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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 $3$.

Game theory studies decisions whose consequences depend on what other decision makers choose. A bidder's gain depends on competing bids in an auction; a driver's travel time depends on other drivers in a congestion game. This strategic interdependence makes choosing well a different problem from optimization against a fixed environment: each participant's best response depends on the others' choices.

A normal-form game specifies the players, their available pure strategies, and their payoffs for each strategy profile. Larger payoffs mean more preferred outcomes for that player. Unless stated otherwise, a two-player table below lists $(u_1,u_2)$: the row player's payoff first, the column player's second. Each player maximizes their own expected utility; the numbers need not represent money, and comparing different players' numerical scales is unnecessary.

The central question is whether everyone's choices fit together. A Nash equilibrium consists of mutual best responses: no unilateral deviation improves a player's expected payoff. It can involve definite choices, or mixed strategies that deliberately randomize. Matching pennies shows why randomization can be essential; the coordination game shows how players can share an interest in coordinating while disagreeing over which outcome to choose. Such shared gains are absent from a two-player zero-sum game, where one player's gain is exactly the other's loss.

These ideas are useful both for analyzing incentives and for designing them. Mechanism design asks how changing the rules changes participants' best responses; evolutionary game theory studies how successful strategies spread. The mathematical appeal is that local comparisons of payoffs can reveal a globally consistent pattern of choices. Consistency alone does not imply Pareto efficiency, uniqueness, or that actual play will converge to a Nash equilibrium.

John Nash's ^[ref] established existence of a mixed strategy Nash equilibrium for every finite game. His ^[ref] developed the theory beyond two-player zero-sum games. The distinction between conflicting and shared interests is fundamental to the subject.

The iterated prisoner's dilemma is a repeated game with a prisoner's dilemma at each round. Earlier choices can change later best responses: a present gain from defection may be outweighed by lost future cooperation. A pure strategy now specifies an action after every possible observed history, including histories that its own prescribed play would never reach. The horizon, observations, initial actions, and payoff convention must be specified before assessing a Nash equilibrium or subgame perfect equilibrium.

For the following worked example, repeat the prisoner's dilemma above infinitely at rounds $t=0,1,\dots$. Assume public perfect monitoring: after each round both players observe both actions exactly, without error. They share a discount factor $0\le \delta <1$ and evaluate a path of stage payoffs $u_i(t)$ using the normalized discounted payoffThe round-zero weight is one before normalization, including when $\delta =0$. A larger discount factor gives future payoffs more weight; the geometric series makes a constant stage payoff have the same normalized value. These assumptions and normalization follow ^[ref]. The numerical payoffs here are our $3,5,0,1$ example; the lecture uses a different table.

Suppose both players use grim trigger: play $C$ initially and while every past action of both players was $C$; after any observed $D$, play $D$ forever. With no unilateral deviation, each gets normalized payoff $3$. A first unilateral deviation to $D$ earns $5$ now, then at most $1$ each round because the opponent switches permanently to always defect. The deviator attains that continuation payoff by choosing $D$ thereafter. At any history of mutual cooperation, the comparison is thereforeCooperation is a best response precisely whenEquivalently, without normalization the geometric series givesMultiplying both unnormalized payoffs by the same positive factor $1-\delta$ leaves the best response comparison unchanged.

For example, with discount factor $\delta =1/4$, the first unilateral deviation gives normalized payoff $4>3$, so mutual grim trigger is not a Nash equilibrium. At $\delta =3/4$, it gives $2<3$, so cooperation is preferable. At $\delta =1/2$ the two payoffs tie: the incentive to cooperate is weak, which is sufficient for a subgame perfect equilibrium.

The punishment is credible after every history containing a defection: the opponent's pure strategy then prescribes $D$ forever, and $D$ remains the stage best response, earning $1$ rather than $0$. Together with the cooperative-history comparison, this establishes mutual grim trigger as a subgame perfect equilibrium for $1/2\le \delta <1$. Delaying the first unilateral deviation merely delays the same gain or loss; further cooperation during punishment only lowers the deviator's payoff. Mutual always defect also remains a subgame perfect equilibrium, so patience permits cooperation without making it inevitable.

This calculation assumes an infinite horizon and $\delta <1$; it does not define a discounted payoff at $\delta =1$. With a finite, commonly known final round and positive weight on each round, backward induction instead gives mutual defection in every round of this stage game. Observational errors change the model too: one mistaken defection permanently activates grim trigger. Historical dependence, initial memories, and the payoff time convention also affect whether a cooperative rule resists or invades other pure strategies; the existing reactive strategy and Tit for tat discussions concern those additional questions.

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 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.

Two people can each have a clear incentive to choose an outcome that leaves both worse off. The prisoner's dilemma isolates this tension between individual best responses and Pareto efficiency. For example, two partners can each save effort by withholding a promised contribution, yet both would prefer a successful joint project to mutual withholding. The model applies when each person's payoffs really have the ordering below; the names “cooperate” and “defect” alone do not establish a prisoner's dilemma.

In the symmetric two-player normal-form game, each player chooses cooperation $C$ or defection $D$ simultaneously, without observing the other's current choice. The four payoffs are the mutual-cooperation reward $R$, mutual-defection punishment $P$, sole-defector temptation $T$, and sole-cooperator payoff $S$, with $T>R>P>S$. Defection is a strictly dominant strategy: it beats cooperation against either opposing pure strategy. This definition and the resulting one-shot Nash equilibrium are discussed in Steven Kuhn, Prisoner's Dilemma, Stanford Encyclopedia of Philosophy, section 1.

Take the concrete payoffs $T=5$, $R=3$, $P=1$, $S=0$. Each entry below gives the row player's payoff followed by the column player's payoff:

For the row player, switching from $C$ to $D$ improves the payoff from $3$ to $5$ against $C$, and from $0$ to $1$ against $D$. The column player has the same comparisons, so $D$ is a strictly dominant strategy for each. Even against a mixed strategy choosing $C$ with probability $q$, the expected payoff advantage is $2q+(1-q)=1+q>0$. Thus cooperation cannot appear with positive probability in a best response, and $(D,D)$ is the unique one-shot Nash equilibrium, including mixed strategies.

At $(D,D)$, a unilateral deviation to $C$ lowers the deviator's payoff from $1$ to $0$. A joint change to $(C,C)$ instead raises both payoffs from $1$ to $3$: mutual cooperation Pareto dominates mutual defection, making the Nash equilibrium fail Pareto efficiency. A Nash equilibrium tests unilateral deviations, so it need not maximize shared gains. This is the useful lesson: improving the available outcome may require changing incentives, rather than telling each participant to choose a better best response.

The additional inequality $2R\ge T+S$ says that mutual cooperation gives each player at least as much average payoff as equal-time alternation between the two asymmetric outcomes. It is a further assumption, not a consequence of $T>R>P>S$. Here it holds strictly: $6>5$, and alternating exploitation gives each player average payoff $5/2<3$. This comparison concerns equal-time averages, not every discounted sequence. See Kuhn, section 3, on the additional cardinal-payoff condition.

A unilateral deviation changes only one player's strategy, holding the other players' strategies fixed. It is profitable if it strictly increases that player's payoff, or expected payoff for mixed strategies. The comparison does not allow the opponents to adjust in response to the deviation. In a sequential game, however, holding an opponent's complete pure strategy fixed still allows their prescribed actions to differ after different observed histories.