Optimal stake in a token pool game

Two players, $X$ and $Y$, each own $100$ tokens. Simultaneously and in secret, each puts between $1$ and $100$ of their tokens into a common pot; say $X$ puts in $x$ and $Y$ puts in $y$. One token is then drawn uniformly from the pot. If it is one of $X$'s, $Y$ pays $X$ an amount $100-x$; if it is one of $Y$'s, $X$ pays $Y$ an amount $100-y$. Both play optimally to maximise their own expected payoff. How many tokens should $X$ put in?

$50$. $X$'s expected gain is $\dfrac{x(100-x)-y(100-y)}{x+y}=\dfrac{(x-y)(100-x-y)}{x+y}$. Against $y=50$ this is $-\tfrac{(x-50)^2}{x+50}\le0$, maximised at $x=50$; by symmetry $50$ is also $Y$'s best reply, so $(50,50)$ is the equilibrium and the game is fair.