 in the bottom half, but the rating difference in doing that interchange exceeds the standard 80-point-difference limit between players being interchanged. If 6 or 8 had been given estimated ratings in the 1429-1589 range then an interchange would have been plausible, but the pairings in the earlier rounds would have been different.
What Jeff writes is true - if players 7 and 8 received zero-point byes for round 1.
I worked from a position that players 7 and 8 received half-point byes for round 1. If that is true, then both players are on 1.5/2 heading into round 3. Perhaps the OP would be willing to clarify whether those two players got half- or zero-point byes.
Not true. Even if the shoe were on the other foot, and 8 were in the lower score group, he would still get due color, because he is more strongly due his color than is his opponent.
Read rule 29E4. … Pairing players due the same color. The “higher ranked gets due color” rule kicks in only if both players have had the same identical color sequence throughout the tournament. In this case, they haven’t. The player who needs white in order to equalize is more strongly due white than his opponent who needs white merely in order to alternate.
The same applies even to color differences weaker than equalization-vs-alternation. For example, WBWB gets white over BWWB, regardless of which player is higher-scoring or higher-rated.
How can you talk about a rating difference being too big, when you can’t talk about a rating difference at all? Since player 6 is unrated, there is no such thing as a rating difference between 3 and 6.
In fact, you can transpose an unrated (within his score group) pretty much any way you want. Look at rule 29E5g:
29E5g. Unrateds and color switches. “If a player is switched to or from an unrated opponent to improve color allocation, this is not in violation of the 80- or 200-point rules for transpositions and interchanges.”
There is no logical reason to consider an unrated player to be near the bottom of his score group. In fact, with the possible exception of the perfect-score group and the zero-point score group, the most logical assumption is that he is near the middle of his score group. He has, after all, played well enough (or poorly enough) to have landed in this score group to begin with.
Thus, the concept of “transposition”, or even “interchange”, has no meaning when an unrated player is involved. You can’t calculate a rating difference, nor declare a top-half-vs-bottom-half violation. The TD might as well pair the unrated, within his score group, in whatever way most conveniently solves color problems (and other problems) within the entire score group.
Nobody has yet brought up what ought to be the most important point here – the dreaded Small Tournament Effect.
If colors work perfectly in every round, in effect the players have been divided into two camps – those who started with white, and those who started with black. With perfect colors, every pairing will be an inter-camp pairing, none will be an intra-camp pairing.
But, in a small tournament, you will quickly run out of inter-camp pairings. Eventually, you’ll have to implement intra-camp pairings, with bad colors.
It is better to have bad colors in odd-numbered rounds, where alternation is the only issue, than in even-numbered rounds, where there are equalization problems.
Accordingly, in a small tournament (fewer than, say, 20 players), I like the idea of transposing only to equalize colors, not to alternate colors. In tournament software, you can change the 80-point alternation limit to zero, while leaving the 200-point equalization limit as is.
If you don’t believe any of this, try making pairings for a 6-player, 4-round tournament, alternating all colors in both rounds 2 and 3. See what you can come up with for pairings in round 4.
This part of pairing theory has never made sense to me.
From 29E4 Equalization, Alternation and Priority of Color, Subheading of “Pairing players due the same color” (btw, these subheadings should be numbered somehow) it states:
To my way of thinking the (larger) general principle at work in 3. contradicts the general principle at 4. I agree with 3.
Here is my thought. Players “expect” that to the extent possible their colors will alternate, and I think that is reasonable.
So when faced with two players due the same color, a simpler rule is equalize the number of alternations.
In 3 we have WWB gets white over WBW. I agree but for a more general reason: WWB is ONE alternation, WBW is two alternations. Assigning the first player W gives WWBW (two alternations) and WBWW (two alternations). Both players are kept as close to the maximum of 3 alternations in 4 rounds as they can be.
So I disagree with the result in 4, example 1: “Example 1: WBWB gets white over BWWB, because the first player had black in round two, the latest round in which colors differed.”
I would say that WBWB has 3 alternations, and BWWB has two alternations, so the SECOND player should get White, again keeping both players as close to alternating as possible.
I think in “Example 2: BWxBW gets white over BWBxW, because the first player had black and the second had no color in round 4, the latest round in which colors differed.” - BOTH players have the same number of alternations: 3.
In this case I again disagree. If the number of total alternations is the same, I would suggest that we apply the same equalization/maximization of alternations rule to the most recent sequence, but on a PERCENTAGE basis of alternations to rounds. This has to be on a percentage basis, because the most recent sequence won’t necessarily be the same length for both players.
So, looking at BWxBW, if assigned black, they have 2 alternations in the last 3 rounds or 2/3, whereas if we assign black they have 1 alternation in 3 rounds = 1/3.
For the second player we would have WW = 0 or WB = 1/2.
My thought is that we can determine the better color allocation through addition: the first choice is 2/3+0 = 2/3. The second is 1/3 + 1/2 = 5/6. So the second pairing is better because on a percentage basis the players will have the greatest perception of alternation.
So I would replace 3. and 4. with:
If both players are due the same color, preference for due color will be given to the player with the fewest number of alternations.
If both players are due the same color, and have the same number of total alternations, then look at the most current alternation sequence (note that this could extend back to round 1, or back to a bye, or back to two consecutive colors) and take a ratio of the number of alternations relative to the number of rounds in the sequence for each player, look at both color allocations, and select the allocation with the sum of the highest ratios for both players, as this will have the greatest perceived alternation.
Those rules seem simpler because the principles are easy and relate directly to the number and length of alternation sequences (even though the final step in 4 requires some calculation.) They also seem to more clearly meet what players would expect as alternation. (Although both players had equal colors, the one with the shortest current sequence will have the greatest perception of alternation with respect to that sequence.)
The original poster asked why WinTD would make the pairings that it did. My posts were designed and intended to answer that question. The explanation I gave is consistent with what I have seen both WinTD and SwissSys do in comparable situations. Jeff Wiewel and Bill Smythe seem to be addressing the question of what the “best” or “correct” pairings are. This is, while related, definitely not the same question.
I have frequently seen computer generated pairings where Boards 1 through X seem logical, but pairings from Board (X+1) on down are messy. If I investigate the issue, I often find that the source of the mess is that the players from Board (X+1) and below are often treated as being in a group of their own. So I have to go in and break up one or more of the pairings on Boards 1 through X, and force the program to make one or two particular pairings, usually on Board X or Board (X-1). The computer then re-pairs everyone else, and the resulting pairings are a lot less messy. I think this happens because, once a program makes a certain pairing, it considers that pairing to be set in stone, and then works down to the next board. (I can’t prove this, since I don’t have the source code, but I see it all the time, even in big tournaments.)
Also, my experience has been that pairing programs put unrated players at the bottom of a score group, and treat them as having a rating of 0 when it comes to transposition limits. In most events, this is reasonable from a practical point of view, as the unrated player is likely the weakest player in the score group.
(Of course, this is not always true. In the Reserve section of the 1991(?) Illinois Open, I won my first three games, and then had Tim Just tell me that I’d lucked out in getting an unrated opponent for my fourth game. The unrated player happened to be Steve Arlinsky, who was playing his first tournament in the US, and went on to represent Illinois in the Denker twice. Thanks, Tim. )
I think you can actually set unrateds to be treated as having a certain rating for 80/200 purposes, but I may be wrong about that, and I don’t have a copy of either program in front of me to verify.
I’m aware of the “Small Tournament Effect”, but I don’t think it’s a big deal in an eight-player, four-round event. Also, I worry more about color distribution in master- or open-level events than I do in amateur events. On a side note…I prefer the FIDE pairing algorithm for Swisses. It incorporates logic for color distribution that is left to TD discretion in USCF rules.
[EDIT: Changed some phrasing to avoid possible confusion.]
One counter-argument to Kevin’s points:
WBWB should get black versus BWWB? Assume that in this case the top half players received an almost free point in the round one large mismatch. Then the first player would have black in three of the four more difficult rounds while the second player would have white in three of those four rounds (depending on the players the perception of the second player’s advantage from this ranges from trivial to massive). Once simple alternation and equalization are found to be the same, the current rulebook logic treats the earliest rounds as less important and looks just at the more recent rounds.
BWxBW should get black versus BWBxW? Similarly the first player would get two blacks and a white in the last three rounds while the second player would get two whites and a bye.
Is the principle about predicting who will get which color in different future rounds, or is the principle about equalizing and where possible alternating?
The principle is about perceived fairness. Because different people perceive fairness differently, there will always be arguments about the best way of doing something (and people with opposite views will each be absolutely certain that their view is best). I figured you might like to hear one reason for the logic behind the current full color history algorithm.
If you follow the book, giving WBWB white and BWWB black, you end up with WBWBW and BWWBB. This equalizes colors for both players during the most recent four rounds. The opposite approach ends up with WBWBB and BWWBW, putting both players out of balance 3-1 in the most recent four rounds.
Another way to see it is that, once a player has a WW non-alternation, it is not so bad to give him a subsequent BB non-alternation (or vice versa) to make up for it.
Still another way is to notice that BWWBB restores the player’s “expected” color sequence (in this case odd rounds black, even rounds white) that he started the tournament with.
Finally, attaching less weight to the round 1 color than to more recent colors recognizes the “automatic blowout” effect often associated with round 1. (Of course, this effect is more pronounced in single-section events, or in events with just two or three sections, than in class tournaments with section cut-offs every 200 points.)
From the rulebook:
On the contrary, 3. and 4. do not contradict each other. In fact, 3. is just a special case of 4. Technically, therefore, rule 3. is not even necessary. Pedagogically, however, 3. leads the reader gently by the hand into the more general case covered by 4.
This idea, I suppose, contains the initial seeds of logic, but it needs work.
In effect, you are assigning a value of +1 to each non-alternation, whether WW or BB, and attempting to keep each player’s non-alternation total to a minimum.
Most players, however, would regard WW as a plus, and BB as a minus, if it happened to them. It might make more sense to assign a value of +1 to WW, and -1 to BB. It might then be a reasonable goal to keep each player’s non-alternation total (taking the plus and minus signs into account) as close to zero as possible. That way, in WBWBW vs BWWBB, both players would have a non-alternation total of 0.
Continuing your quote from the rulebook:
I shudder any time the concept of proportion or percentage is brought up in a context like this. Let’s keep it simple with just addition and subtraction. (As Chevy Chase once said while imitating Gerald Ford, “it was my understanding that there was to be no math …”)
I would say neither of the above, or, in the case of a small tournament, a little of the first and none of the second. Of course, we all agree that equalizing and alternating are desirable goals in general. The question is what to do when those goals are not possible.
Can’t say for sure, but it seems to me that current versions of both WinTD and SwisSys are more sophisticated than that. What you describe sounds more like a human mindset, especially for a TD that makes the hideous mistake of writing down each pairing as he makes it, rather than making all the pairings first, then writing them all down afterwards.
Problems in the lowest score groups can be caused by the higher-ranked-gets-due-color rule. Players with low ratings in the lowest score groups may often have to be paired against each other, and will often both have bad colors like WW and WW. Then, somebody gets a third white. I’ve even heard of a case where correct pairings pitted two players with zero points, both with WWWW, against each other in round 5.
I like to put adult unrateds in at 1399, although I’ll vary this if I know something about a player. I list them as UNR rather than 1399 on the wall chart, but they’re not at the end of the wall chart. I also reserve the right to change estimates (for pairing purposes) in the middle of the tournament, and never to worry about the 80- and 200-point rules for unrated players.
I disagree with this statement in general, although it may be true in the first round or two, and in the zero-score group in later rounds. But for higher-scoring players in the later rounds, it is probably more logical to assume that an unrated belongs smack-dab in the middle of his group. After all, he must have won a game or two to earn his way into the group in the first place.
On the contrary, it can be a huge deal. Consider, for example, the “theoretical” 4-round, 8-player tournament (no byes, no upsets, no drop-outs):
Player 1: W5 -1- B4 -2- W2 -3-
Player 2: B6 -1- W3 -2- B1 -2-
Player 3: W7 -1- B2 -1- W5 -2-
Player 4: B8 -1- W1 -1- B6 -2-
Player 5: B1 -0- W8 -1- B3 -1-
Player 6: W2 -0- B7 -1- W4 -1-
Player 7: B3 -0- W6 -0- B8 -1-
Player 8: W4 -0- B5 -0- W7 -0-
(Note that transpositions were made in round 2 to equalize colors.)
Now what do you do, in round 4? The only way to make the colors work would be:
Player 1: W5 -1- B4 -2- W2 -3- B7
Player 2: B6 -1- W3 -2- B1 -2- W8
Player 3: W7 -1- B2 -1- W5 -2- B4
Player 4: B8 -1- W1 -1- B6 -2- W3
Player 5: B1 -0- W8 -1- B3 -1- W6
Player 6: W2 -0- B7 -1- W4 -1- B5
Player 7: B3 -0- W6 -0- B8 -1- W1
Player 8: W4 -0- B5 -0- W7 -0- B2
– pairing a 3-pointer against a 1-pointer, and a 2-pointer against a zero.
Of course, any reasonable TD would forget about colors at this point, and pair by score, resulting in bad colors on all four boards.
Here’s where the Small Tournament Effect comes in. If we had arbitrarily (and capriciously, at first glance) reversed colors on the middle two boards in round 3, we would have:
Player 1: W5 -1- B4 -2- W2 -3-
Player 2: B6 -1- W3 -2- B1 -2-
Player 3: W7 -1- B2 -1- B5 -2-
Player 4: B8 -1- W1 -1- W6 -2-
Player 5: B1 -0- W8 -1- W3 -1-
Player 6: W2 -0- B7 -1- B4 -1-
Player 7: B3 -0- W6 -0- B8 -1-
Player 8: W4 -0- B5 -0- W7 -0-
– which would lead, in round 4, to:
Player 1: W5 -1- B4 -2- W2 -3- B3
Player 2: B6 -1- W3 -2- B1 -2- W4
Player 3: W7 -1- B2 -1- B5 -2- W1
Player 4: B8 -1- W1 -1- W6 -2- B2
Player 5: B1 -0- W8 -1- W3 -1- B7
Player 6: W2 -0- B7 -1- B4 -1- W8
Player 7: B3 -0- W6 -0- B8 -1- B5
Player 8: W4 -0- B5 -0- W7 -0- B6
Perfect pairings, perfect colors. Deliberately skrewing up some of the colors in round 3 saved the day in round 4.
As I recall, somebody posted that the FIDE “algorithm” is not an algorithm at all, any more than USCF “algorithms” are. However, the FIDE logic may include a directive, in the event of bad colors, to assign the wrong color to the player least due his color. If so, this could be a Good Thing.
For example, in an 8-player group, if 5 are due white and 3 due black, and the 5 due white are as follows:
WBWB
BBWB
BWWB
xBWB
WBWB
– then the “least due” white would be BWWB, who would then be removed from the due-white group and added to the due-black group, for pairing purposes in round 5.
The vast majority of my recent pairing experience is with pairing moderate to large sections, mostly with SwissSys, but some WinTD too. As I said before, these programs do some weird things, even in good-sized score groups, or in situations where one pairing seems utterly forced. This post detailed the most egregious example of this I’ve come across.
If this was a Swiss event, it seems likely that this event had enough players to make the given color history incorrect. I’d be curious to see a wallchart from this tournament, though.
I should have been clearer. In most open events, the unrated player isn’t going to be in a high score group. The instances where the unrated is in a high score group can be handled on a case-by-case basis.
(example deleted for space)
I don’t like to fudge colors in earlier rounds, because that anticipates certain results to make the colors lay out correctly at the end. I feel it’s better - and easier to explain to players - to make the correct pairings each round. If colors are FUBAR in later rounds because certain pairings are forced, I don’t have a problem with that. But this is a matter of personal preference, and I acknowledge that your method has definite advantages…unless things don’t go according to Hoyle.
I’d have to disagree with that someone, at least on a practical level.
WinTD paired rd 3 as follows 1 vs 3 ; 8 vs 2 ; 7 vs 4 ; 6 vs 5 (Players 4 and 7 needed to play early. Giving players #1. WBW player #2. BWB #3. WB #4. WBB #5. BWB #6. WBW #7 BB # 8 BWW
So alternation rule to follow, Board # 3 was changed from 7vs 4 to 4 vs 7.
Causes an color issue in rd 4 due to 4 vs 1 giving # 4 WBWW.
First of all you have a couple of typos (corrected in red below):
1831 rd1 (W5) +1) rd2 (B2) +2) rd3 (W3)
1788 rd1 (B6) + 1) rd2 W1) +1 ) rd 3 (B8)
1741 — rd 2 W4) +1) rd 3 (B1)
1509 rd 1 (W8 1-0) rd 2 B3) +1) rd 3 (W7) +2)
1474 rd 1 (B1 0) rd2 W7) 0) rd 3 (B6)
938 rd 1 (W2 0) rd2 B8) 0) rd 3 (W5)
UNR ---- rd 2 B5) +1) rd 3 (B4) +1)
UNR rd1 (B4 0) rd 2 W6 +1) rd 3 (2W
Those pairings are decent. In the 1-point group the “natural” pairings, 2 vs 7 and 4 vs 8, won’t work because 4 has already played 8. The alternatives are 8 vs 2 and 7 vs 4 despite the bad colors, or 4 vs 2 and 7 vs 8 which is a top-half-vs-bottom-half violation. The latter (also called an “interchange”) wouldn’t have been so bad here, because unrated players can be considered to belong anywhere in the score group for pairing purposes. But WinTD chose the former, which is also not so bad.
That was an error, and might cause you to fail your Local TD exam. Equalization should take precedence over alternation. 7 needs white to equalize, 4 needs white only to alternate, so WinTD was correct to give the nod to 7 over 4.
It’s as though the pairing gods punished you for your error. Giving 4 the correct color in round 3 would have smoothed things in round 4.
Either way, it’s another case of the Small Tournament Effect, and the undesirability of having the colors work too well early on. It’s just as well WinTD avoided the interchange (with good colors) in round 3. With two bad colors in round 3, pairings were smoother in round 4.
Mr. Just, glad you I would wonder if SwissSys would pair our tournament the same as WinTD did? George Mirijanian, thought that the pairings for rd 3 were not correct on how WinTD had made the pairings, giving 7 vs 4. George Mirijanian thought that it was correct to change to 4 vs 7.
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