Prisoner's dilemma ($T>R>P>S$)

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 social game in which cooperation costs its actor $c$ and benefits its partner by $b$, with $b>c>0$. Mutual cooperation yields $b-c$ to each, unilateral defection yields $b$ to the defector and $-c$ to the cooperator, and mutual defection yields zero.

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.

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.

A reactive strategy cooperates with probability $p$ after the opponent cooperated and probability $q$ after the opponent defected. It ignores its own previous action. Independent behavioural randomization gives a four-state joint-action Markov chain. The reduced marginal recurrences depend on the prescribed initial action distribution.

Always defect chooses defection independently of the opponent's actions.

Always cooperate chooses cooperation independently of the opponent's actions.

Tit for tat cooperates initially and then repeats the opponent's previous action. Two such players cooperate forever without mistakes under this initial convention. It retaliates against defection and returns to cooperation after the opponent cooperates.

In a symmetric two-action stage game write $R,P,T_{\mathrm{tempt}},S$ for mutual cooperation, mutual defection, sole defection and sole cooperation payoffs. Suppose $T_{\mathrm{tempt}}\ge R$, $P\le R$ and $T_{\mathrm{tempt}}+S\le 2R$. Against an initially cooperative Tit for tat, every action sequence has limiting upper average payoff at most $R$.

For a deviator's actions $a_t$, put $a_{-1}=C$, $h(C)=0$ and $h(D)=T_{\mathrm{tempt}}-R$. The opponent at time $t$ plays $a_{t-1}$. Check the four transitions: $C$ to $C$ and $C$ to $D$ satisfywith equality; $D$ to $C$ uses $S\le 2R-T_{\mathrm{tempt}}$, and $D$ to $D$ uses $P\le R$. A telescoping sum gives the displayed finite-horizon bound. Divide by the horizon and let it increase. Mutual Tit for tat yields $R$ every round, so it is a Nash equilibrium under the long-run average payoff criterion. The argument is pathwise and therefore covers arbitrary deviations representable by finite-state machines, as well as action sequences outside that class.

For a reactive resident $(p,q)$ with $|p-q|<1$, let $z=q/(1-p+q)$ and $g(z)=R{z}^2+(S+T)z(1-z)+P(1-z{)}^2$. Its self-payoff and both ordered cross-payoffs against Tit for tat equal $g(z)$. At introduced Tit for tat frequency $ϵ$, its fitness difference is therefore the displayed expression. If $2R\ge S+T$, it is positive for $z<1$ because $R-g(z)=(1-z)[R-P+(R+P-S-T)z]$. Fully cooperative reached histories tie; the deterministic opposite-response strategy needs a separate cycle calculation. Initial advantage is order frequency rather than a positive invasion exponent at zero.

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.

Under cooperative initial memories, matches involving only Tit for tat and always cooperate contain only cooperation, so all four payoffs equal $R$. The two-condition criterion for evolutionary stability has both differences zero. Thus Tit for tat is not an evolutionarily stable strategy against a strategy space containing unconditional cooperation, even if it is a Nash equilibrium. This counterexample is valid for finite horizons and discounting as well as infinite averages.

With initially independent actions, same-round independence is preserved by reactive updates: each next action uses the other independent previous action and fresh independent randomization. Marginal cooperation follows $x^{\prime }=q_1+(p_1-q_1)y$, $y^{\prime }=q_2+(p_2-q_2)x$. If the product of response slopes has modulus below one, these recurrences converge and the joint distribution is their product. Averaging $R,S,T,P$ over that distribution gives the long-time payoffs. At deterministic response slopes of modulus one, initial-memory-specific cycles must instead be averaged directly.