Antisymmetric zero-sum game

A square game with antisymmetric payoff matrix has value zero.

For an antisymmetric payoff matrix, such a vector guarantees that an opponent playing rows cannot win positively against it. Antisymmetry also gives $p^{T}A\ge 0$, so the same mixed strategy guarantees nonnegative payoff as a row strategy and proves game value zero. Furthermore $p^{T}Ap=0$ forces $(Ap{)}_i=0$ whenever $p_i>0$. To prove uniqueness, use one certified strategy $q$: any other optimal $p$ satisfies $q^{T}Ap=-(Aq{)}^{T}p$, so a strict negative component of $Aq$ forces the corresponding probability in $p$ to vanish.

In this game the higher of two chosen integers wins one unit if they are consecutive, but pays two units if their difference is at least two. Ties pay zero. For any choices $1,\dots ,n$ with $n\ge 3$, both players can optimally choose $1,2,3$ with probabilities $1/4,1/2,1/4$ and assign zero probability to larger numbers. If $p$ is this vector, $Ap$ is zero in its first three entries, $-5/4$ in its fourth when present, and $-2$ in all later entries. Thus $Ap\le 0$ and $p^{T}A\ge 0$, proving the zero value and both security guarantees.