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Represent satisfied and unsatisfied goals by payoffs one and zero. The second player's goal is an exclusive or: it is true precisely when $q\ne r$, independently of $p$. The first goal is false only when all three Boolean variables are false. The resulting payoff matrix is below; each boxed entry marks that player's best response to the opposing pure strategy, including ties:For player 1, choosing $r$ satisfies the goal against every opposing pure strategy. Choosing $\bar{r}$ does equally well except against $\bar{p}\bar{q}$, where it does worse. Thus $r$ is player 1's unique dominant strategy in the weak sense, and there is no strictly dominant strategy. The ties are why most first-player entries in both rows are boxed.

For player 2, either pure strategy with $q=0$ is a best response against $r$, while either with $q=1$ is a best response against $\bar{r}$. No one choice of $q$ works against both rows, so player 2 has no dominant strategy, and there is no dominant-strategy equilibrium. The value of $p$ does not affect player 2's own payoff.

The cells with both entries boxed give exactly four pure strategy Nash equilibria:All four give payoffs $(1,1)$. In particular, the weakly dominated choice $\bar{r}$ can occur in a Nash equilibrium because player 1 is already satisfied there and cannot strictly improve.

Against Tit for tat, let $a_t$ be an arbitrary deviator's action at round $t$, and set $a_{-1}=C$ to encode the opponent's initial cooperation. The opponent then plays $a_{t-1}$. Define $h(C)=0$ and $h(D)=3$. The four possible transitions giveThus every stage payoff is at most $3+h(a_t)-h(a_{t-1})$. The telescoping sum yields the uniform payoff bound against Tit for tat:After division by $T$, no long-run average payoff can exceed three. The bound applies to every action sequence, so in particular it covers every finite-state machine representing a unilateral deviation. Mutual Tit for tat produces $C,C$ in every round and gives each player three. Neither player can improve, proving $(\mathrm{TFT},\mathrm{TFT})$ is a Nash equilibrium.

Another pair is $(\mathrm{TFT},\text{grim trigger})$. Here grim trigger initially cooperates and continues cooperating until the opponent first defects, after which it defects forever. Both strategies are two-state finite-state machines: Tit for tat records the opponent's previous action, while grim trigger records whether defection has ever occurred. Their prescribed play again consists entirely of cooperation and gives $(3,3)$.

The player using grim trigger faces Tit for tat, so every unilateral deviation is bounded by the same telescoping sum and cannot improve on three. The player using Tit for tat faces grim trigger. If that player never defects, the payoff remains three. If the first defection occurs at a finite round $\tau$, it gives a one-time payoff six; thereafter the opponent always defects, against which each subsequent payoff is at most one. Any finite initial prefix has zero contribution to the long-run average payoff, so such a unilateral deviation has limiting upper average at most one. It cannot improve on three either. Both players are therefore best responses, proving the second claimed Nash equilibrium.

For deterministic finite-state machines, the joint state path is eventually periodic, so the displayed averages have a limit. The first bound also controls the limiting upper average for arbitrary action sequences, making the no-improvement argument stronger than an enumeration of finite-state deviations. Notice that equality in the bound permits alternating exploitation to average three: equilibrium requires absence of a strict improvement, not uniqueness of a best response.

Read the lower-left entry of each cell as player 1's payoff and the upper-right entry as player 2's payoff, as indicated by the source diagram. Against $L$, player 1 strictly prefers $B$ because $2\pi >1$; against $R$, player 1 strictly prefers $T$ because $3>2$. Against $T$, player 2 strictly prefers $R$ because $6>0$; against $B$, player 2 strictly prefers $L$ because $2\sqrt{2}>2$. Thus the pure strategy Nash equilibria are

For a fully mixed strategy Nash equilibrium, write $p=\mathrm{Pr}(T)$ and $q=\mathrm{Pr}(L)$. Player 1's expected payoffs from the two pure strategies areEquality, required when both pure strategies have positive probability, givesPlayer 2's corresponding expected payoffs areso equality givesBoth probabilities lie strictly between zero and one. Hence the unique fully mixed strategy Nash equilibrium iswith complementary probabilities on $B$ and $R$. Its expected payoffs are

These are all three Nash equilibria. There are no equilibria with just one player genuinely mixing: against each opposing pure strategy, the other player's best response is unique, so that player cannot mix. When both players mix, the two linear indifference equations have the unique solution just obtained. This is support enumeration for a bimatrix game with all possible supports checked.

Measure all payoffs in millions of dollars, in the order (Ann, Bob). Since both players maximize expected money, those numbers are already their expected utility scale.

In the extensive-form game, Ann moves first. Staying out terminates at $(2,6)$. Entering leads to Bob's observed decision between fighting, with terminal payoffs $(1,1)$, and acquiescing, with terminal payoffs $(3,3)$. Each decision node is its own information set, so this is a perfect-information game.

The corresponding normal-form game gives Ann the pure strategies $O$ (stay out) and $E$ (enter), and Bob the complete contingent pure strategies $F$ (fight if entry occurs) and $A$ (acquiesce if entry occurs). The payoff matrix isThe pure strategy Nash equilibria are $(O,F)$ and $(E,A)$. Against $F$, Ann prefers $O$; against $A$, Ann prefers $E$. When Ann stays out, Bob's contingent response does not affect his payoff, so $F$ is a best response there.

There is exactly one subgame perfect equilibrium, $(E,A)$. In the subgame following entry, Bob strictly prefers $3$ to $1$, so backward induction selects acquiescence; Ann then strictly prefers entering for $3$ to staying out for $2$. The threat supporting $(O,F)$ is not optimal in that one-shot subgame.

For infinite repetition, the paper specifies no discount factor. First take the long-run average payoff convention, with all past actions observed. The feasible payoff set is the convex hullAnn can guarantee $2$ in every stage by staying out; Bob can guarantee at least $3$ in every stage by acquiescing whenever entry occurs. These guarantees are also their minmax payoffs: fighting holds Ann's best possible stage payoff to $2$, and entering holds Bob's best possible stage payoff to $3$.

Therefore the sustainable Nash equilibrium average payoffs are exactly the feasible, weakly individually rational payoffs in a repeated game:This is the closed triangle with vertices $(2,3),(2,6),(3,3)$.

Here is a direct sufficiency argument for this particular game. Choose a publicly known sequence of stage outcomes whose frequencies give the target point. Before any observed deviation, follow that sequence. After an Ann deviation, Bob fights forever, so Ann can thereafter obtain at most $2$ per stage. After a Bob deviation, Ann enters forever, so Bob can thereafter obtain at most $3$ per stage. A finite prefix has no effect on long-run average payoff, so no unilateral deviation improves a target with $a\ge 2,b\ge 3$. Rational frequencies can be implemented periodically; arbitrary real frequencies can be implemented by a deterministic sequence with convergent frequencies. Thus even the weak boundary is covered here. In particular, staying out forever and entry followed by acquiescence forever are sustainable; fighting forever is not.

For discounted payoffs $(1-\delta ){\sum }_{t\ge 0}{\delta }^t{u}_i^t$, the folk theorem for repeated games instead gives the usual patience qualification: every feasible point with $a>2,b>3$ can be sustained for all sufficiently large $\delta <1$. It does not classify the whole region for an unspecified fixed $\delta$. The two one-shot Nash equilibrium outcomes remain sustainable for every $\delta$.^[ref]

Finally consider the particular history-dependent pair in the paper. Initially Bob has never acquiesced, so Ann stays out. That preserves the same condition forever, giving payoffs $(2,6)$. The pair is a Nash equilibrium. Holding Bob's strategy fixed, Ann cannot induce an acquiescence: he fights whenever she enters, and his rule never changes unless he has previously acquiesced. Every entry would give her $1$ instead of $2$, so no unilateral deviation helps. Holding Ann's strategy fixed, Bob never reaches a decision node because she stays out; changing his strategy alone cannot create a history in which he actually acquiesces. Hence he also cannot improve.

This direct argument works for both discounted and long-run average payoffs. The outcome gives Ann exactly her minmax payoff and Bob more than his, so it is compatible with the folk theorem for repeated games. A version requiring strict gains above both minmax payoffs is a sufficient condition, and cannot be used to rule out this boundary example.

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 pure strategy is a definite, complete plan for a player's choices. In a game with one simultaneous choice, it selects one available action with certainty: heads or tails in matching pennies, for example. In a sequential game it specifies an action at every situation where that player could be called upon to move, including situations never reached when the plan is followed. The plan can use only information available to the player: indistinguishable decision situations must prescribe the same action. An action at one such situation is only part of the plan.

If $S_i$ is player $i$'s set of pure strategies, the payoff $u_i(s_i,s_{-i})$ depends on the whole strategy profile. Thus the same pure strategy can be a best response against one opposing choice and perform poorly against another. A dominant strategy is the stronger case in which it is a best response against every opposing strategy profile.

With values $v_i>0$ and effort costs $b_i^2$, the unique pure Nash equilibrium has the efforts displayed above. Interior first-order conditions give $v_i{b}_j/(b_1+b_2{)}^2=2b_i$, hence $b_1/b_2=\sqrt{v_1/v_2}$ and $(b_1+b_2{)}^2=\sqrt{v_1{v}_2}/2$. Against a positive rival effort, each payoff is a strictly concave function, so these conditions identify global best responses. On the axes, lowering a positive uncontested effort or adding a sufficiently small effort at the zero tie rules out equilibrium.

Assume $2\le k<m$ and ${\bigcup }_i{S}_i=S$. Construct a two-player game with row $i$ for $S_i$ and columns $0,1,\dots ,n$. Column $0$ pays both players one; column $j>0$ pays $(1,0)$ when $j\in S_i$, and $(0,k/(k-1))$ otherwise. There is a Nash equilibrium whose row player's support of a mixed strategy has size $k$ exactly when $k$ rows cover the universe. A uniform mixture on a cover makes column $0$ a best response. Conversely, if some element is uncovered by the support, the column player's best responses are uncovered elements and the row player's current payoff is zero. The union promise provides a profitable row deviation. Without the union promise the equivalence is false: a globally uncovered element is a best response against every row mixture.

// Target: mathematical-optimization.bigb

A signalling equilibrium is a stable pattern of signal production and response in a game where one participant has information another lacks. A sender's strategy selects a signal according to its type; the receiver's strategy chooses a response after observing that signal. Equilibrium requires both choices to be best responses under consistent beliefs or a specified evolutionary stability criterion.

A separating pattern uses different signals for different types, while a pooling pattern uses the same signal. A costly signal can support separation only if each type prefers its assigned signal and response to mimicking another type. In evolutionary game theory, stability concerns resistance to invading strategies; it is a stronger, separately specified condition than merely observing a correlation between signals and types.

A sequential model in game theory in which a leader commits to an action before a follower observes it and chooses a best response. The leader anticipates the follower's response when optimizing. With private follower information, the leader maximizes its expected payoff over follower types. The resulting solution is a Stackelberg equilibrium.

A solution of Stackelberg competition in which the follower chooses a best response after each observed leader action and the leader optimizes given those responses. Indifference may require a specified response or leader-selection convention; it cannot be silently treated as a strict inequality. In a sequential private-value all-pay contest, the leader's objective is $vF(b)-b$.

The stag hunt is a coordination game with a mutually beneficial risky action and a safer action. A representative two-player payoff matrix isBoth mutual stag and mutual hare are Nash equilibria: neither player can improve by deviating alone. Mutual stag improves both players' payoffs relative to mutual hare, but choosing stag against an uncertain partner can be risky. If the partner chooses stag with probability $p$, stag yields $4p$ while hare yields three, so stag is a best response exactly when $p\ge 3/4$. The example separates mutual gain from assurance that the other player will cooperate.

The support of a finite mixed strategy consists of the pure strategies assigned positive probability. At a Nash equilibrium, every action in the support is a best response. In a game satisfying nondegeneracy of a bimatrix game, the two equilibrium supports have equal size.

The support of a mixed strategy consists of the pure strategies used with positive probability. At a Nash equilibrium, every supported pure strategy is a best response, while unused strategies may give no higher payoff.

// Target: computer-science.bigb

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.

A strategy is weakly dominated if another gives at least as much payoff against every opponent strategy and strictly more against at least one. Unlike a strictly dominated strategy, it may remain a best response to opponents who never choose a strategy where the comparison is strict. Thus weak domination does not exclude equilibrium strategies.