Variance of the optimal pick in a two-below game

Two players each secretly choose an integer from $1$ to $100$, and the loser pays the winner $\text{\char36}1$. Normally the higher number wins and equal numbers are a draw, with one exception: a number exactly $2$ less than the opponent's beats it (e.g. $80$ beats $82$). Under optimal play each player uses a mixed strategy, choosing a random integer $X$. What is $\text{Var}(X)$?

$2/3$. Choosing $n+3$ is always at least as good as choosing $n$, so play collapses onto $98$, $99$, $100$, which form a rock-paper-scissors cycle ($99$ beats $98$, $100$ beats $99$, $98$ beats $100$). The equilibrium is uniform on these three values, whose variance is $\tfrac{1+0+1}{3}=\tfrac23$.