# Bye or same player twice
**URL:** https://forum.uschess.org/t/bye-or-same-player-twice/682
**Category:** Running Chess Tournaments
**Created:** [October 27, 2005, 2:16am UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682 "2005-10-27T02:16:27Z")
**Posts on this page:** 14
**Page:** 2
### Author:  [@Smythe\_Dakota](https://forum.uschess.org/u/Smythe_Dakota)
#### Post date: [November 1, 2005, 7:25am UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/21 "2005-11-01T07:25:06Z")
> [@rfeditor](#):
>
> … 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. …
Pairing players twice, and giving two byes to the same player, are BOTH violations of the letter of the rules, so in this case there’s no way NOT to let the TD do it.
> [@rfeditor](#):
>
> … On a more technical note, should half-point and full-point byes be treated the same in this context? …
I would agree that a full-point bye on top of a half-point bye is slightly less serious than two full-point byes, if those are the only options.
> [@rfeditor](#):
>
> … 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. …
When you have a tiny number of players, and especially when that tiny number is odd, the whole tournament is, inherently, one long string of major pairing distortions. It’s not a question of justifying it or avoiding it, it’s just THERE.
Bill Smythe
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### Author:  [@rfeditor](https://forum.uschess.org/u/rfeditor)
#### Post date: [November 1, 2005, 7:55am UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/22 "2005-11-01T07:55:53Z")
> [@Smythe Dakota](#):
>
> > [@rfeditor](#):
> >
> > … 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. …
>
> Pairing players twice, and giving two byes to the same player, are BOTH violations of the letter of the rules, so in this case there’s no way NOT to let the TD do it.
>
> Bill Smythe
But you’re arguing in a circle here – assuming what you set out to prove. It seems to me that the prohibition on pairing players twice is explicitly described as a “basic swiss system rule” of “highest priority.” The rules on assigning byes (including 28L3) are administrative detail on _how_ to make pairings that are in accordance with the basic rules. These details should not be interpreted in such a was as to contradict the underlying laws, any more than a statue can overrule a constitutional provision.
If what you are arguing is that the two rules are of “equal weight,” as is implied by “…there’s no way NOT to let the TD do it,” you are asserting that the pairing rules are inherently not algorithmically executable. That’s certainly a defensible argument, but I don’t believe it was the intention of the people who wrote them.
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### Author:  [@Smythe\_Dakota](https://forum.uschess.org/u/Smythe_Dakota)
#### Post date: [November 1, 2005, 8:15am UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/23 "2005-11-01T08:15:56Z")
> [@rfeditor](#):
>
> … you’re arguing in a circle here – assuming what you set out to prove. …
All I was trying to prove was that a violation of one rule or the other was unavoidable. And all I was assuming was the information (pairings and results) presented by Donna. Since my conclusion follows from my assumption (please review my puzzle-solving reasoning in a previous post), there is no circular argument here.
> [@rfeditor](#):
>
> … If what you are arguing is that the two rules are of “equal weight,” as is implied by “…there’s no way NOT to let the TD do it,” …
Whether or not the rules are of equal weight has nothing to do with the conclusion that violating ONE OR THE OTHER was unavoidable.
Somehow I have the feeling that you have never had to pair a 4-round 5-player Swiss.
Bill Smythe
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### Author:  [@rfeditor](https://forum.uschess.org/u/rfeditor)
#### Post date: [November 1, 2005, 9:24am UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/24 "2005-11-01T09:24:56Z")
> [@Smythe Dakota](#):
>
> > [@rfeditor](#):
> >
> > … you’re arguing in a circle here – assuming what you set out to prove. …
>
> All I was trying to prove was that a violation of one rule or the other was unavoidable.
I agree with that. (27A1, “Even this most basic of pairing rules must be violated when …”)
> [@](#):
>
> > [@rfeditor](#):
> >
> > … If what you are arguing is that the two rules are of “equal weight,” as is implied by “…there’s no way NOT to let the TD do it,” …
>
> Whether or not the rules are of equal weight has nothing to do with the conclusion that violating ONE OR THE OTHER was unavoidable.
But that is crucial to my point. Pairings _ought_ to be a determinate operation – the TD follows the rules in order of precedence, and every TD should in principle come up with the same pairing. Of course it doesn’t work that way in practice, since the real world is messy, but a construction which _in principle_ allows two TDs following the rules to come up with different pairings should always be avoided.
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### Author:  [@relyea](https://forum.uschess.org/u/relyea)
#### Post date: [November 1, 2005, 5:19pm UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/25 "2005-11-01T17:19:05Z")
What am I missing? 28L4 seems clear in this case. Assuing that B and C have had byes already, then the bye must go to D. All other players in the score group have already had a bye or a no-show forfeit win, right?
Alex Relyea
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### Author:  [@dmforsythe](https://forum.uschess.org/u/dmforsythe)
#### Post date: [November 1, 2005, 6:29pm UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/26 "2005-11-01T18:29:16Z")
Would it not have been very simple for the director, just place herself into the tournament, than pair the next round. It would take care of the problem of the bye, with someone new in the pairings: it can take care of the problem – the players being paired twice.
Do love when adults’ clobber the realistically and elementary way to emend the dilemma. It is like having a round peg, you could violate the round peg into the star or diamond slot, it would destroy the round peg, but it is nice to see evolved individuals use confrontation over how to butcher the dilemma. Very delighted they do not have a locked house door and the house key, as they would be belligerent over what window needs to be shattered.
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### Author:  [@rfeditor](https://forum.uschess.org/u/rfeditor)
#### Post date: [November 1, 2005, 10:22pm UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/27 "2005-11-01T22:22:25Z")
> [@relyea](#):
>
> What am I missing? 28L4 seems clear in this case. Assuming that B and C have had byes already, then the bye must go to D. All other players in the score group have already had a bye or a no-show forfeit win, right?
>
> Alex Relyea
That’s essentially Bill’s argument (though I think his argument from equity is stronger). The problem with it is that you are arguing that the procedure for assigning a bye (a minor subroutine in 28L4) is stronger than the _highest_ of the “Basic swiss system rules” in 27A.
Another way to put it is to parallel your sentence – “27A1 seems clear in this case. A has already played B, C, and D, so he must be paired against E.” Bill is quite correct that one of these rules must be violated – but which one?
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### Author:  [@relyea](https://forum.uschess.org/u/relyea)
#### Post date: [November 1, 2005, 11:25pm UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/28 "2005-11-01T23:25:42Z")
> [@rfeditor](#):
>
> Another way to put it is to parallel your sentence – “27A1 seems clear in this case. A has already played B, C, and E, so he must be paired against D.” Bill is quite correct that one of these rules must be violated – but which one?
I’m sorry. I thought that A had played B, C, and D. Am I misreading?
Alex Relyea
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### Author:  [@rfeditor](https://forum.uschess.org/u/rfeditor)
#### Post date: [November 2, 2005, 1:43am UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/29 "2005-11-02T01:43:36Z")
> [@relyea](#):
>
> > [@rfeditor](#):
> >
> > Another way to put it is to parallel your sentence – “27A1 seems clear in this case. A has already played B, C, and E, so he must be paired against D.” Bill is quite correct that one of these rules must be violated – but which one?
>
> I’m sorry. I thought that A had played B, C, and D. Am I misreading?
>
> Alex Relyea
Yes, that’s correct. Note, however, that the discussion has moved to the more general question “What happens when the _only_ choices are to give a bye to someone who has already had a (half-point) bye, or pair two people who have already played?” Obviously the best answer is “don’t let this situation arise,” but sometimes it does.
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### Author:  [@timjust](https://forum.uschess.org/u/timjust)
#### Post date: [November 2, 2005, 4:11am UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/30 "2005-11-02T04:11:17Z")
> [@rfeditor](#):
>
> Yes, that’s correct. Note, however, that the discussion has moved to the more general question “What happens when the _only_ choices are to give a bye to someone who has already had a (half-point) bye, or pair two people who have already played?” Obviously the best answer is “don’t let this situation arise,” but sometimes it does.
Exactamondo! The reason pairing programs and TDs can come up with legal yet different pairings is that we “law givers” have yet to agree on what carries greater weight in conflicting situations.
Tim
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### Author:  [@DACP](https://forum.uschess.org/u/DACP)
#### Post date: [November 20, 2005, 6:35pm UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/31 "2005-11-20T18:35:09Z")
Hi! Sorry, didn’t realize all these replies had come in…my notification must have gone off in advertently.
Since someone asked what I eventually did, here is the final wall chart:
A 1608 B3 W4 B5 B2
1 2 3 4
B 1580 W5 B3 B4 W1
0 .5 1 1
C 1509 W1 W2 bye B4
0 .5 1.5 2
D 1425 bye B1 W2 W3
1 1 1.5 2
E 1259 B2 bye W1 bye
1 1.5 1.5 2.5
Player A won first place, Player E won second place.
Entry fee was $25…first was $40, second was $30
Player B was okay with playing Player A since he was the only one who hadn’t played him. He was more concerned about why he didn’t get the bye because he is a club TD himself but hasn’t had much experience in pairings and wanted to understand the rules better.
Player E was disappointed to not have a game and is studying to become a club TD. He was also okay with the pairings.
All three of us pretty much agreed that the fairest thing was not to have a player play the same player twice in an event…particularly in this small a section.
The reason I didn’t “enter myself in” is that would be grossly unfair since my rating is hovering around 1770…also, I was playing in the Open section for this event. And yes, I do play two games at once to have my players avoid byes, but in this particular case, player E showed late since the advance pairings showed him as having the bye.
And there is a reason why we have a small section like this…in the past I had one section. However, by vote of my club members, the lower rated players would prefer to play in a section by themselves every once in a while so that they could play others more at their own level instead of having to play 7 players rated over 1750 every month.
However, in November, I did use the Round Robin pairings…a lesson was learned from October. 🙂
A short comment to those who suggested I ask the players what they wanted to do…I firmly believe that the rules should be applied as best as the TD can interpret them. If the rationale is explained to most players and it is logical, the vast majority of players will agree even when disappointed. However, once a TD starts asking players what they “want to do”, then it infers that the players have discretion in the pairings in the event and that opens a Pandora’s Box of what each individual “thinks is more fair” and whether the TD is “playing favorites”.
Personally, I think the rulebook is fantastic. It’s just virtually impossible to cover every single situation. Each TD may have a different interpretation and has to do the best he/she can when on the hot seat.
Thanks to all who helped me with the sanity check. 🙂
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### Author:  [@Smythe\_Dakota](https://forum.uschess.org/u/Smythe_Dakota)
#### Post date: [November 23, 2005, 12:47pm UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/32 "2005-11-23T12:47:54Z")
> [@DACP](#):
>
> … Player B was … more concerned about why he didn’t get the bye because he is a club TD himself … and wanted to understand the rules better. …
So you’re raising new club TDs by introducing them to the small-section concept. I’d say you’re a tough teacher! 🙂
> [@DACP](#):
>
> … here is the final wall chart …
Obviously the A,B,C,D,E in your wall chart aren’t the same as the A,B,C,D,E in your original post. I think they were listed by score the other time, and by rating this time. Apparently the translation table is something like this:
A → A
B → C
C → D
D → E
E → B
> [@DACP](#):
>
> … A short comment to those who suggested I ask the players what they wanted to do … once a TD starts asking players what they “want to do”, then it infers that the players have discretion in the pairings in the event and that opens a Pandora’s Box …
True, but sometimes the TD might have two or three reasonable options in mind. In the case of a very small section, the TD could lead a discussion about these options, presenting them, as it were, as multiple choice. After the discussion, the TD could then make (and explain) his decision, rather than simply accepting the will of the majority.
Bill Smythe
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### Author:  [@sloan](https://forum.uschess.org/u/sloan)
#### Post date: [November 26, 2005, 7:31pm UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/33 "2005-11-26T19:31:59Z")
> [@martinak](#):
>
> > [@Smythe Dakota](#):
> >
> > 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.
>
> 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)
> …
I prefer the method outlined in the rulebook. Essentially, you run a subset of a round robin. In the first round, you pair normally. After you pair, you assign numbers to the players so that the pairing you just made is the same as the first line in the appropriate round-robin table. In every subsequent round, you pair the top score group to find the correct pairing for the top-ranked player. Find that round in the round-robin table and use those pairins. Modify colors, as necessary.
Alas, the original poster could not do this because he was running a 4-week event during which (I assume) it was possible for players to enter at any time (and disappear for a given round).
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### Author:  [@system](https://forum.uschess.org/u/system)
#### Post date: [April 22, 2024, 8:39am UTC](https://forum.uschess.org/t/bye-or-same-player-twice/682/34 "2024-04-22T08:39:09Z")
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