Bye or same player twice

Running Chess Tournaments
1

Sorry if this is a repeat and asked before but just came up and I thought I’d ask for opinions since I don’t think this is clear cut in the rule book…

I have a two section tournament running and coincidentally, my lower rated section only has five players. Unfortunately it’s a 4 week event so not a round robin. Week 4 of course is the issue now…

Player A = 3 points
Player B = 1.5 points
Player C = 1.5 points
Player D = 1.5 points
Player E = 1 point

Player A has already played B,C, D.
Player D had a half point bye in Round 2.

Does Player A play Player E, in which case Player D gets a full point bye?

OR

Does Player A play Player B who he has already played and Player E get the full point bye?

Swiss Sys decided on one pairing and I decided to force a different pairing. Computer programs aren’t always great with small pair groups and circular references and conflicting rules so now I’m just doing a sanity check. Thanks for the help!

2

I’m curious as to what justification you have for pairing 2.
27A, Basic Swiss System rules. “The following rules are listed in order of priority.”
27A1, Avoid players meeting twice (highest prority).

3

And that’s why I asked…28L2. Determination of full point bye: Given to the lowest-rated player in the lowest score group. Doesn’t say anything about you can give it to another player if necessary to avoid players playing each other twice, etc.

And like I said, Swiss Sys did the pairing giving the bye to Player E and having Player A play Player B again.

My gut feeling was that not playing each other twice was most important, but one of my players questioned the projected pairing which made me want to verify with someone experienced - like you! :slight_smile:

4

Why can’t we give the bye to players B or C?

Alex Relyea

5

We can give the bye to someone other than Player E. I just wanted to be sure I was interpretting the rules correctly.

It’s Player E however, who wanted to know why he isn’t getting the bye since he’s the lowest rated player in the lowest score group.

I do believe that the correct pairing is to give it to Player D, the lowest rated player in the next lowest score group.

In that way, Player A does not play the same player twice. Player E now understands that and is okay with it. I just had wanted to make sure I was doing the right thing.

6

I was referring to rule 28C6, which says that you shouldn’t give a full point bye after a half point bye, if possible.

E must play A because there is no other way to pair the 3.0 score group, as unlikely as that sounds.

Alex Relyea

7

Unfortunately, there is only some of the information given that is needed to understand what is happening.

Which players have already played each other and which players have already received byes?

Have B, C and D also all played each other? If so, then even if you pair A and E to avoid having A play somebody a second time then you will still have another pair playing each other a second time. In that case, the SwissSys pairing of A and B makes sense. I can’t think of any way that could happen though (even with cross-round pairings), so I’d be interested in having all the information.

8

Donna Alarie

Since you are going to have to give someone a bye, what would happen with SwissSys if you place yourself into the tournament. You would not have a bye in the next round, if you place yourself into the tournament, you should not have the problem with the same players being paired twice.

If you are in the tournament, what would the pairings be like if you withdraw from the event. You would not have the problem with the bye, you should not have the problem with the same players being paired twice.

Having yourself in the next round, or withdrawing yourself from the event, can be the best way to settle the pairing problem.

9

Before we wander off on a tangent here, this is explicitly covered in 29D1.

“The lowest-rated player … is generally treated as the odd player. … Care must be taken … that the remaining members of both affected score groups can be paired with each other.”

It is true that this refers to the odd man rahter than the bye, but a bye is simply an odd man who has no lower score group to drop to. You cannot pair players twice if there is any other possible pairing. This is more important than anything else, including scoregroup or three in a row.

10

When there are an odd number of players and the number of players is close to the number of rounds, as here, I recommend that the TD talk to the players before pairing the first round.

There are several choices, none of them fully satisfactory. However, my experience has been that players appreciate being involved in the decision, whatever it is. (I usually talk to the players about things even if there are an even number of players.)

Here are some options, there could be others:

  1. Turn the event into a round robin. (I realize that was not possible in this situation.)

  2. Start at the bottom and give the lowest rated player the bye in round 1,
    the next lowest in round 2, etc. This could result in the highest rated player getting a full point bye in the last round.

  3. If all of the players in the lowest score group have already gotten a bye, move up to the next highest score group where there is a player who has not yet gotten a bye.

  4. If there are unrated players, they may have to get a bye if they are in a low score group and all rated players in that score group have already had a bye.

  5. If the event is paired as a Swiss, it may STILL be necessary or desirable to pair some players against each other more than once.

11

Donna, you have run afoul of the infamous Dennis Keen trap.

Dennis Keen was a middle-level TD in the 1970’s, now deceased. One fateful weekend, the top section of his 5-round Swiss drew only six players. Blissfully, he paired the first 3 rounds as a Swiss, only to discover that there were NO possible pairings in round 4.

Many TDs, both more and less illustrious than Dennis Keen, have fallen into this same trap many times.

The trap lurks whenever there are six players in either a 4- or a 5-round Swiss. It also lurks whenever there are five players, because then the sixth player, Mr. Bye, is supposed to be paired according to the same rules as the other five – do not pair him against the same opponent twice.

In your tournament, you have created two “camps”, B-C-D and A-E-bye, who have played exclusively inter-camp pairings, no intra-camp pairings. So if now B plays C, whom does D play? Or if B plays D, whom does C play? Or if C plays D, whom does B play? And similarly for the other camp.

The following theorems are true of six-player tournaments of either 4 or 5 rounds. The proofs are left as exercises for the reader. By extension, the theorems are also true of five-player tournaments, with Mr. Bye playing the role of the sixth player.

Theorem 1. The first 2 rounds may be paired any way you wish, without causing problems in later rounds.

Theorem 2. After round 3, there is trouble for round 4 if, and only if, there now exist 2 three-player camps who have played exclusively inter-camp pairings, no intra-camp pairings.

Theorem 3. The easiest way to fall into the trap is to make all the colors alternate for all the players in both rounds 2 and 3. (In that case the two camps are those who started with white, and those who started with black.)

Theorem 4. Immediately after making tentative round 3 pairings, check to see whether round 4 pairings exist. (Don’t worry about score or color, just make sure nobody is facing the same opponent twice.) If round 4 pairings don’t exist, change the round 3 pairings.

Theorem 5. If round 4 pairings exist, then round 5 pairings will automatically exist. (Just pair each player against the one opponent he has not yet faced.)

Theorem 6. After 2 rounds, there are exactly four ways to pair round 3 (not including color reversals). Three of these are OK. Only one leads to trouble in round 4.

Theorem 7. After 2 rounds, draw a regular hexagon on paper. Write the names of the six players at the vertices, in such a way that two players are at adjacent vertices if and only if they have already played each other. Then, the round 3 pairing that leads to trouble is the one with three diagonal (diametrically opposite) pairings. The OK pairings are those with one diagonal and two non-diagonal pairings.

Try it, you’ll like it.

Bill Smythe

12

Another way to handle this is to make sure you switch to a RR-table for the 3rd round.

It also generalizes:
4 rounds - 5 or 6 players (2 groups of 3 players)
6 rounds - 9 or 10 players (2 groups of 5 players)
8 rounds - 13 or 14 players (2 groups of 7 players)
…

13

I think it’s a bit pedantic to apply such rules so religiously. 27A is the first rule listed under “The Swiss System Tournament”, and is intended as a general introduction to the most important concepts. Accordingly, the rule “Avoid giving a bye to the same player twice” isn’t even on the list. If it were, it would probably be about equal with “Avoid players meeting twice”.

I think in this case, Donna, you’d be FAR better off pairing the same players twice than giving a second bye to the same player, even though the first was a half-point bye. Better PR, and all that.

What you really might consider, though (now that the Dennis Keen trap has already happened), is a creative solution.

For example: Contact players B, C, and D, and ask them if they’d be willing to play an extra game or two during the week, perhaps even at the homes of some of them. If so, you might be able to have, for example, B play D on Monday, then C play D on Tuesday, then B play C at the regular meeting on Wednesday. A would also play E on Wednesday. That way, you eliminate all byes (past and future) and turn the event into a perfect round-robin, where everybody plays exactly 4 games, one against each player in the section.

Or, here is another possibility: Does your 1770 rating qualify you for the bottom section (i.e. is the cut-off at 1800)? If so, you could jump in and play a game on Wednesday. (Maybe even 2 or 3 games, simultaneously?)

Or, is there a (reasonably low-rated) player in the top section who is scheduled for a full-point bye? If so, maybe he’d be willing to play a game or two in the lower section.

Give the players what they are paying for!

Bill Smythe

14

I’m sorry, Bill, but I think you’re completely wrong on the first point. Not pairing players twice is the most important Swiss System rule. It is more important than scoregroup or color. This is explicitly stated in 27A1, and repeated thoughout the rules (for example, in 29C2 and 29D1b). I am not a fan of appeals, but pairing players twice when there is any alternative is one of the few things for which a TD can and should be overruled. The TD is a constutional monarch, not a despot.

It is certainly possible to argue that it is a “better pairing” to pair two players a second time rather than, say, give the leader a full-point bye in the last round. But your argument here is with the rules themselves. You can’t ignore them just because it’s convenient to do so. If you don’t like the result (for which I wouldn’t blame you), compose a rule (or variation) which will prevent it. You can’t interpret the rules to mean something other than what their plain language says.

I agree entirely with your last point, that it is highly desirable to do something other than give a bye in this situation. But, sometimes you just have to bite the bullet.

15

Hmm, interesting. So far we apparently have:

Theorem 1. In a tournament with 4N+2 players, it is possible to run afoul of the Dennis Keen trap (no possible pairings) as early as round 2N+2.

Conjecture 2. In a tournament with 4N players, the Dennis Keen trap will never happen (as long as the number of rounds is 4N-1 or less).

Conjecture 3. In a tournament with 4N+2 players, no matter how rounds 1 through 2N are paired, the Dennis Keen trap can be avoided with judicious pairings for the rest of the tournament (as long as the number of rounds is 4N+1 or less).

Theorem 4. In a tournament with either 4N+2 players or 4N players, if the first 4N rounds or the first 4N-2 rounds (respectively) have been paired successfully, then pairings will always exist for round 4N+1 or round 4N-1 (respectively).

Obviously, I am making some simplifying assumptions, such as that there are no byes or forfeits, and that we don’t care about colors, etc.

Can anybody prove or disprove conjectures 2 and 3? I have labeled 1 and 4 as theorems, either because I can envision a proof (Theorem 1) or because it is obvious (Theorem 4).

Bill Smythe

16

Even if you adopt a dogmatic, tunnel-vision, letter-of-the-law approach to rule 27A1, it is not entirely clear that your argument is airtight, especially in view of some of the discussions in rules 29K and 29L, which deal with small swisses:

On top of that, if one can rise above dogmatism, one can readily see that 27A is intended as an introduction to the operation of a Swiss system, with details to follow.

Well, maybe my beef IS with the rules themselves, at least slightly. It might have been better to soften 27A, or to add a clause to 27A1 something like “or to receive more than one bye in a tournament”.

But one should always consider intent, and should never abandon common sense. The tournament in question is obviously a “fun” tournament, probably with low entry fees and no prizes (or small prizes), where the primary goal is to provide an enjoyable experience for the players. Giving a player 2 byes out of 4 rounds just doesn’t cut it.

Bill Smythe

17

I don’t believe that is a correct reading of the paragraph. What it’s saying is that pairing it as a Swiss (rather than a round robin) may lead to a situation in which you must pair players twice, but the writer nevertheless thinks this may be the bettter choice.
If there is no other possible pairing, then of course you have to pair people again, following the other rules as far as possible. (This has happened a few times in some of those multi-section-multi-schedule jobs.)

Legislative intent governs when interpreting an ambiguous or contradictory passage, but it can’t be argued against the plain language of the statute.

I agree with your last point, but that’s an argument for giving the bye to B or C instead of D (28L4), not for tinkering with the other pairings. Since I don’t have the whole crosstable I can’t be certain what’s going on, but I would give a second bye to D only if there were no other legal pairing.

18

Now, if you want to restate this as a pure hypothetical – “What if the only choices are to pair two players a second time, or give a full-point bye to someone who has already had a half-point bye” – we have a question worthy of debate. This sets up a conflict between two rules of nearly equal importance. I think I’d still have to come down on the side of not pairing the players twice, but in this case Bill’s “intent” argument cannot be easily dismissed.

19

Except that B and C have also received byes during the tournament.

From the clues Donna provided, I was able to deduce not only the pairings, but also the results, for the first three rounds (but not the order in which the games were played). See below.

Which was exactly the case. Since Donna had already fallen into the Dennis Keen trap, there was no longer any choice OTHER than to pair two players again (counting Mr. Bye as a “player”). So we’re back to square one – should the repeat pairing be one which involves Mr. Bye, or not?

Then I guess we have a legitimate debate.

The details, for those who like puzzles:

Donna’s first two clues are that A has already played B, C, D, and that D has had a bye:

A: has played B, C, D
B: has played A
C: has played A
D: has played A, bye

Then Donna asks “does player E get the bye?” which implies that E has not already had a bye. E has also not yet played A, since A’s three games are already accounted for. So E’s three opponents must have been B, C, D (the only remaining possibilities):

A: has played B, C, D
B: has played A, E
C: has played A, E
D: has played A, E, bye
E: has played B, C, D

Next question: How do we account for the remaining two byes? (We know there must be three total, since there have been three rounds.) There are only two spots left:

A: has played B, C, D
B: has played A, E, bye
C: has played A, E, bye
D: has played A, E, bye
E: has played B, C, D

It is also possible to deduce the results of each game, but (as we math people say) I’ll leave that as an exercise to the reader.

Bill Smythe

20

Unfortunately, the question is probably not resolvable, since it is a matter of interpretation. My fundamental objection to letting the TD make this sort of decision (making a pairing in violation of the letter of the rules because it is “right”) is that if you let one do it, you have to let all do it. While I (naturally) believe that I could make such a decision responsibly, I would rather deny myself that discretion than allow it to the many, many TDs who would misuse it.

On a more technical note, should half-point and full-point byes be treated the same in this context? Assigning a second “involuntary” bye would certainly be an injustice, but it is arguable that a (voluntary) half-point bye should not be considered “strong enough” to justify a major pairing distortion.