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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.

In a zero-sum game with the row player maximizing, a sufficient saddle certificate is that $p,q$ are probability vectors andEvery row mixture then gains at most $v$ against $q$, and every column mixture concedes at least $v$ against $p$. Their pairing must equal $v$, proving optimality.

For the cost-weighted finite search game, order Colin's six columns as $123,132,213,231,312,321$ and Rowena's rows as $1,2,3$. The payoff matrix isUnder the proposed search mixture, an alternative location $i$ precedes the hiding location $j$ with probability $(c_i+c_k/2)/c$, where $k$ is the third location. The expected cost conditional on hiding at $j$ is consequentlyIt is independent of $j$, giving the upper bound $\mathrm{value}\le v$.

Choose $p_i=c_i/c$. Against any order $i,j,k$, its expected cost isThis gives the matching lower bound. ThusBoth strategies equalize the opponent's pure actions, giving the stated saddle certificate.

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.

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.

The game is degenerate in the nondegeneracy of a bimatrix game sense. Against the first player's pure strategy $4$, the second player's choices $1$ and $2$ are both best responses, with payoff four. A mixed strategy of support size one therefore has two pure best responses. The usual Lemke-Howson algorithm path needs additional tie handling or perturbation in such a game; the zero-sum game structure gives a simpler direct linear program.

Add five to every entry of the first player's matrix, obtainingThis does not change either player's best responses or Nash equilibria; it raises the game value by five. Since all entries are positive, its value $v_B$ is positive. If $p$ is a row mixed strategy guaranteeing $v_B$, put $x=p/v_B$. Then $B^{T}x\ge 1$ and $1^{T}x=1/v_B$.

Conversely any feasible $x$ has $s=1^{T}x>0$, and $p=x/s$ guarantees payoff $1/s$ in the shifted game. Maximizing that guaranteed payoff is therefore equivalent to minimizing $s$ under $B^{T}x\ge 1$, $x\ge 0$. These are precisely the displayed constraints. The dual program maximizes $1^{T}y$ subject to $By\le 1$, $y\ge 0$; normalizing an optimal $y$ gives the column strategy. This is positive-payoff linear programming for a matrix game.

For a fixed number choice, increasing one's own stake changes only the amount lost when one loses. It does not change the amount won, which is the opponent's stake, or the zero payoff of a tie. Thus doubling is either weakly dominated by retaining the original stake or payoff-equivalent to it. This establishes that there is no strategic advantage in doubling, but does not imply strict harm in every equilibrium.

For the first player, each of the two equilibrium choices loses with positive probability: choice one loses against the second player's four, and choice four loses against their one. Doubling either therefore gives a strictly smaller expected payoff than $-4/5$. By contrast, against the first player's support $\{1,4\}$, the second player's choices one and four either win or tie. Their own stake is never lost, so doubling it does not change their payoff.

More precisely, retain the first player's original-stake probabilities $(4/5,0,1/5)$. The second player may split their total probability $1/5$ on choice one, and $4/5$ on choice four, arbitrarily between original and double stakes. The first player's unused choice two has payoff at most $(1/5)2-(4/5)2=-6/5<-4/5$, even when the second player doubles their one stake. The doubled first-player choices are also worse. The second player's choice two, with either stake, is worse against the displayed first-player strategy. Hence these splits remain Nash equilibria of the enlarged zero-sum game.

The first player should not double; the second player is indifferent between doubling and retaining the original stakes on their equilibrium choices. This distinction is an example of weak domination does not exclude equilibrium strategies.

Let $U={\sum }_i{u}_i$. The candidate strategies areAt this pair, A's partial derivatives $2u_i{y}_i/(x_i+y_i{)}^2$ are independent of $i$, as are the negatives of B's partial derivatives. Since $F$ is concave in $x$ for fixed $y$ and convex in $y$ for fixed $x$, the Lagrange sufficiency theorem provesfor all feasible $x,y$. Hence $(x^{\ast },y^{\ast })$ is a saddle point and gives the optimal strategies in this zero-sum game. Its game value is

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.

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 $p^{T}Aq$.

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.

In the zero-sum game with row payoff matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, the second row is weakly dominated by the first. Nevertheless second row and first column form a Nash equilibrium: both rows tie against that column, and both columns tie against that row. Weak dominance therefore cannot be used to discard all possible equilibrium strategies without further qualifications.