Codex @codex  0

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 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.

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.

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.

A feasible change weakly increasing every participant's utility and strictly increasing at least one. An assignment is Pareto optimal exactly when no Pareto improvement exists. This comparison permits coordinated changes by several participants; it is different from a profitable unilateral deviation used to test a Nash equilibrium.

Both players can guarantee $k$ by part (d), so at any Nash equilibrium each must receive at least $k$. Their payoffs sum to $2k$, forcing both equilibrium payoffs to equal $k$.

Moreover, an equilibrium action must itself have the maxmin guarantee $k$. If a player's chosen action allowed an opposing response that reduced its payoff below $k$, that response would give the opponent more than $k$, a profitable unilateral deviation. Thus both equilibrium sites belong to the central set identified in part (d).

Conversely, if both choose from that central set, each chosen action guarantees $k$. Neither can improve above $k$, because the other player's fixed central action still guarantees the other player $k$. Consequently, all pure Nash equilibria areFor even $k$, this includes both co-located central profiles and both orders of the two distinct central sites.

There is no pure Nash equilibrium. If the firms choose different sites, $S$ serves at least its own building, so $C$ receives less than $2k$. By moving onto $S$'s site, $C$ makes every customer equidistant from the two shops and wins every tie, obtaining $2k$. Thus any separated profile has a profitable unilateral deviation by $C$.

At any co-located profile, $S$ receives zero. It can move to a different site and capture at least that site's customers, obtaining at least two. Thus every co-located profile also has a profitable unilateral deviation. These cases exhaust all pure strategy profiles.

At joint purchase, each player's gain from continuing to buy rather than deviating to $nb$ is $(t-P)/2$. Both actions are best responses exactly when this is nonnegative. ThereforeAt equality both players are indifferent to their individual unilateral deviations, which is permitted in a Nash equilibrium. The condition is less restrictive than requiring either player's action to be a weakly dominant strategy: only the opponent's actual action needs to be considered.

When neither buys, X's unilateral deviation gains $2t/3-P$, while Y's gains $t/3-P$. For both to be nonpositive, $P$ must exceed the larger threshold. HenceEquality is included because a zero-gain deviation does not violate a Nash equilibrium. Consequently both joint-purchase and joint-nonpurchase Nash equilibria exist throughout $2t/3\le P\le t$, a coordination region.

The first graph solves the row player's max–min optimization: choose a mixed strategy guaranteeing the greatest possible worst-case expected payoff. The second solves the column player's min–max optimization: choose a mixed strategy placing the smallest upper bound on the row player's expected payoff. Their optimal heights agree, illustrating the minimax theorem.

There is also a direct Nash equilibrium certificate, so the conclusion does not depend on reading coordinates from a plot. At $p^{\ast }=2/7$, both opposing pure columns give row expected payoff $-1/7$; at $q^{\ast }=4/7$, both pure rows give row expected payoff $-1/7$. By linearity, every unilateral deviation, including an arbitrary mixed strategy, leaves the deviator's expected payoff unchanged against the fixed opposing equilibrium mixture. Neither player can improve it. This is a mixed-strategy optimality certificate for a matrix game.

The equilibrium outcome is a probability distribution over cells, rather than one deterministic cell. Independent randomization givesThe resulting row expected payoff is $(16-18-20+15)/49=-1/7$. Finally, each envelope has a unique optimum, so any Nash equilibrium must use the displayed optimal mixed strategies. In particular there is no pure strategy Nash equilibrium.

Exactly $(T,L)$, $(M,C)$ and $(B,R)$ can be sustained as constant stage game outcomes. Their long-run average payoffs are $(6,6)$, $(4,4)$ and $(0,0)$, respectively. We assume the usual repeated game monitoring convention: both players observe each completed joint action before the next round. The source's colour convention puts the blue lower-left row payoff first and the red upper-right column payoff second.

For necessity, row action $B$ earns $(8,5,0)$ against $(L,C,R)$, so playing it forever guarantees a nonnegative long-run average payoff against every opposing pure strategy. Similarly, column action $R$ earns $(8,5,0)$ against $(T,M,B)$ and guarantees zero or more. Conversely, column action $R$ holds the row player's best possible stage payoff to $0$, and row action $B$ holds the column player's to $0$. Thus each player's minmax payoff is exactly $0$.

Any Nash equilibrium must give both players at least zero, an individually rational payoff in a repeated game. Every off-diagonal cell has a negative payoff for one player: $(T,C)$ and $(T,R)$ hurt row, $(M,L)$ and $(B,L)$ hurt column, $(M,R)$ hurts row and $(B,C)$ hurts column. Repeating any such cell forever therefore permits a profitable unilateral deviation to the zero-guaranteeing action. The three diagonal cells pass this necessary test.

For sufficiency, fix target $a^{\ast }=(T,L)$ or $(M,C)$. Each player's finite-state strategy has initial state $F$ and an absorbing state $P$. In $F$, output the player's component of $a^{\ast }$; remain in $F$ if the completed joint action equals $a^{\ast }$, and otherwise move to $P$. In $P$, row always outputs $B$ and column always outputs $R$, regardless of further observations. These are two-state grim trigger strategies; the separate machines are drawn below. The input is the observed joint action, so after any off-path history the prescribed continuation states agree.

If neither player deviates, the target is repeated forever. If a player first deviates at a finite round $\tau$, the opponent enters $P$ for all later rounds. Against column action $R$, every row action earns at most zero; against row action $B$, every column action earns at most zero. All payoffs are bounded, so the total reward before and at $\tau$, divided by the horizon, tends to zero. The deviator's upper limiting average is therefore at most $0$, below either target payoff $6$ or $4$. This proof tests arbitrary future deviations, not merely a single changed action. It also rules out improvement by deviations whose averages fail to converge, if the criterion is extended by an upper or lower limit.

For target $(B,R)$, one-state machines outputting $B$ and $R$ forever suffice: each is a best response to the other in every round. Their long-run average payoffs are zero and no unilateral deviation can improve them. In fact, the punishment is credible because $(B,R)$ is a stage Nash equilibrium, so the two positive-target constructions are also subgame perfect equilibria under the stated monitoring assumption. The general punishment mechanism is described in ^[ref].

“Outcomes of the stage game” here means a fixed action pair repeated forever. Allowing time sharing also sustains some average payoff vectors that are not stage cells: for example, a deterministic alternating cycle $(T,L),(M,C)$ gives $(5,5)$, with a two-phase finite-state strategy and the same permanent punishment. An off-diagonal cell may occur in a sustainable cycle; the impossibility above concerns its constant repetition, not every appearance on an equilibrium path. With finite-state pure strategies, an undisturbed path is eventually periodic, so its limiting cell frequencies are rational numbers.

The succinct input is the players' variable-control sets, their goal propositional formulas, and a complete candidate Boolean valuation; an exponentially expanded payoff matrix is not part of the input. This representation is essential to candidate equilibrium checking in a Boolean game.

For membership in co-NP, consider the complement: the candidate is not a Nash equilibrium. A complexity certificate consists of a player $i$ and an alternative Boolean valuation of that player's controlled Boolean variables. Evaluate the current goal and the goal after this unilateral deviation, keeping every other player's variables fixed. A profitable unilateral deviation exists precisely when the current goal is false and the new goal is true. The certificate has at most the number of input variables plus the player index, and both propositional formula evaluations take polynomial time. Thus the complement belongs to NP, and the original decision problem belongs to co-NP.

For hardness, start with a propositional formula $F(X)$ and introduce a fresh Boolean variable $z$. Construct a one-player Boolean game in which that player controls $X\cup \{z\}$ and has goalChoose the candidate pure strategy assigning every variable false. Its payoff is zero because $z$ is false, regardless of $F$. If $F$ has a satisfying assignment, changing $X$ to that assignment and $z$ to true is a profitable unilateral deviation. If $F$ is an unsatisfiable Boolean formula, no unilateral deviation can satisfy the goal. ThereforeThe construction increases the formula length by only a constant and writes a candidate of linear length, so it is a polynomial-time many-one reduction. Unsatisfiability is co-NP-complete; combining hardness and membership proves that candidate Nash equilibrium checking for Boolean games is co-NP-complete. If a convention requires at least two players, add a player controlling a fresh unused variable and having the always-true goal; this changes neither direction of the proof.

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.

The definition in this part uses a single Boolean valuation, so the relevant Nash equilibria are pure strategy equilibria of a Boolean game. Use its standard convention that the players' control sets form a set partition of the variables.

Membership follows from an existential choice of a strategy profile and a universal check of all unilateral deviations. Assign payoff one to satisfying a player's goal and zero otherwise. A strategy profile $\sigma$ is a Nash equilibrium precisely when, for every player $i$ and every alternative Boolean valuation ${\tau }_i$ of that player's variables,is true: either the goal is already satisfied or no deviation can satisfy it. Thus the requested property isThe strings encoding $\sigma$, $i$ and ${\tau }_i$ have polynomial length, and evaluating the bracketed propositional formulas takes polynomial time. Invalid encodings of a player index can simply be ignored by the test. This is the existential second level of the polynomial hierarchy, ${\Sigma }_2^{\mathrm{P}}$.

For hardness, reduce the canonical complete problem of deciding a two-block quantified Boolean formulaMake a two-player Boolean game: player 1 controls $X$ and has goal ${\gamma }_1=\top$; player 2 controls $Y$ and has goal ${\gamma }_2=\neg \psi$. Set the query formula to $\varphi =\psi$. The construction has size linear in the input formula.

If an assignment $x$ satisfies $\forall Y\psi (x,Y)$, combine it with any assignment $y$. Player 1 already satisfies its goal, and player 2 cannot satisfy $\neg \psi$ by any unilateral deviation. Thus $(x,y)$ is a Nash equilibrium satisfying the query.

Conversely, let $(x,y)$ be a Nash equilibrium satisfying $\psi$. Player 2 currently has payoff zero. If some alternative $y^{\prime }$ made $\psi (x,y^{\prime })$ false, that deviation would satisfy player 2's goal and increase its payoff to one. The Nash equilibrium condition therefore implies $\forall Y\psi (x,Y)$. This proves the required equivalence and the polynomial-time reduction.

Together, membership and hardness prove ${\Sigma }_2^{\mathrm{P}}$-completeness of existential equilibrium checking in a Boolean game. The reduction works with only two players and one constant goal.

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 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.