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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 $3$.

In a finite perfect-information game without chance moves, solve terminal decision points first. At each earlier decision point, its player chooses an action maximizing its own payoff among the already solved continuations. Record a choice in every subgame, including those not reached by the root choice. This constructs a subgame perfect equilibrium. Tied choices may yield several equilibrium strategies and several outcomes. The name refers to Ernst Zermelo; the computational procedure is an application of backward induction.

A binary game is a finite normal-form game in which each player's payoff at every pure strategy profile is either zero or one. This describes the numerical payoffs, not the number of players or available actions. A mixed strategy can have an intermediate expected payoff. If a weighted average of binary payoffs is at least one, every positive-weight term must equal one; this observation makes pure-by-mixed domination unusually simple in binary games.

A finite game represented succinctly by propositional logic. Each player controls a disjoint collection of Boolean variables and chooses their truth values. Together those choices determine a Boolean valuation. A player's goal is a propositional formula, with payoff one if satisfied and zero otherwise. The control sets normally form a set partition of all variables.

A pure strategy strategy profile is a Nash equilibrium if every unsatisfied player remains unsatisfied after every change to that player's controlled variables. Checking a candidate can therefore require solving a Boolean satisfiability problem, even though evaluating its current payoffs is easy.

Given a succinct Boolean game and a complete pure strategy strategy profile, determine whether it is a Nash equilibrium. The problem is co-NP-complete for goal propositional formulas, even with one player. This differs from existential equilibrium checking in a Boolean game, which searches for some qualifying Nash equilibrium.

A negative instance has a short complexity certificate: a currently unsatisfied player and new truth values for that player's controlled Boolean variables that make the goal true. Evaluating the current and changed propositional formulas takes polynomial time, proving membership in co-NP.

For hardness, given a propositional formula $F(X)$, let one player control $X$ and a fresh variable $z$, with goal $z\wedge F(X)$. The all-false candidate has payoff zero. It is a Nash equilibrium exactly when $F$ is an unsatisfiable Boolean formula, since any satisfying assignment permits an improvement by also setting $z$ true. This is a polynomial-time many-one reduction from unsatisfiability. The fresh guard ensures that the candidate loses independently of the input formula's value there.

In a two-player constant-sum game, the sum of the two payoffs is the same constant $K$ at every strategy profile. Replacing each player's payoff by $u_i-K/2$ gives a zero-sum game without changing best responses or Nash equilibria. If both players can guarantee $K/2$, then any Nash equilibrium gives each that value. An equilibrium in pure strategies must use pure strategies with that guarantee: otherwise the opponent could improve on $K/2$ by a unilateral deviation.

Let a proposer divide a pie of size $M>0$ by continuously offering its responder $q\in [0,M]$. Rejection gives fixed continuation payoffs $b$ to the proposer and $c$ to the responder, with $0\le c<M$ and $b<M-c$. Then a subgame perfect equilibrium must have the responder accept $q=c$ with probability one. The proposer offers exactly $c$, retaining $M-c$.

Sequential rationality forces rejection for $q<c$ and acceptance for $q>c$. If acceptance at $c$ had probability $a<1$, that offer would give the proposer $a(M-c)+(1-a)b<M-c$. Every larger offer gives less than $M-c$ and can be improved by a smaller offer still above $c$; every smaller offer gives $b<M-c$. The supremum $M-c$ would therefore be unattained. Randomizing offers cannot attain it either: the nonnegative gap from $M-c$ would be strictly positive almost surely and hence have positive expectation. No best response would exist, contradicting equilibrium. Acceptance at $c$ removes this gap and makes $q=c$ the unique optimal proposal. The assumptions $c<M$ and $b<M-c$ exclude degenerate indifference of the proposer.

For a monotone simple cooperative game with empty value $0$ and grand-coalition value $1$, the core of a cooperative game consists exactly of the nonnegative unit-total payoff vectors supported on veto players.

For the proof, singleton core conditions give $x_i\ge 0$. Every winning coalition must receive at least $1$, the full available total, so all players outside it receive zero. A positively paid player therefore belongs to every winning coalition and is a veto player. Conversely, a unit-total nonnegative allocation supported on veto players gives every winning coalition exactly $1$ and every losing coalition at least $0$. These are precisely the core inequalities. The core is thus nonempty if and only if a veto player exists.

A discrete location game can model firms choosing sites from $\{1,\dots ,k\}$ while customers at those sites choose a nearest firm. The customer masses, distance rule, permission to share a location and rule for ties are part of the model. For two firms, customer value two per site and equal splitting of ties, the payoff to a firm at $x$ against one at $y$ isFor $x<y$, counting sites strictly left of the midpoint and assigning half of an equidistant site's value gives $x+y-1$, whether $x+y$ is even or odd. The other firm receives the remainder, so this is a constant-sum game with total $2k$. Changing tie priority can fundamentally change its Nash equilibria, especially when co-location is permitted.

A value assigning a vector of payoffs $\psi (v)$ to a transferable utility game has cooperative-game efficiency when it distributes exactly the grand-coalition value. This is a budget-total condition; it does not assert membership in the core of a cooperative game.

Write column $j$ of a two-row binary game as $((a_j,c_j),(b_j,d_j){)}^T$, with the row player's payoff first. If there is no pure Nash equilibrium, the column player's maximum payoff in each row must be one. Otherwise every column would be a best response in a zero-payoff row, and excluding an equilibrium there would force the opposite row to dominate in every column, giving an equilibrium in that opposite row.

A column maximizing the column player's top-row payoff must then have $(a_j,b_j)=(0,1)$, or its top cell would be an equilibrium. Its bottom column-player payoff must be zero, or its bottom cell would be one. Thus a column $((0,1),(1,0){)}^T$ occurs. The same argument for the bottom row produces $((1,0),(0,1){)}^T$. No column can give the column player payoffs $(1,1)$, since a row-player best response within that column would be a pure equilibrium.

Consequently, if neither pure equilibrium is found, mix equally over the two rows and equally over those two special columns. The row player's two expected payoffs are $1/2$; every column gives the column player at most $1/2$, and the supported columns achieve $1/2$. This is a Nash equilibrium. Computing the two row maxima, scanning for pure equilibria and locating the two special columns takes $O(n)$ running time for $n$ columns.

Evolutionary game theory studies how a population's composition changes when the success of a pure strategy depends on which other pure strategies it encounters. The central feedback is a frequency-dependent payoff: becoming common changes the opponents a strategy meets, which can change whether it does well. In a coordination game, common behaviour can reinforce itself; in the coexistence example below, being rare is an advantage. This makes evolutionary game theory useful for asking which behaviours can persist and which initial compositions lead to different outcomes, without assuming that each participant solves an optimization problem.

In the basic model, a large, well-mixed population has finitely many inherited pure strategies; interactions sample opponents according to their population frequencies. Assume that a strategy's expected payoff determines its per capita net reproductive growth rate, with a common baseline allowed. These are modelling assumptions: reproduction must transmit the strategy, encounters must match the proposed sampling rule, and the conversion from payoff to growth must be justified for an application. The replicator equation below describes the resulting continuous-time frequencies. Taylor and Jonker's ^[ref] explicitly derive this growth model.

For learned behaviour, reproduction can instead be replaced by an assumed copying rule: an individual samples another individual uniformly and copies a better-performing pure strategy at a rate proportional to the positive payoff difference. At frequencies $x_i,x_j$, the net flow from $j$ to $i$ is $x_i{x}_j(f_i-f_j)$, because the two directional positive parts subtract to $f_i-f_j$. Summing over $j$ gives $x_i(f_i-\bar{f})$, the replicator equation. This derivation states a particular mechanism; an arbitrary rule for learning or imitation need not give the same dynamical system.

Three questions must be distinguished. A symmetric Nash equilibrium tests whether an individual can improve its expected payoff against the resident composition. An evolutionarily stable strategy tests whether each distinct rare mutant is outperformed after it is introduced. Dynamical stability tests the response of an actual dynamical system to small changes of composition, and asymptotic stability additionally requires nearby trajectories to converge. For a finite symmetric bimatrix game under the continuous-time replicator equation, every evolutionarily stable strategy is a symmetric Nash equilibrium and a locally asymptotically stable rest point; the converse from asymptotic stability to an evolutionarily stable strategy fails in general. See Ross Cressman and Yi Tao, The replicator equation and other game dynamics (2014), Theorems 1–2 and Example 1. These statements refer to stability on the full probability simplex, including perturbations introducing currently absent pure strategies.

A concrete experiment makes the frequency feedback visible. In 2009, Gore, Youk and van Oudenaarden competed two yeast strains: a producer of the enzyme invertase, which enables sucrose hydrolysis, and a mutant lacking the corresponding SUC2 gene. Released sugars can feed neighbours, so avoiding production can be advantageous. They used shaken 5 mL cultures with 5% sucrose at $30{}^{\circ }\mathrm{C}$, diluted daily after 23 hours of growth. Different labelled proteins allowed the two strains to be counted separately by flow cytometry; selective plating periodically checked the counts. The cooperator required externally supplied histidine, so restricting that amino acid imposed an adjustable growth disadvantage. See Snowdrift game dynamics and facultative cheating in yeast, pp. 253 and 255–256, Methods Summary.

Starting with rare producers and, separately, rare nonproducers tested both invasion directions. Under the reported coexistence conditions each rare strain increased; changing histidine or added glucose shifted the eventual composition. Their explanation used nonlinear growth benefits, rather than a fixed pairwise payoff matrix. This motivates testing both rarity regimes before inferring an evolutionarily stable strategy or fitting a replicator equation; it does not identify the illustrative tables below with measured payoffs. See Gore, Youk and van Oudenaarden, Figures 1–3.

An early motivation was understanding why animal contests can involve restraint even when escalation might win an individual encounter. Maynard Smith and Price's ^[ref] introduced the evolutionarily stable strategy in this setting. Taylor and Jonker's ^[ref] then studied explicit continuous and discrete dynamics. Static resistance to invasion and the paths populations follow are related questions, but they require separate definitions.

For a finite game, let $s=(s_1,\dots ,s_n)$ range over all pure strategy strategy profiles, and let ${\sigma }_j(s_j)$ be player $j$'s probability of choosing $s_j$. Under independent mixed strategies, the probability of $s$ is ${\prod }_j{\sigma }_j(s_j)$. The displayed sum is the expected value of player $i$'s payoff. Holding everyone else's mixed strategy fixed makes $U_i$ a linear function of ${\sigma }_i$; this is why checking pure strategy deviations suffices to check all mixed strategy deviations.

A game described by a game tree, with the acting player and available actions specified at each decision node, payoffs at terminal nodes, and information sets indicating which decision nodes a player can distinguish. A pure strategy specifies an action at every information set belonging to that player, including ones not reached during realized play.

The convex hull of the stage game's attainable payoff vectors. Public randomization or deterministic sequences with prescribed limiting frequencies produce such average payoffs. It can be larger than the set attainable using independent mixed strategies in a single stage.

For $k\ge 3$, suppose two distinct occupied sites are $\ell <r$, a final firm must choose a different site, customers have value two per site and nearest-firm ties are split equally. A final firm left of both incumbents maximizes its payoff uniquely at $\ell -1$, when that site exists, obtaining $2(\ell -1)$. To the right it maximizes uniquely at $r+1$, obtaining $2(k-r)$. At every available interior site it obtains the same payoff $r-\ell$. Thus all best responses are obtained by comparing these three values, omitting regions with no available site and retaining every tied maximizing choice.

More generally, three distinct locations $a<b<c$ give their left, middle and right firms payoffs $a+b-1$, $c-a$ and $2k-b-c+1$. Counting the two midpoint boundaries proves these formulas; the middle firm's demand is the difference between the cumulative demands at the two boundaries. The formulas retain all even/odd midpoint ties automatically.

In an $L$-round iterated prisoner's dilemma, Tit for tat against always defect gets $S$ once and $P$ thereafter; the defector gets $T$ once and $P$ thereafter. Their self-payoffs are $R,P$. At introduced frequency $ϵ$, subtracting mixed-population payoffs gives $ϵ[(R-P)-(T+S-2P)/L]-(P-S)/L$. The displayed positive threshold applies when its denominator is positive and the threshold is less than one. Taking $L\to \infty$ first removes this finite entry cost, so the horizon and rare-mutant limits cannot be casually interchanged.

A family of results describing the large set of Nash equilibrium payoffs available in indefinitely repeated interaction. For a finite game with observed past play, the discounted Nash version supports any feasible, strictly individually rational payoff in a repeated game when players are sufficiently patient.^[ref]

The mechanism is to prescribe a path and punish the first deviator by holding that player's future payoff to its minmax payoff. With long-run average payoffs, a finite gain disappears from the average; with discounting, the promised future loss must outweigh the gain. Weak boundary points require care when a stated theorem assumes strict inequalities. A construction proving a Nash equilibrium need not prove a subgame perfect equilibrium, because the punishers' continuation incentives may fail.

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.

A grim trigger is a history-dependent pure strategy in a repeated game that switches permanently to punishment after a trigger event. In the iterated prisoner's dilemma above, the trigger is any player's observed defection, and punishment is always defect. Unlike Tit for tat, grim trigger never restores cooperation after the opponent cooperates again. Whether mutual grim trigger is a subgame perfect equilibrium depends on the stage payoffs, discount factor, and monitoring assumptions; it is not an unconditional property of the rule. See Ozdaglar, lecture 15, slides 12–15.