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In a zero-sum game, an entry $a_{ij}$ is the net amount received by the row player when the players choose actions $i,j$; the column player receives its negative. A mixed strategy $p$ has nonnegative entries summing to one. Against column strategy $q$, expected row payoff is $p^{T}Aq$. An optimal row strategy maximizes its guaranteed payoff ${\min }_q{p}^{T}Aq$; an optimal column strategy minimizes ${\max }_p{p}^{T}Aq$.

The betting payoff must subtract the player's own stake: a winner receives both stakes but has already paid its own. Its net gain is the opponent's stake, and a loser loses its own. Ties give zero. Thus the row-payoff matrix isIt is antisymmetric. Interchanging players therefore changes the sign of the game value while leaving the game unchanged, so its value is zero. An optimal column strategy $p$ must prevent any pure row choice from gaining positively, namely $(Ap{)}_i\le 0$. Antisymmetry then also gives $p^{T}A\ge 0$, so the same vector is an optimal row strategy. This is the support certificate for an antisymmetric matrix game.

Seek a mixture supported on actions $1,3,4$ making those three opposing choices indifferent. Their zero-payoff equations areWith $p_1+p_3+p_4=1$, these giveCheck every opponent action, including the omitted second one:This verifies optimality rather than simply assuming that a support choice is valid.

It is also unique. Let $r$ be any other optimal strategy, so $Ar\le 0$. For the displayed $p$, antisymmetry gives $p^{T}Ar=-(Ap{)}^{T}r=(7/8)r_2$. The left side is nonpositive and the right side nonnegative, forcing $r_2=0$. Since the other three entries of $p$ are positive, $p^{T}Ar=0$ also forces $(Ar{)}_1=(Ar{)}_3=(Ar{)}_4=0$. Their equations and normalization recover the same probabilities. Thus both players' unique optimal strategy uses bets one, three and four in the proportions $3:4:1$.