# Can all teams in a tournament finish with the same points? If yes, how can there also be fewest drawn matches?

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
draws

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### 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, $$d \ge \frac n2. \quad\quad\checkmark$$

• 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. Commented Mar 19, 2022 at 3:26