n teams play pairwise matches. If the match is a draw, both teams get 1 point.
The winner gets 3 points, and the loser gets 0 points otherwise. Is it possible that all n teams
finish with the same total points, and how? Could we devise an algorithm for it that
includes the fewest drawn matches among such a possibility?

My Thoughts
there will be n(n-1)/2 matches. And each team will play with other n-1 teams. There must
be a draw in the tournament. Because if there is no draw, that means for each match, one team gets 3 points, and the other gets 0 points. Suppose team i wins wi matches, so the total score of the team i = 3 * wi. Since the total score for all teams is the same, they
must have won the same number of matches and lost the same number of matches.
I am not sure how to proceed with this argument and whether showing that there will be

  • $\begingroup$ Was my answer helpful? Have you considered upvoting and accepting my answer? Please comment if my answer can be improved. (This comment will be deleted upon feedback.) $\endgroup$
    – John L.
    Commented May 29, 2022 at 16:17

1 Answer 1


A simple algorithm that let all teams finish with same points

Let $n$ teams sit around a circle evenly.

  • If $n$ is odd, let each team win over the next $(n-1)/2$ teams clockwise.
  • Otherwise $n$ is even. Let each team win over the next $n/2-1$ teams clockwise and draw with the team sitting opposite to it.

Why does the algorithm above also include the fewest draws?

In the case when $n$ is odd, each team gets $3(n-1)/2$ points and no draws. So the algorithm includes the fewest draws.

In the case when $n$ is even, each team gets $3(n/2-1)+1$ points. The total number of draws is $n/2$. The following lemma shows the algorithm achieves the desired goal.

Lemma. Assume that $n$ is even and at the end, each team gets the same point. Then there are at least $\frac n2$ draws.

Proof. Let $d$ be the number of draws.
A match contributes $2$ points if it ends in a draw. Otherwise, it contributes $3$ points. The total number of points is, $$ d \times 2 + (n(n-1)/2 - d) \times 3 = n(\frac{3n}2-1)-(\frac n2+d),$$ which means the total number of points is smaller than $n(\frac{3n}2-1)$.
Since each team scores the same points, the total number of points must be a multiple of $n$. So it is at most the largest multiple of $n$ that is smaller than $n(\frac{3n}2-1)$, i.e., it is at most $n(\frac{3n}2-2)$. $$ n(\frac{3n}2-1)-(\frac n2+d) \le n(\frac{3n}2-2)$$ which means, $$ d \ge \frac n2. \quad\quad\checkmark$$

  • $\begingroup$ Exercise. Suppose each team gets the same points. Show that the set of possible numbers of draws is $\{n, 2n, \cdots, ((n-1)/2)n\}$ if $n$ is odd. The set of possible numbers of draws is $\{n/2, 3n/2, \cdots, (n-1)n/2\}$ if $n$ is even. $\endgroup$
    – John L.
    Commented Mar 19, 2022 at 3:26

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