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The row player maximizes the payoff and the column player minimizes it. Row three is weakly dominated by row one, so the row player can discard it. On the remaining two rows, column two is weakly dominated by column three, so the column player can discard column two. The reduced zero-sum game has payoff matrixIf the first row is chosen with probability $r$, the expected payoffs against its two columns are $3-r$ and $2+2r$. Equalizing them gives $r=1/3$ and value $8/3$. If the first column is chosen with probability $s$, the payoffs of the two rows are $4-2s$ and $2+s$. Equalizing them gives $s=2/3$.

Restoring the original indexing, the optimal mixed strategies and value of a zero-sum game areTo verify these against every original strategy, the row mixture gives the three column payoffs $(8/3,3,8/3)$, each at least $v$. The column mixture gives the three row payoffs $(8/3,8/3,7/3)$, each at most $v$. Thus each mixed strategy guarantees the stated value, including against the discarded strategies. The strict inequalities for column two and row three also force those probabilities to be zero in any optimal strategy, and the remaining equalities determine the displayed mixtures uniquely.

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.

Put $p=\mathrm{Pr}(T)$ and $q=\mathrm{Pr}(L)$, so the complementary probabilities are $1-p$ and $1-q$. All four expressions are the row player's expected payoff, including those indexed by a column action. Direct expected value calculations giveThe column player's expected payoff is the negative of the row player's because this is a zero-sum game.

There is no perfect recall. Player 1's later information set joins histories following both of her initial actions, so she has forgotten whether she chose $L$ or $R$. It also hides player 2's intervening action, but failure to observe another player's action is imperfect information, whereas forgetting one's own earlier action is the particular violation of perfect recall. Player 2's two nodes form a single information set because player 2 does not observe the initial choice.

For a behavioural strategy, let $p$ be player 1's initial probability of $L$ and $r$ her later probability of $L$. These are separate information sets, so $p$ and $r$ need not be equal; their random draws are independent. Name player 2's unlabeled branches left and right, and let $q$ be the probability of taking the left branch. The only unit-payoff paths are left–left–left and right–right–right. ConsequentlyBecause this is a zero-sum game, the natural meaning of the claimed best behavioural strategy is its maximum guaranteed expected payoff. Player 2 can choose either branch, so the guarantee at $(p,r)$ isFor nonnegative numbers, their minimum is at most their geometric mean. Also $p(1-p)\le 1/4$ and $r(1-r)\le 1/4$, henceTaking $p=r=1/2$ makes both possible branch-conditioned success probabilities $1/4$, so this upper bound is attained against every $q$. Thus the optimal guaranteed behavioural strategy success probability is $1/4$.

A mixed strategy can instead select the complete plan $LL$ with probability $1/2$ and the complete plan $RR$ with probability $1/2$. The two choices within a sampled plan are correlated. If player 2 chooses left, exactly the sampled plan $LL$ succeeds; if player 2 chooses right, exactly $RR$ succeeds. Therefore the success probability isagainst every opposing mixed strategy. It is optimal among mixed strategies too: player 2's uniform mixture gives success probability $1/2$ to each of $LL,RR$ and zero to $LR,RL$, so no mixture of the four plans can exceed $1/2$ against that opponent. This establishes mixed and behavioural security levels without perfect recall directly.

The guarantee qualification matters: if player 2's branch were known to be left, the deterministic plan $LL$ would succeed with probability one. The number $1/4$ is not an upper bound against every fixed opposing choice separately.

Against $q$, choosing row $i$ gives Player I expected payoff $(Aq{)}_i={\sum }_j{a}_{ij}{q}_j$. The best response maximizes this number; randomizing among rows cannot exceed the largest row payoff. Player II therefore minimizes that maximum, exactly the printed optimization problem.

The minimax theorem for a finite zero-sum game statesThis common number is the value of a zero-sum game: Player I can guarantee an expected payoff of at least $v$, and Player II can hold it to at most $v$. At optimal mixed strategies their expected transfer is $v$.

A mixed strategy for Player II is a probability vector $q\ge 0$ with ${\sum }_j{q}_j=1$, specifying the random choice of a column. If Player I chooses row $i$, the expected payment is $(Aq{)}_i$. A mixture of rows cannot exceed the maximum of these row expectations, so the smallest payment Player II can guarantee isAn optimal strategy is a vector attaining that minimum, which exists by compactness of the probability simplex and continuity. The minimax theorem identifies this guaranteed loss with Player I's maximal guaranteed gain, the value of the finite zero-sum game.

A payoff is the number assigned to a player's outcome at a strategy profile, with larger numbers preferred. In the usual expected utility model, expected payoffs are probability-weighted averages of these numbers. If a payoff is specified as money and a player has preferences involving risk, first apply the relevant utility function; maximizing expected money assumes risk neutrality. In a two-player zero-sum game, one player's payoff is the negative of the other's. General bimatrix games instead require a payoff matrix for each player.

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 function taking the lesser value at every common-domain point. It forms the lower envelope of two graphs. For finitely many affine functions it is a concave function: each affine branch is at least the corresponding convex combination of the two endpoint minimum values, so taking the minimum preserves that lower bound. A maximizing mixed strategy in a two-player zero-sum game maximizes such a lower envelope of opposing pure-choice expected payoffs.

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.

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 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.

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.