Body: <p><strong>Background</strong></p>
  <p>A few weeks ago I was asked to be the arbiter for one of a sequence of 15 round Swiss blitz tournaments. At the pre-tournament briefing a few days before we were told that if the number of players was close to 16 (specifically 24 or fewer) then we should change the format to an all-play-all to avoid the possibility of a pairing jam.</p>
  <p>As two of the other attendees were mathematicians and higher level arbiters I tried asking about the probabilities of pairing jams varied with numbers but got a bit of a <a href="https://en.wikiquote.org/wiki/Pierre_de_Fermat" rel="noreferrer">Pierre de Fermat answer</a>.</p>
  <p><strong>What is a pairing jam?</strong></p>
  <p>A pairing jam occurs in a Swiss tournament where the number of players is close to that which would better suit a round robin. In the last round (possibly before) a valid pairing cannot be made without pairing two players who have played before.</p>
  <p>A pairing jam is not a problem with a 2 or 4 player tournament but can be with a 4 or 5 round 6 player Swiss.</p>
  <p>In a 2 player tournament there is only one pairing, A v B, so a Swiss is identical to a round robin.</p>
  <p>Similarly for a 4 player tournament there are only 3 possible pairings, one for each round of RR or 3 round Swiss:</p>
  <ol>
  <li>A v B, C v D</li>
  <li>A v C, B v D</li>
  <li>A v D, B v C</li>
  </ol>
  <p>With 6 players the number of permutations opens up and it is quite easy to have a pairing jam in a 5 round Swiss. There are 5 ways to pair player A in the first round then 3 ways to pair the remaining 4 players giving 3 x 5 = 15 possible first round pairings.</p>
  <p><strong>Constructing a 6 player pairing jam in a 4 or 5 round Swiss</strong></p>
  <p>It is actually quite easy. The method is to construct two groups of roughly equal strength players in each group such that players within a group will always draw but players in one group will always beat players in the other group. Eventually one group will look (almost) like a RR and require pairings with the other group but there will not be enough players in the other group to satisfy the pairing requirements.</p>
  <p><a href="https://chess-results.com/tnr681071.aspx?lan=1&amp;art=4" rel="noreferrer">Here is an example</a> I constructed in Swiss Manager and put online. The first 3 rounds work OK</p>
  <p><a href="https://i.sstatic.net/2aPk1.png" rel="noreferrer"><img src="https://i.sstatic.net/2aPk1.png" alt="enter image description here" /></a></p>
  <p>but then the 4th round can't be paired.</p>
  <p>A needs to play C or E<br />
  B needs to play D or F<br />
  C needs to play A or E<br />
  D needs to play B or F<br />
  E needs to play A or C<br />
  F needs to play B or D</p>
  <p>E and F (bottom group) have played each other so each have to play someone from the top group (A, B, C, D) but whichever valid pairing of E and F you make leaves invalid pairings for those left:</p>
  <p>E v A, F v B but C and D have already played<br />
  E v A, F v D but B and C have already played<br />
  E v C, F v B but A and D have already played<br />
  E v C, F v D but A and B have already played</p>
  <p>So we don't even reach the 5th round. Already we have a pairing jam in the 4th round. There is almost no information that I could find on the internet, not even a definition of &quot;pairing jam&quot;. The one thing I did find was a suggestion that when the number of players is 2 x odd number it is more likely to jam than 2 x even number. I used this hint in my 6 player pairing jam by having one group of 4 and one of 2.</p>
  <p>What I'm asking is how does the probability of a pairing jam in a Swiss tournament, where the number of rounds is the same as would fit a round robin, change as the number of players increases. I've shown here that if n = 2 or 4 then p = 0 but if n = 6 then p &gt; 0.</p>
  <p>Just to spell out a bit more what is going on with a 6 player event. There are 15 possible pairings. A 5 round Swiss is going to use 5 out of those 15 pairings. I guess that means that there are 5!x10!/15! possible combinations of 5 pairings. Some of those will be illegal, avoidable pairings in early rounds. Exclude those. How many of those remaining result in a pairing jam? The answer, I suspect, gives you the probability of a 6 player pairing jam. Repeat for 8 players, 12 players, 16 etc. See if there is a pattern.</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-10-02T20:11:20.060Z
Comments.Text: There are two challenges I see for giving a mathematical answer. (1) The intuitive idea of a Swiss tournament is easy to grasp - split into two groups by points, have one group play the other. However, beyond that, there's lots of fiddly things the algorithm does to choose between many equally-good options early on, and to balance playing color. A probabilistic model would have to take all of those into account. (2) It would be unrealistic to assume that all games are equally likely to have any outcome, but the actual probabilities depend on the strength of players in a hard-to-quantify way.
Comments.UserDisplayName: Misha Lavrov
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-10-02T20:32:15.900Z
Comments.Text: @MishaLavrov In a 5 round, 6 player Swiss there are 5!x10!/15! possible different combinations of pairings. Eliminate illegal pairings (for later rounds) then see how many of them result in a pairing jam?
Comments.UserDisplayName: Brian Towers
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-10-02T21:23:20.703Z
Comments.Text: So you're just suggesting modeling a Swiss tournament by saying that at every round, we pick a random pairing that does not pair two players that have already played before?
Comments.UserDisplayName: Misha Lavrov
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-10-02T21:35:47.350Z
Comments.Text: @MishaLavrov Yes and see how many such lead to pairing jams and how many don't.
Comments.UserDisplayName: Brian Towers
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-10-04T08:48:19.913Z
Comments.Text: @BrianTowers: Interesting problem. The answer will strongly depend on which rating distribution we impose on the players (Gaussian? All same Elo?). Did you try a Monte Carlo run? (This would need some FIDE-approved algorithm, to which you could "prepend" a Monte Carlo manager to repeat the calculation often enough. With standalone Swiss programs, this seems unfeasable to me.)
Comments.UserDisplayName: Hauke Reddmann
Tags: <swiss-pairing><round-robin>
Title: How does the likelihood of a pairing jam change as the number of players increases?
ThreadText: # How does the likelihood of a pairing jam change as the number of players increases?
  
  (Asked by: username_0 on 2022-10-02)
  
  **Background**
  
  A few weeks ago I was asked to be the arbiter for one of a sequence of 15 round Swiss blitz tournaments. At the pre-tournament briefing a few days before we were told that if the number of players was close to 16 (specifically 24 or fewer) then we should change the format to an all-play-all to avoid the possibility of a pairing jam.
  
  As two of the other attendees were mathematicians and higher level arbiters I tried asking about the probabilities of pairing jams varied with numbers but got a bit of a [Pierre de Fermat answer](https://en.wikiquote.org/wiki/Pierre_de_Fermat).
  
  **What is a pairing jam?**
  
  A pairing jam occurs in a Swiss tournament where the number of players is close to that which would better suit a round robin. In the last round (possibly before) a valid pairing cannot be made without pairing two players who have played before.
  
  A pairing jam is not a problem with a 2 or 4 player tournament but can be with a 4 or 5 round 6 player Swiss.
  
  In a 2 player tournament there is only one pairing, A v B, so a Swiss is identical to a round robin.
  
  Similarly for a 4 player tournament there are only 3 possible pairings, one for each round of RR or 3 round Swiss:
  
  1. A v B, C v D
  2. A v C, B v D
  3. A v D, B v C
  
  With 6 players the number of permutations opens up and it is quite easy to have a pairing jam in a 5 round Swiss. There are 5 ways to pair player A in the first round then 3 ways to pair the remaining 4 players giving 3 x 5 = 15 possible first round pairings.
  
  **Constructing a 6 player pairing jam in a 4 or 5 round Swiss**
  
  It is actually quite easy. The method is to construct two groups of roughly equal strength players in each group such that players within a group will always draw but players in one group will always beat players in the other group. Eventually one group will look (almost) like a RR and require pairings with the other group but there will not be enough players in the other group to satisfy the pairing requirements.
  
  [Here is an example](https://chess-results.com/tnr681071.aspx?lan=1&art=4) I constructed in Swiss Manager and put online. The first 3 rounds work OK
  
  [![enter image description here](https://i.sstatic.net/2aPk1.png)](https://i.sstatic.net/2aPk1.png)
  
  but then the 4th round can't be paired.
  
  A needs to play C or E  
  B needs to play D or F  
  C needs to play A or E  
  D needs to play B or F  
  E needs to play A or C  
  F needs to play B or D
  
  E and F (bottom group) have played each other so each have to play someone from the top group (A, B, C, D) but whichever valid pairing of E and F you make leaves invalid pairings for those left:
  
  E v A, F v B but C and D have already played  
  E v A, F v D but B and C have already played  
  E v C, F v B but A and D have already played  
  E v C, F v D but A and B have already played
  
  So we don't even reach the 5th round. Already we have a pairing jam in the 4th round. There is almost no information that I could find on the internet, not even a definition of "pairing jam". The one thing I did find was a suggestion that when the number of players is 2 x odd number it is more likely to jam than 2 x even number. I used this hint in my 6 player pairing jam by having one group of 4 and one of 2.
  
  What I'm asking is how does the probability of a pairing jam in a Swiss tournament, where the number of rounds is the same as would fit a round robin, change as the number of players increases. I've shown here that if n = 2 or 4 then p = 0 but if n = 6 then p > 0.
  
  Just to spell out a bit more what is going on with a 6 player event. There are 15 possible pairings. A 5 round Swiss is going to use 5 out of those 15 pairings. I guess that means that there are 5!x10!/15! possible combinations of 5 pairings. Some of those will be illegal, avoidable pairings in early rounds. Exclude those. How many of those remaining result in a pairing jam? The answer, I suspect, gives you the probability of a 6 player pairing jam. Repeat for 8 players, 12 players, 16 etc. See if there is a pattern.
  
  **Tags:** swiss-pairing, round-robin
  
  > **Comments:**
  > **username_1**: There are two challenges I see for giving a mathematical answer. (1) The intuitive idea of a Swiss tournament is easy to grasp - split into two groups by points, have one group play the other. However, beyond that, there's lots of fiddly things the algorithm does to choose between many equally-good options early on, and to balance playing color. A probabilistic model would have to take all of those into account. (2) It would be unrealistic to assume that all games are equally likely to have any outcome, but the actual probabilities depend on the strength of players in a hard-to-quantify way.  
  > **username_0**: @MishaLavrov In a 5 round, 6 player Swiss there are 5!x10!/15! possible different combinations of pairings. Eliminate illegal pairings (for later rounds) then see how many of them result in a pairing jam?  
  > **username_1**: So you're just suggesting modeling a Swiss tournament by saying that at every round, we pick a random pairing that does not pair two players that have already played before?  
  > **username_0**: @MishaLavrov Yes and see how many such lead to pairing jams and how many don't.  
  > **username_2**: @BrianTowers: Interesting problem. The answer will strongly depend on which rating distribution we impose on the players (Gaussian? All same Elo?). Did you try a Monte Carlo run? (This would need some FIDE-approved algorithm, to which you could "prepend" a Monte Carlo manager to repeat the calculation often enough. With standalone Swiss programs, this seems unfeasable to me.)  
  
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  ## 0 answers
  
  