Bimatrix game

A bimatrix game is a finite two-player game specified by one payoff matrix for each player.

Both players choose heads $H$ or tails $T$ without seeing the other's choice. The row player receives $1$ when the choices match and $-1$ when they differ; the column player receives the opposite payoff. Thus matching pennies is a zero-sum game, with ordered payoff pairs:

At $(H,H)$ the column player improves by switching to $T$; at $(H,T)$ the row player improves by switching to $T$; at $(T,H)$ the row player improves by switching to $H$; at $(T,T)$ the column player improves by switching to $H$. Every pure strategy strategy profile therefore has a profitable unilateral deviation, so none is a Nash equilibrium.

Let $p$ be the row player's probability of heads and $q$ the column player's probability of heads. Their random choices are independent. Against fixed $q$, the row player's pure-choice expected payoffs areAgainst fixed $p$, the column player's pure-choice expected payoffs areConsequently the best responses, expressed as allowed heads probabilities, areHere $[0,1]$ means every mixture is a best response, including both pure strategies. If $q>1/2$, the row player must choose $p=1$, which makes the column player's best response $q=0$, a contradiction. If $q<1/2$, the row player must choose $p=0$, which forces $q=1$, also a contradiction. Hence $q=1/2$; for the column player to mix at this $q$, the two column payoffs must tie, forcing $p=1/2$. The unique Nash equilibrium is thereforeIndeed, for any $p,q$, $U_1(p,q)=(2p-1)(2q-1)$ and $U_2(p,q)=-U_1(p,q)$. Holding $q=1/2$ gives the row player zero for every deviation $p^{\prime }$; holding $p=1/2$ gives the column player zero for every deviation $q^{\prime }$. A player is indifferent against the equilibrium mixture, but choosing an arbitrary best response to it need not preserve the opponent's best response. This is why that indifference does not produce additional Nash equilibria.

For a teaching treatment with the same payoff convention, see MIT 17.810, lecture 3, slides 29–38, on mixed strategies and matching pennies.

A coordination game rewards compatible choices. Players can have a common interest in coordinating while having different preferred coordinated outcomes, so their interests need not be opposite as in a zero-sum game.

Two players independently choose venues $A$ or $B$. Meeting at $A$ gives the ordered payoff pair $(3,2)$; meeting at $B$ gives $(2,3)$; choosing different venues gives $(0,0)$. As throughout this article, the row player's payoff comes first:

Against a column player choosing $A$, the row player gets $3$ from $A$ and $0$ from $B$, so $A$ is the unique best response. Against $B$, the row player gets $0$ from $A$ and $2$ from $B$, so $B$ is the unique best response. The column player likewise prefers to match the row choice. Thus $(A,A)$ and $(B,B)$ are pure strategy Nash equilibria, while each mismatched strategy profile has a profitable unilateral deviation.

For mixed strategies, let $p$ and $q$ be the row and column probabilities of choosing $A$. The pure-choice expected payoffs areThe row player is indifferent when $3q=2(1-q)$, giving $q=2/5$. The column player is indifferent when $2p=3(1-p)$, giving $p=3/5$. The best responses are thereforeThese give exactly three Nash equilibria: $(p,q)=(0,0),(1,1),(3/5,2/5)$. If either player uses both choices, their indifference condition places the other player's probability strictly between zero and one, so both conditions must hold; otherwise both choices are pure and must match. Notice whose equation determines whose probability: the row player's indifference fixes the column player's $q$, and the column player's indifference fixes the row player's $p$.

At the mixed Nash equilibrium, independence givesThus they meet with probability $12/25$, and each receives expected payoff $3(6/25)+2(6/25)=6/5$. Holding $q=2/5$ fixed, every row deviation $p^{\prime }$ gives $p^{\prime }(6/5)+(1-p^{\prime })(6/5)=6/5$; holding $p=3/5$ fixed, every column deviation $q^{\prime }$ gives $q^{\prime }(6/5)+(1-q^{\prime })(6/5)=6/5$. This directly verifies the Nash equilibrium inequalities.

Both coordinated pure strategy outcomes strictly improve both players' expected payoffs over $(6/5,6/5)$: $(3,2)$ and $(2,3)$ each do so. The mixed Nash equilibrium therefore fails Pareto efficiency, even though no unilateral deviation helps. Switching together can help when switching alone cannot. Neither coordinated outcome improves both players over the other, and both are Pareto efficient, even when a lottery over joint action profiles is allowed: any positive weight on another outcome reduces the payoff of the player receiving $3$.

A probability distribution over joint action profiles implements a convex combination of their two-player payoff vectors. The attainable expected payoffs form the convex hull of the pure joint payoffs. The distribution need not factor into independent mixed strategies, and a freely enforceable lottery need not satisfy the incentive constraints of a Nash equilibrium.

Enumerate pairs of strategy supports, solve the equations making each supported action indifferent, then check strict positivity of supported probabilities and all omitted-action payoff inequalities. These conditions suffice for a Nash equilibrium. Under nondegeneracy of a bimatrix game, only pairs of equal support size need be tested.

A square bimatrix game is symmetric when exchanging players exchanges their payoffs: its payoff matrices are $P,Q$ with $Q=P^T$. If $(x,y)$ is a Nash equilibrium, then $(y,x)$ is also a Nash equilibrium. A symmetric bimatrix game can still have asymmetric equilibria, so the two strategy vectors must be kept separate in the Lemke-Howson algorithm.

A bimatrix game is nondegenerate if every mixed strategy with $k$ positive coordinates has at most $k$ pure best responses for the other player. This condition prevents ties in the leaving-variable choice along the usual Lemke-Howson algorithm path and allows unique continuation once the dropped label has been selected.

A Nash equilibrium is a strategy profile in which every player's strategy is a best response to the others. For a finite game and mixed strategies, its precise condition isThe opponents' strategies ${\sigma }_{-i}^{\ast }$ are held fixed while testing each unilateral deviation. Equality is allowed: a player may have other equally good best responses. Because expected payoff is a linear function of the player's own mixture, it suffices to test every pure strategy. Equivalently, every choice in the support of a mixed strategy must attain the largest pure-choice expected payoff, and no unused choice may exceed it.

The two worked examples illustrate different uses of this condition. Matching pennies has no pure strategy Nash equilibrium, but its unique mixed strategy Nash equilibrium removes each player's vulnerability to predictable play. The asymmetric coordination example has three Nash equilibria, and its mixed one fails Pareto efficiency. A Nash equilibrium tests individual incentives, whereas Pareto efficiency compares feasible outcomes that may require several players to change together. In the prisoner's dilemma, for example, mutual defection is a Nash equilibrium even though mutual cooperation gives both players a larger payoff because $R>P$.

Existence in a finite game is guaranteed by Nash's theorem. The theorem does not select one Nash equilibrium, promise Pareto efficiency, or assert convergence of a learning process. Nash's ^[ref] used the Kakutani fixed-point theorem; his ^[ref] gave a proof using the Brouwer fixed-point theorem. The connection to a fixed point captures the idea of choices that remain mutually optimal when considered together.

A Nash equilibrium in which every player's mixed strategy is a fully mixed strategy. Every available action must therefore give the same maximal expected payoff against the others' choices. These indifference equations, together with strictly positive probabilities summing to one, suffice to verify such a Nash equilibrium.

Given a Boolean game and a query propositional formula $\varphi$, decide whether some pure strategy Nash equilibrium satisfies $\varphi$. The problem is complete for the existential second level of the polynomial hierarchy.

For membership, existentially choose a Boolean valuation and universally check every player's every unilateral deviation. The goal must remain false under all such deviations for each player whose goal is currently false. Each individual check takes polynomial time.

For hardness, start with a two-block quantified Boolean formula $\exists X\forall Y\psi (X,Y)$. Give player 1 control of $X$ and the constant true goal. Give player 2 control of $Y$ and goal $\neg \psi$, and use $\psi$ as the query. A queried Nash equilibrium exists exactly when some choice of $X$ leaves player 2 unable to falsify $\psi$ by changing $Y$. This is precisely the input quantified statement, and the construction is a polynomial-time reduction.

There are $mq$ distinctly ranked students and $m$ labelled universities with $q$ seats each. Each student applies to one university; each university selects the greatest ranks, and admitted students receive their admitted cohort's average rank. At a pure Nash equilibrium, every student is admitted: any rejected student could move to a university with a vacancy. The university with the largest average must contain the top $q$ ranks, since any omitted higher rank could displace its lowest rank and improve. Repeat on the remaining students to obtain consecutive rank blocks. Assigning these blocks to the labelled universities gives exactly $m!$ pure equilibria.

In a game invariant under permuting players, a symmetric Nash equilibrium gives every player the same strategy, which may be a mixed strategy. Independent randomization does not require realized resource loads to be equal. When resource delays are affine, a player's expected delay depends on their own certain resource use plus the expected use by the other players; replacing all loads by unconditional total expected loads loses that own-player correction.

A finite nondegenerate bimatrix game has a finite odd number of Nash equilibria. After shifting to positive payoffs, fully labelled vertex pairs in the bounded best-response polytopes correspond to equilibria plus one artificial zero pair. Dropping a fixed label gives a finite path graph whose endpoints are exactly these pairs. The number of endpoints is even, leaving an odd number of genuine equilibria. This is the endpoint-parity argument underlying the Lemke-Howson algorithm.

A symmetric equilibrium of a symmetric game has every player use the same strategy or the same distribution of strategies. Symmetry between players does not require identical behavior across contests with different prizes.

In a nondegenerate symmetric bimatrix game, player interchange pairs every nonsymmetric Nash equilibrium with a distinct swapped equilibrium. The equilibrium oddness theorem gives an odd total; the remaining fixed points, which are the symmetric equilibria, therefore also number an odd amount.

The continuous gain map takes a probability simplex to itself. The Brouwer fixed-point theorem supplies a fixed point, whose positive gains would contradict the zero strategy-weighted average payoff excess. Thus all gains vanish, giving a strategy that is a best response to itself. In a symmetric bimatrix game, this produces a symmetric equilibrium.

For a symmetric finite game, pure-action expected payoffs $e(i,p)$ against independent identical opposing mixed strategies are continuous. The displayed map takes the probability simplex continuously into itself. At a Brouwer fixed-point theorem fixed point, put $r_i=[e(i,p)-e(p){]}_{+}$ and $S={\sum }_i{r}_i$. The fixed-point equations give $r_i=p_{i}S$. If $S>0$, every supported action has strictly positive excess payoff, contradicting ${\sum }_i{p}_i(e(i,p)-e(p))=0$. Hence $S=0$ and every supported action is a best response, proving a symmetric Nash equilibrium.

For a symmetric bimatrix game with payoff matrices $P,P^{\mathsf{T}}$, any nonzero $x$ satisfying these equations gives the symmetric Nash equilibrium $(x/{\sum }_i{x}_i,x/{\sum }_i{x}_i)$. Every positive coordinate attains the same maximal payoff. This one-vector construction differs from a two-vector Lemke-Howson algorithm path, which can return asymmetric equilibria.

The Lemke-Howson algorithm follows an almost completely labeled path in the product of two best-response polytopes of a bimatrix game. Drop one label at the artificial zero pair, then alternately pivot away the duplicated label until the dropped label returns. A nonzero completely labeled endpoint, after normalization, gives a Nash equilibrium. The two players' vectors must be kept separate even in a symmetric game.

Complementary pivoting follows adjacent bases while maintaining all but one label of a complementarity system. The Lemke-Howson algorithm for a bimatrix game follows the duplicated label between two tableaux until it restores the dropped label at a nonzero completely labelled pair. Normalization then gives a Nash equilibrium.

Nash's theorem states that every finite game has at least one Nash equilibrium in mixed strategies.

For each player in a finite bimatrix game, add the positive pure-strategy gains to its current mixed strategy and normalize. This gives a continuous self-map of the product of probability simplices. At a Brouwer fixed-point theorem fixed point, the gain coordinates equal the current strategy times their total gain. If that total were positive, every supported action would beat the current average payoff, contradicting that average. Thus every pure deviation has nonpositive gain for both players, proving a Nash equilibrium.

Normalize each player's positive pure-strategy payoff improvements to define a continuous self-map of the product of their mixed-strategy simplices. At a Brouwer fixed point every positive improvement must vanish, giving a Nash equilibrium.