Number of game won in a round-robin tournament

combinatorics

Sixteen players participated in a round-robin tennis tournament. Each of them won a different number of games. How many games did the player finishing sixth win?

I got some idea here:Round Robin Tournament

The post here suggests that if there are $2^{n}$ players, then at least one player won $\dfrac{2^{n}(2^{n}-1)/2}{2^{n}}$ (and round up) games.

I believe this can also be applied any $n$ players.

So for us, at least one player won 8 games. However, by hypothesis that each player cannot win the same number of games. We can conclude that there must be a unique player who won 8 games.

I do not think this attempt has any help to this question, and I don't know how to do the next.

Also, how could I use the fact that "player finishing sixth"?

Thank you!

Best Answer

Define a player's score as the number of games won by that player.

By hypothesis, there are $16$ distinct scores.

Consider the maximum score.

Each player plays exactly $15$ games, so the maximum score can't be more than $15$.

But the maximum score can't be less than $15$ either, else there would be at most $15$ distinct scores.

Thus, the maximum score is $15$.

It follows that the $16$ scores are just $$0,1,2,...,15$$ hence the player who finished in $6$-th place had a score of $10$.