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new to puzzles here. A real life situation led me to consider this. We have a small league of 6 people playing in teams of two. I would like to rotate the teams in such a way that everyone gets an equal chance to play with every other person and against every other person.

e.g. in the first game, it's A+B vs C+D and E+F sit out.

How do I arrange rotations in such a way that everyone gets an equal number of chances to play with every other player and against every other player?

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Each person plays the same number of matches with each of the other 5, so the number of matches each person plays must be a multiple of 5.

If each person plays 5 matches, then there are a total of 30 players across the matches. But this is impossible because 4 players are in each match and 30 isn't divisible by 4. So each player must play at least 10 matches, with a total of 15 matches.

You can make five sets of three games each: for each set, divide the players into three pairs, and have each pair play the other two once. This is one possible way to divide the pairs:

  1. AB/CD/EF
  2. AC/BE/DF
  3. AD/BF/CE
  4. AE/BD/CF
  5. AF/BC/DE

You can check that for every two players, they are together in two games (in the set where they are paired) and against each other in four games (once in each of the other sets).

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  • $\begingroup$ I came to the same conclusion - that the most efficient way to get a fair set of games is to play 30 games. It's the trick of assigning players in a systematic way that has stumped me. Any guidance on how to do that? $\endgroup$ – user436157 Nov 12 '15 at 3:31
  • $\begingroup$ There are 15 possible pairs of players, and this strategy relies on dividing them into five groups so that every pair appears once. There are several ways to do this division; I constructed the one in my answer by pairing each person with the earliest other person that wouldn't cause a repeat. $\endgroup$ – f'' Nov 12 '15 at 3:39
  • $\begingroup$ For example, B is paired with E in set 2 because pairing B with D would make EF repeat. $\endgroup$ – f'' Nov 12 '15 at 3:39
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    $\begingroup$ If these groups are drawn on a hexagon, they make this symmetrical pattern: i.stack.imgur.com/1cU3S.png $\endgroup$ – f'' Nov 12 '15 at 3:43
  • $\begingroup$ I think you misunderstood the scenario. The games are played in teams of two, so each game involves 4 people with 2 people that need to sit out (imagine a doubles tournament). The twist is that the teams change every round. You have effectively produced a list of the 15 team makeups possible for the tournament. I am looking for a way to match up those teams in a way that allows everyone to play the same number of 2v2 games against and with every player. $\endgroup$ – user436157 Nov 12 '15 at 6:33
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Well, for four players (A,B,C and D), there are three possible pairings:
A&B v C&D
A&C v B&D
A&D v B&C

Those games exclude E&F.

Then there are $\binom{6}{2}$ = 15 different ways to have two people sitting out.

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  • $\begingroup$ This produces every possible pair vs. pair matchup, but I'm not sure that's what the question was asking for specifically. If the question only requires each person to play with each other person equally and against each other person equally, this still works, but there might be a way with fewer matches. $\endgroup$ – f'' Nov 12 '15 at 2:38
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What you've asked for here is similar to an Individual-Pairs tournament. This link has many schedules for varying number of players if you are interested.

The 6 person schedule is asymmetric, meaning some people play an unfair number of times.

The schedules proposed so far are not strictly "Individual-Pairs" tournaments since they don't meet the critia:

  • each partners each other once
  • each opposes each other twice

In the solutions posted so far, people parter with each other twice, and play against each other four times. However, these solutions are fair and given that you accepted the answer, this is probably what you are looking for. However, for some numbers of people (4, 5, 8, 9, 12, 13, etc) there exist very nice minimal tournaments which are quite handy.

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