# Equal scores in a round-robin tournament implies wins=losses?

Suppose you have a round robin tournament with $n$ teams where each team get in every single 1-1 match:

• $3$ points for a win,
• $1$ point for a tie,
• $0$ points for a loss.

I would like to decide the truth of this conjecture:

If the tournament ends with all the teams with the same score then every team had as much wins as losses.

It is certainly true for $n=3$ but even checking the case n=4 is nontrivial.

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There's a counterexample with 8 teams: A, B, C, X, Y, Z, M, S

• S wins against M
• M wins against A, B, and C
• A wins against Y and Z
• B wins against Z and X
• C wins against X and Y
• X wins against M and A
• Y wins against M and B
• Z wins against M and C
• Everything else ties.

Teams ABCXYZ each win twice and lose twice for a score of $2\times0+3\times1+2\times3=9$.

Team M wins against ABC and loses to XYZS, for a score of $4\times0+0\times1+3\times3=9$.

Team S wins once and ties 6 times, for a score of $0\times 0+6\times1+1\times3=9$.

How I found this: First, for easier counting I changed the rules by subtracting $n-1$ points from each team such that ties give $0$ points, and wins and losses each $2$ and $-1$. That makes it easier to see which possible combinations of wins and losses add up to the same. Then I'm looking for a directed graph with no 2-loops such that the value of each node is the same.

Since there are as many wins as losses, in a counterexample there must be at least one team that win more times than they lose. But that team cannot possibly have less than $2$ points, so let's see which way we can make a node with value 2: They are: $1W+0L$, $2W+2L$, $3W+4L$ and so forth -- so if we have one of $1W+0L$ and $3W+4L$ the number of wins/losses add up right.

From there it was just a matter of puzzling out where to add $2W+2L$ nodes to the graph such that we don't need more than one match to take place between the same two teams.

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Nice solution! So is $n=8$ actually the minimum $n$ such that the statement is false? –  Marco Sep 25 '12 at 19:34
Yes, $n=8$ is minimal. A counterexample has to contain a team that lose more times than they win. If they are to make at least 2 points on my modified scale, they can't possibly have played less than non-tied 7 matches. That team together with all its opponents make at least 8 teams in the tournament. –  Henning Makholm Sep 25 '12 at 20:04