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In a two-player zero-sum game, the row player chooses a row, the column player chooses a column, and the row player's gain is the column player's loss. If their independent mixed strategies are probability vectors $x\in {\mathbb{R}}^m$, $y\in {\mathbb{R}}^n$, the row player's expected payoff is $x^{T}Py$.

The optimal mixed strategies solveThe minimax theorem equates these two guaranteed values. Equivalently, for a value $v$, $x_{\ast }^{T}P\ge v{1}^T$ and $P{y}_{\ast }\le v1$ componentwise.

For a row strategy $(p,1-p)$, the payoffs against the two pure columns are $2-6p$ and $-4+6p$. Their minimum is maximized at their intersection, $p=1/2$, where both are $-1$. Symmetrically the column player's maximum of the two row payoffs is minimized at $q=1/2$. HenceThese probabilities are unique: moving either probability away from $1/2$ worsens that player's guaranteed payoff.

Let $p$ be the row player's mixed strategy and $q$ the column player's mixed strategy. The fourth row is strictly dominated by the first. A useful candidate mixes the first two rows so that their first and third column payoffs agree. If the first row has probability $t$, these payoffs are $7t-2$ and $2-7t$, giving $t=2/7$ and common value zero.

To prove that these are optimal mixed strategies, use the two explicit mixed strategiesTheir payoff vectors satisfyThus the row player guarantees a payoff at least zero against every column mixed strategy, and the column player guarantees a payoff at most zero against every row mixed strategy. This matching pair proves that the value of the zero-sum game is zero.

These optimal mixed strategies are also unique. Since $Aq$ is strictly negative in rows three and four, any optimal row strategy gives them probability zero. Against columns one and three it must then satisfy both $7t-2\ge 0$ and $2-7t\ge 0$, so $t=2/7$. Similarly, $p^{T}A$ is strictly positive in column two, forcing an optimal column strategy to avoid that column; its constraints against rows one and two then force equal weights on columns one and three.

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.

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.