Zero-sum game (${\sum }_i{u}_i(s)=0$)

In a zero-sum game, the players' payoffs sum to zero at every pure strategy strategy profile. For two players this means $u_2=-u_1$: maximizing the column player's payoff is minimizing the row player's. A single payoff matrix $A$ therefore specifies a finite two-player matrix game, and independent mixed strategies $x,y$ give expected payoffs $x^{\mathsf{T}}Ay$ and $-x^{\mathsf{T}}Ay$.

Matching pennies and the usual rock paper scissors are two-player zero-sum games. The asymmetric coordination example is not: its total payoff is $5$ when choices match and $0$ otherwise. Applying a two-player minimax theorem to general bimatrix games would discard this possibility of shared gains. For a two-player zero-sum game, however, optimal mixed strategies guarantee the value of a zero-sum game, and pairing any two opposing optimal strategies gives a Nash equilibrium.

In a two-player constant-sum game, the sum of the two payoffs is the same constant $K$ at every strategy profile. Replacing each player's payoff by $u_i-K/2$ gives a zero-sum game without changing best responses or Nash equilibria. If both players can guarantee $K/2$, then any Nash equilibrium gives each that value. An equilibrium in pure strategies must use pure strategies with that guarantee: otherwise the opponent could improve on $K/2$ by a unilateral deviation.

With choices ordered rock, scissors, paper and winner payoff $1$, the row-player payoff matrix is $\left(\begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ 1& -1& 0\end{array}\right)$. Both the row and column sums vanish. Thus the uniform mixed strategy guarantees expected payoff zero against every opposing strategy; the value of a zero-sum game is zero. A player responding to a known nonuniform mixture can instead optimize its linear expected payoff over pure responses and their ties.

For a finite matrix zero-sum game, the value is the payoff that the maximizing player can guarantee and the minimizing player can prevent the payoff from exceeding, using mixed strategies. The common value of the two displayed optimizations exists by the minimax theorem. A pair of strategies can be verified directly by checking that every pure column pays the row mixture at least $v$ and every pure row earns at most $v$ against the column mixture.

In a hide-and-search zero-sum game with positive inspection costs $c_i$ and $c={\sum }_i{c}_i$, hiding with probabilities $c_i/c$ gives the same expected search cost for every complete search order. For three locations, beginning at $i$ with probability $c_i/c$ and randomizing the remaining two orders also equalizes every hiding location. The common value is the displayed quantity.

A matrix game is a finite two-player zero-sum game whose entry $A_{ij}$ is the row player's payoff when the players choose row $i$ and column $j$.

When every entry of a payoff matrix $B$ is positive, its game value $v$ is positive. Setting $x=p/v$ converts the row player's probability vector and guaranteed value into the displayed linear program, whose optimum is $1/v$. Conversely normalizing a feasible $x$ gives a mixed strategy guaranteeing $1/(1^{T}x)$. The dual maximizes $1^{T}y$ under $By\le 1$, $y\ge 0$ and recovers the column strategy by normalization. Adding a constant to every payoff preserves Nash equilibria and lets this positive-payoff formulation handle general finite zero-sum games.

A payoff matrix records the row player's gains when row $i$ meets column $j$. In a zero-sum game the other player's payoff is the negative. For mixed strategies $p,q$, the expected row payoff is $p^{T}Aq$.

The row player solves

If the column player has an optimal strategy at value $v$ and a row probability vector $p$ satisfies $p^{T}A\ge v{e}^T$, then $p$ is optimal: it guarantees $v$, while the optimal column strategy prevents every row strategy from exceeding $v$.

If a symmetric invertible payoff matrix satisfies $A^{-1}e\ge 0$, both players may useand the game value is $(e^T{A}^{-1}e{)}^{-1}$.

Forboth players optimally choose cards $1,2,3$ with probabilities $41/50,2/25,1/10$, and the row player's expected payoff is $57/25$.

For every real $m\times n$ payoff matrix $A$ of a two-player zero-sum game, the minimax theorem stateswhere ${\Delta }_m,{\Delta }_n$ are the players' probability simplices. Both outer optima are attained. The row player's security level payoff is the largest payoff it can guarantee against all opposing choices; the column player can hold that payoff to the same number $v$, the value of a zero-sum game. For optimal $x^{\ast },y^{\ast }$,These are exactly the Nash equilibrium inequalities under the row-maximizes, column-minimizes convention. Conversely, a Nash equilibrium of this two-player zero-sum game supplies both guarantees. This equality explains why independent worst-case optimization solves the opposing incentives in this special class.

Allowing mixed strategies is essential. For matching pennies, $A=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$. Restricting both players to pure strategies gives ${\max }_i{\min }_j{A}_{ij}=-1$ and ${\min }_j{\max }_i{A}_{ij}=1$. Allowing mixed strategies closes the gap: each fair mixture guarantees zero, so $v=0$. These guarantees hold even if the opponent knows the mixture's probabilities, provided the opponent does not observe the sampled choice before choosing.

John von Neumann proved the two-player zero-sum game result in Zur Theorie der Gesellschaftsspiele, Mathematische Annalen 100, 295–320 (1928); the original article's footnote 7 explicitly discusses the equality of max–min and min–max. Nash's ^[ref] states that existence of a Nash equilibrium is equivalent to this result in the two-player zero-sum game case, and extends existence to all finite games.

An optimal mixed strategy guarantees the value of the game against every opposing pure strategy.

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.