Equal but different

Suppose that a round-robin tournament can be played amongst $n$ teams so that each team scores the same number $S$ of points, but not every team has the same number of wins. Suppose that team $i$ wins $W_i$ games, draws $D_i$ games and loses $L_i$ games, with $W_i+D_i+L_i = n-1$ for each $1 \le i \le n$ . Since the number of won games must equal the number of lost games, we deduce that $\sum_i W_i \,=\, \sum_iL_i$ .

If $W_i \le L_i$ for all $1 \le i \le n$ , the fact that $\sum_i W_i = \sum_iL_i$ would imply that $W_i = L_i$ for all $1 \le i \le n$ . Then team $i$ would score $3W_i + (n-1-W_i-L_i) \,=\, W_i + n-1$ and hence all teams would have the same number of wins, which is not possible. Thus there must exist $1 \le \alpha \le n$ such that $W_\alpha > L_\alpha$ . This team (and hence all teams) scores $S \,=\, 3W_\alpha + (n-1-W_\alpha-L_\alpha) \,=\, 2W_\alpha - L_\alpha + n-1$ , and hence $S > W_\alpha + n-1 > n$ . Thus $S \ge n+1$ .

Similarly, if $W_i \ge L_i$ for all $1 \le i \le n$ , the fact that $\sum_iW_i = \sum_iL_i$ would imply that $W_i = L_i$ for all $1 \le i \le n$ , which we have already seen is impossible. Thus there must exist $1 \le \beta \le n$ such that $W_\beta < L_\beta$ . This team scores $S = 2W_\beta + n-1 - L_\beta$ , and hence $\begin{array}{rcl} 2W_\beta + n-1 - L_\beta & \ge & n+1 \\ 2W_\beta - L_\beta & \ge & 2 \\ 2W_\beta & \ge & L_\beta + 2 \; > \; W_\beta + 2\\ W_\beta & > & 2 \end{array}$ so that $W_\beta \ge 3$ , and hence $L_\beta \ge 4$ . Thus team $\beta$ plays at least $7$ games, and hence $n \ge 8$ .

On the other hand, it is possible to have a round-robin tournament of $8$ teams where all the conditions are satisfied, as the following table of results shows: $\begin{array}{|c|cccccccc|ccc|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & W & D & L & P \\ \hline 1 & \times & D & D & L & W & W & L & D & 2 & 3 & 2 & 9 \\ 2 & D & \times & D & W & L & W & L & D & 2 & 3 & 2 & 9 \\ 3 & D & D & \times & W & W & L & L & D & 2 & 3 & 2 & 9 \\ 4 & W & L & L & \times & D & D & W & D & 2 & 3 & 2 & 9 \\ 5 & L & W & L & D & \times & D & W & D & 2 & 3 & 2 & 9 \\ 6 & L & L & W & D & D & \times & W & D & 2 & 3 & 2 & 9 \\ 7 & W & W & W & L & L & L & \times & L & 3 & 0 & 4 & 9 \\ 8 & D & D & D & D & D & D & W & \times & 1 & 6 & 0 & 9 \\ \hline \end{array}$ Thus $8$ is the smallest number of teams in a round-robin tournament in which it is possible for all teams to end up with the same point score, but for at least two teams to have different numbers of wins.

Firstly, there is a possible solution with $N=\boxed{8}$ : crosstable linked

Now we will show that there is no possible smaller solution. Suppose $N<8$ .

Let $w$ be the number of matches won, out of the total number of matches which is $N(N-1) \over 2$ . Consider the total score (sum of all teams' scores). A won game contributes $3$ and a tie contributes $2$ . So the total score is $\begin{aligned} &3w - 2( {N(N-1) \over 2} - w ) \\ = &w - N(N-1) \\ = &N({w \over N} - (N-1)) \end{aligned}$

Since each team finishes on the same score, each team's score must be $k + N - 1$ where $k = {w \over N}$ .

Further, the mean number of wins for each team is $k$ . Since it's stipulated that not all teams finish on the same score, there must be a team with more than $k$ wins (let's call that team $W$ ) and a team with fewer than $k$ wins, called $L$ . If there are more than one possible team in each case here just choose one of them.

Suppose that the difference in number of wins between $W$ and $L$ is more than $2$ . Then their difference in score from wins is at least $9$ , so $L$ must tie at least $9$ games to match $W$ 's score. Since $N<8$ , this is not possible. So we can conclude that $W$ wins $k+1$ games and $L$ wins $k-1$ games.

Let $d$ be the number of games tied by $W$ . To make their scores match, $L$ must tie $d+6$ games. So team $L$ has: $k-1 \mbox{ wins} \\ d+6 \mbox{ ties} \\ (N-1) - (k+d+5) \mbox{ losses}$ The last line lets us conclude that $N \ge k+d+6$

Now, the score for $L$ is $3(k-1) + (d+6) = 3k + d + 3$ . But we already worked out that each player's score is $k + N - 1$ . So

$\begin{aligned} k+N-1 &= 3k+d+3 \\ N &= 2k + d + 4 \\ k + d + 6 &\le 2k + d + 4 \\ 2 &\le k \\ 8+d &\le N \end{aligned}$

Since $d \ge 0$ we have proven that $N \ge 8$ .

These values gave me the hint to start building the crosstable: team $W$ has $3$ wins and $4$ losses; and $L$ has $1$ win, $6$ ties. Therefore $L$ must have beaten $W$ and the other teams are easy to fill out by inspection.