Actually, I'm not even sure this is mathematics. I don't even know what discipline it is. I realize this might be trivial to a lot of you, but my brain has never gotten along with numbers. (We all have weaknesses. The key to success in life is to know how to ask for help with the things we're not good at.)

Say you have four pool players—call them 1, 2, 3 and 4. The person who pays breaks, and each player pays and breaks in a fixed sequence, 1-2-3-4. Each game is two against two, but the teams rotate so that each person takes turns playing with the others. How do you arrange this so both sequences work?

It seems to me that there are four possible breakers, 1-2-3-4, but only three possible teams—that is, 1 plays with 2, then with 3, then with 4. How can both sequences continue? Sitting at the Moose the other day trying to bodge up a score sheet while the others were saying "come on, it's your turn!", I couldn't quite get there. To me it's like one of those odd time signatures that only jazz drummers know how to play.

*Mike*

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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

Posted by: Speed | Saturday, 17 August 2019 at 10:16 AM

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!

Posted by: Nick Prior | Saturday, 17 August 2019 at 10:31 AM

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.

Posted by: C.R. Marshall | Saturday, 17 August 2019 at 10:32 AM

Mike,

You have 6 teams as follows:

1-2

1-3

1-4

2-3

2-4

3-4

Posted by: Fred Kinder | Saturday, 17 August 2019 at 10:51 AM

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:

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

Posted by: Tommy Williams | Saturday, 17 August 2019 at 11:00 AM

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...

Alan

Posted by: Alan | Saturday, 17 August 2019 at 11:02 AM

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.

(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3).

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

Posted by: Dale N | Saturday, 17 August 2019 at 12:15 PM

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

Posted by: Bruce Norikane | Saturday, 17 August 2019 at 12:26 PM

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.

Posted by: Mark | Saturday, 17 August 2019 at 01:20 PM

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

Posted by: Globules | Saturday, 17 August 2019 at 01:40 PM

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.

Posted by: Damien | Saturday, 17 August 2019 at 01:56 PM

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:

12

23

34

41

13

24

31

42

14

21

32

43

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

Posted by: Peter Norvig | Saturday, 17 August 2019 at 02:17 PM

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

Posted by: Peter Norvig | Saturday, 17 August 2019 at 02:25 PM

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

Posted by: Dogman | Saturday, 17 August 2019 at 03:57 PM

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

Posted by: Richard Parkin | Saturday, 17 August 2019 at 04:31 PM

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

Isn't it nice that they all agree?

Posted by: C.R. Marshall | Saturday, 17 August 2019 at 09:36 PM

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 ;)

Posted by: Cliff | Sunday, 18 August 2019 at 07:35 AM

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.

Posted by: Julian Lynch | Sunday, 18 August 2019 at 04:01 PM