# Ask Uncle Colin: A round-robin

Dear Uncle Colin

I’m organising a tournament with 16 teams, and wanted to arrange it in five stages, each consisting of four groups of four teams. However, I found that after three rounds, it wasn’t possible to find any groups without making teams play each other again! Why is that?

Dumb Rematches Aren’t Wanted

Hi, DRAW, and thanks for your message!

### The problem

Suppose you have 16 teams, names A, B, C… all the way up to P. Arrange them in a grid like this:

A B C D E F G H I J K L M N O P

The first two rounds are easy enough: without loss of generality, we can say that the first-round groups go along the rows (so, for instance, A play B, C and D), and the second-round groups go down the colums (A play E, I and M). In the third round, we can go diagonally (so A, F, K and P are in the same group).

In fact, any satisfactory arrangement of the teams in the first three rounds can be relabelled the same way.

Now look at who team A could possibly be grouped with in round 4:

A (B) (C) (D)
(E) (F) G H

(I) J (K) L
(M) N O (P)

See the problem? Each of A’s possible opponents only have two teams in the list they haven’t played – and those two teams have already played each other! (The proof is left as an exercise).

### How I’d run the tournament

If we’re going to insist that all 16 teams play each possible opponent once, a total of 240 games, which seems a lot, here’s how we can do it.

In the first round, A plays B, C plays D and so on. We have eight two-team groups.

In rounds 2 and 3, each team in the first group plays a team from the second group; each team in the third group plays a team from the fourth, and so on; we’ve effectively folded the eight two-team groups into four four-team groups, each of which have played among themselves but not any other teams.

We repeat the trick in rounds 4-7: each team from A, B, C and D now play a team from E, F, G and H in turn. Every team from A to H will have played each other; each team from I to P will have played each other; no team from either group has played anyone from the other.

So rounds 8-15 finish that off: again, in each round, each team from the A-H group plays, in turn, a team from the I-P group. At the end of this, everyone has played everyone else exactly once.

(If you want to mix it up a bit, you can rearrange the order of the rounds so the teams aren’t isolated from each other.)

### How I’d actually run the tournament

Personally, I think round-robins are overrated. I much prefer variants on the Swiss System, where teams in each round play teams with a similar record they haven’t played before. This has the good points of a knockout (lots of competitive games) and a round-robin (everyone gets to play lots of games) with only relative complexity as a drawback.