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A normal-form game specifies each player's strategy set and their payoff for every strategy profile. In a finite two-player game these data form a bimatrix game; if the two payoffs always sum to zero, one payoff matrix describes the matrix game. The representation specifies choices and incentives, without a temporal order of observed moves.

Use net monetary gain as payoff and order each player's pure strategies as $1,2,4$. The payoff matrix for the first player isThe second player's matrix is $-A$. Equal choices return both stakes, so the diagonal is zero; when the first player wins, their net gain is the other's stake, and when they lose it is minus their own stake. This is a matrix game, hence a zero-sum game and a normal-form game.

In a matrix game, entry $a_{ij}$ is the row player's payoff, and the column player's payoff is $-a_{ij}$. A mixed strategy is a probability vector over the available rows or columns. With strategies $p,q$, the row payoff is $p^{T}Aq$. Its optimal mixed strategy maximizes its guaranteed payoff, while the column player minimizes the largest row payoff:The equality is the finite minimax theorem. Equivalently the row player maximizes ${\min }_j(p^{T}A{)}_j$, and the column player minimizes ${\max }_i(Aq{)}_i$. Each optimal strategy guarantees its corresponding bound against every opponent strategy; for the column player's own negative payoff this is also a maximin strategy.

In this consecutive-number antisymmetric game, with the row player's number denoted $i$, the payoff matrix isIt is an antisymmetric matrix, so the matrix game has value zero for every $n\ge 3$. For $n=3$,satisfies $Ap=0$ and $p^{T}A=0$, proving optimality for both players.

For any larger $n$, extend the same strategy by zero probabilities:Indeed, the first three coordinates of $Ap$ are zero. For $i=4$, $(Ap{)}_4=-2/4-2/2+1/4=-5/4$; for $i\ge 5$, all three supported numbers are at least two below $i$, so $(Ap{)}_i=-2$. Hence $Ap\le 0$ and, by antisymmetry, $p^{T}A\ge 0$. The column player using $p$ prevents positive payoff, while the row player using $p$ prevents negative payoff, a mixed-strategy optimality certificate for a matrix game. This explicitly checks all the additional strategies rather than assuming the three-number subgame remains optimal.

An $m\times n$ matrix game with payoff matrix $A$ is a two-player zero-sum game: player I chooses a row and receives $A_{ij}$, while player II chooses a column and loses the same amount. For mixed strategies $p$ and $q$, the expected payoff to player I isPlayer I's optimal mixed strategy maximizes the payoff guaranteed against every $q$, while player II's minimizes the largest payoff obtainable by any $p$. Thusby the minimax theorem. Equivalently, optimal strategies satisfy

Here $A^T=-A$, so this is an antisymmetric zero-sum game. For every probability vector $r$,It follows that the row player's guaranteed payoff cannot exceed zero and the column player's worst loss cannot be below zero. Minimax therefore givesIf $p$ is optimal for player I, thenTransposing and using $A^T=-A$ giveswhich is precisely the optimality condition for player II. Thus every optimal strategy for player I is also optimal for player II.

The condition $Ap\le 0$ explicitly readsThe probability vectorsatisfiesso it is optimal for both players.

To prove uniqueness, let $q$ be any optimal strategy. Since $p$ is optimal for player II and $q$ is optimal for player I,and hence $q_3=0$. Writing $q=(a,b,0,d{)}^T$, the first, second, and fourth inequalities in $Aq\le 0$ giveConsequentlyso equality holds throughout: $a=2d$ and $b=4d$. The normalization $a+b+d=1$ yields $d=1/7$. Therefore the displayed $p$ is the unique optimal strategy.