# rock, paper scissors tournament, win at least z times

Consider a rock, paper scissors (one wins, one loses) tournament with n players where everybody plays against eachother exactly once:

what's the maximum number, $$z=z(n)$$, of games that assure that at least one player has won z times?

The first thing i think is needed is to determine the number of matches, N, which is $$N=\frac{(n-1)n}{2}$$, now what i dont know is how to calculate how many of those $$N$$ matches need to be played before someone has won at least $$z$$ times.

Note that if $$N \geq n(z-1)+1$$, then by the Pigeonhole principle there must be some player who has won at least $$z$$ games. Thus setting $$z = \lfloor (N-1)/n \rfloor + 1 = \lceil (n-1)/2 \rceil$$ is always safe. To show that this is the best we can do, consider the following: all players sit in a circle. For the match between player $$A$$ and player $$B$$, let $$A$$ win if and only if $$B$$ is closer on $$A$$'s right side than on $$A$$'s left. For matches where $$A$$ and $$B$$ are directly across from one another, either $$A$$ or $$B$$ may win (this only happens when $$n$$ is even). If the results of the rock-paper-scissors matches follow this pattern, then the number of matches won by each player is easy to calculate: it is $$(n-1)/2$$ if $$n$$ is odd, and either $$n/2$$ or $$n/2 - 1$$ if $$n$$ is even. In either case, each player wins at most $$z = \lceil (n-1)/2 \rceil$$ games, so we could not have chosen $$z$$ to be any larger.