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Saturday, 17 August 2019


12 combinations.

1&2, 3&4, one breaks
1&2, 3&4, two breaks
1&2, 3&4, three breaks
1&2, 3&4, four breaks

1&3, 2&4, one beaks
1&3, 2&4, two breaks
1&3, 2&4, three breaks
1&3, 2&4, four breaks

1&4, 2&3, one breaks
1&4, 2&3, two breaks
1&4, 2&3, three breaks
1&4, 2&3, four breaks

There are only 3 types of matches to play:
Type A: 1&2 plays 3&4
Type B: 1&3 plays 2&4
Type C: 1&4 plays 3&2
Play Type A 4 times with a different player breaking and paying each time.
Do the same for Type B, then for Type C.

If you wanted to get complicated, play A,B,C, each with player 1 paying, then play A,B,C with player 2 paying etc

Its like rock and roll in three four time!

Six possible teams give you three possible pairings: 12v34, 13v24 and 14v23. Rotate through this 4 times (12 games) and everyone will have paid three times.

I haven't worked it out, but I think that this also puts each player on the team that breaks the same number of times.


You have 6 teams as follows:

You're right that there can only be three teams. So you need to split the table fee up 3/4 and 1/4 in each game.

For a table fee F:

  • Team 1-2 vs Team 3-4: 1 pays 3/4F and 4 pays 1/4F
  • Team 1-3 vs Team 2-4: 2 pays 3/4F and 4 pays 1/4F
  • Team 1-4 vs Team 2-3: 3 pays 3/4F and 4 pays 1/4F

This way, every player pays 3/4 of the table fee.

Hi Mike,

There are four players all of whom are expecting to pay and break. That means at least four games to be played.
However, as you say, there are only three possible team combinations such that by the fourth game you’re going to have to repeat one team combo.
The solution is a dozen games. Each player will pay and break three times and you can rotate the team combinations four times.
Could be a long afternoon...

So you have 4 players playing 1 on 1 matches. This is expressed as 4 things taken 2 at a time, with order important (the first player of each pairing has to pay).

This means we are counting permutations of n=4 things k=2 at a time, ordered. The formula is n! / (n-k)! or 4! / (4-2)! = 4*3*2*1 / 2*1 = 12

Thus you must play 12 separate games, with the first player paying, to achieve equity in pay.

You see each player pays/breaks 3 times over 12 games.

payer-breaker = 1 teams 12 vs 34
payer-breaker = 2 teams 24 vs 13
payer-breaker = 3 teams 32 vs 14
payer-breaker = 4 teams 34 vs 12

I’m not quite sure I understand the setup of the problem, but if I’m interpreting the question correctly, there are six potential teams: 1-2, 1-3, and 1-4; but then 2-3, 2-4, and 3-4 also need to play together. But! 3-4 play together while 1-2 are playing together, 2-4 play together while 1-3 play together, and 2-3 play together while 1-4 play together. So you only need three games to play the combination of six teams.

If you want each player to pay for the same number of games and each team to play together the same number of times, you need to play 12 games (or some multiple of 12), since 12 is the least common multiple of 3 and 4.

Yes, it’s math.

I would say that tournament design is an application of mathematics: probably graph theory, or more generally, combinatorics.

If I understand your criteria correctly, you want a pure round-robin tournament consisting of three rounds, each of which has two games. (They can be played concurrently if you have two tables available.) Since each person plays only once in each round it's not "fair" in terms of who pays and breaks first. But, if you have enough time you could just play each round twice, or spread them out over two nights.

Here's the set of rounds generated using the circle method described on the Round-robin tournament Wikipedia page. Each person plays the one directly below them.

Round 1
  1 2
  4 3
Round 2
  1 4
  3 2
Round 3
  1 3
  2 4

Round 4 would be the same as Round 1, so we're done.

Correct, it seems there are four breakers and three team combinations. But as far as I understand these two sequences are independent: who pays & breaks doesn't impact the team selection. You'll just have to play in multiples of twelve games for everyone to be even!
Mathematically, that is because 12 is the Least Common Multiple of 3 and 4.

Right; there is a "3" and a "4" cycle, so you will need to play 3 times 4 equals 12 games before everything is even. The 12 games, described by the breaker first and then his/her partner are:




After 12 games, each player has had the break 3 times, and has partnered with each of the other players twice.

Oops, I said each player partners with each other twice; I meant four times.

You lost me when you wrote the word "math".

I’m sure I could tell you the answer ... if only I could understand the question :-)

Mike, Isn't it nice to have friends who know combinatorics?

Isn't it nice that they all agree?

4 player three combinations... seems to me, you just need to convince your buddies that since you figured it out you shouldn’t have to pay ;)

I was really hoping to see (n-1) appear here in the comments... next time Mike!!!

Doh, I just saw (n-k), close, but no 8 ball.

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