# Tournament bracket for a 4-players game

for Christmas a friend is trying to organize a tournament for 13 players. Each game will be played by 4 persons, each player will play 4 games and must play against everybody else.

I can find a solution, but I'd like to minimize the number of players playing two games in a row, and my solution is terrible. If someone can help us ... :)

The problem is similar to the social golfer and the schoolgirls, but the games will be played 1 by 1.

• 13 rounds of 1 single game. But, as much as possible, I'd like to avoid people playing twice in a row. My bracket seems to maximize it ... Commented Dec 18, 2018 at 21:15
• Note that player 2 plays player 11 three times. Same for 3 and 12 and 4 and 13, so this cannot be what you want ... indeed, 2 does not play 13, and ... Commented Dec 18, 2018 at 21:20
• Yes, sorry, this was a copy/paste mistake. 1 5 6 7 - 1 8 9 10 - 1 11 12 13 - 2 5 8 11 - 2 6 9 12 - 2 7 10 13 - 3 5 9 13 - 3 6 10 11 - 3 7 8 12 - 4 5 10 12 - 4 6 8 13 - 4 7 9 11 - 1 2 3 4 Commented Dec 18, 2018 at 21:23
• This one has 1 player repeating each rpound ... and 12 different players repeat just once ... player 5 never repeats ..1 2 3 4 - 2 5 8 11 - 3 7 8 12 - 3 5 9 13 - 4 6 8 13 - 3 6 10 11 - 4 7 9 11 - 4 5 10 12 - 2 7 10 13 - 1 5 6 7 - 1 8 9 10 - 2 6 9 12 - 1 11 12 13 - I got this just by playing with your foursomes .. nothing systematic ... surely some kind of systematic way exists to 'spread' the play-times as much as possible (and not just in repeating, but also in terms of games in which you don't play between games you do play .. but I don;t know of any such method Commented Dec 18, 2018 at 21:52
• Thanks, I will keep it :) Commented Dec 20, 2018 at 16:11

An easy way to set up such a tournament makes use of the fact that the projective plane $$PG(2,3)$$ consists of $$13$$ points on $$13$$ lines, each line containing $$4$$ points (for a graphic of that plane, see e.g. https://tex.stackexchange.com/questions/336897/).

So, you can identify the players with the points and the games with the lines, and run them in any order: Since you have a projective plane, any two lines will intersect in a single point, i.e. any two games will have one player in common. That will add up to $$12$$ players having to play two games in a row — I don't know how far this is from the minimum you are trying to achieve.

This one has 1 player repeating each round ... and 12 different players repeating just once (player 5 never repeats) ... which also implies that no one three-peats:

1 2 3 4

2 5 8 11

3 7 8 12

3 5 9 13

4 6 8 13

3 6 10 11

4 7 9 11

4 5 10 12

2 7 10 13

1 5 6 7

1 8 9 10

2 6 9 12

1 11 12 13

I got this just by playing with your foursomes ... I didn't use any systematic method, though I would think that such a method exists.

There are schedules where no one repeats during some round ... but now you will have multiple people repeating some other round ... so I think having exactly 1 person repeat each round, and having those 12 repeats be of 12 different people is about as fair as possible as far as repeating goes.

Still, in addition to worrying about repeats, you could also worry about the spread of games more general. For example, in the schedule above, player 1 plays in games 1, 10, 12, and 13 ... that seems like it is a lot less 'evenly distributed' than, say, player 9, who plays in games 4,7,11, and 12, although at the same time it is more 'spread out' than a player like 3, who plays in games 1,3,4, and 6.

So, you could try and come up with some 'measure' of how 'evenly distributed' or 'spread out' the games are, and a measure of 'fairness' on top of that. Again, I assume that some work has already been done on that, though I am not an expert. Also, for your purposes of having some fun at a Christmas party, this would seem to going far beyond what you were looking for :)