One Game or Many: Why the Prisoner's Dilemma Does Not Forbid Cooperation — Epoche B2
The common belief The prisoner's dilemma is usually told as a story. Two suspects are interrogated in separate rooms; each can stay silent (cooperate with the partner) or confess (defect). Each fares best by confessing while the other stays silent, and worst by staying silent while the other confesses. Stripped to essentials, each of two players chooses C or D, and the four possible outcomes pay off as follows, with Axelrod's standard numbers alongside the traditional letters: Opponent plays C Opponent plays D You play C $R = 3$ (reward for mutual cooperation) $S = 0$ (sucker's payoff) You play D $T = 5$ (temptation to exploit) $P = 1$ (punishment for mutual defection) The defining ordering is $T > R > P > S$: exploiting a cooperator beats mutual cooperation, which beats mutual defection, which beats being exploited. (A second condition, $2R > T + S$, is standard too: it ensures that taking turns exploiting each other, worth $(T+S)/2 = 2.5$ per round on average, pays less than steady cooperation at $R = 3$.) Many people learn this game and conclude that rational, self-interested agents simply cannot cooperate. In the one-shot game this reading is correct, and it is worth seeing exactly why. Check defection against each thing the opponent might do. If the opponent cooperates, defecting yields $T = 5$ against cooperation's $R = 3$. If the opponent defects, defecting yields $P = 1$ against cooperation's $S = 0$. Defection pays strictly more in both cases — the technical phrase is that it strictly dominates cooperation — so a self-interested player defects without needing to predict the opponent at all. A Nash equilibrium is a pair of choices such that neither player can improve their own payoff by changing alone; $(D, D)$ is one, since a lone switch to C drops a player from $P$ to $S$, and it is the only one, since in any outcome containing a cooperator that cooperator gains by switching. The bite of the dilemma is that both players, reasoning impeccably, land on $(D, D)$ with payoff $P = 1$ each, though $(C, C)$ would give each $R = 3$. The mistake is to treat this special case as a general law. Why a known ending changes nothing A natural repair — "just play the game many times" — fails if the number of rounds is fixed and known, and the reason is instructive. Suppose both players know the game lasts exactly 100 rounds. In round 100 there is no future to protect, so the round is an ordinary one-shot dilemma and both defect. But then round 99 cannot influence anything either — round 100's play is already settled — so round 99 is effectively one-shot too, and both defect. The argument marches backwards through every round; this style of reasoning from the last move to the first is called backward induction , and it unravels cooperation completely (Luce & Raiffa, 1957). What matters, then, is not repetition as such but the absence of a known final round. Contrast: one shot versus indefinite repetition So compare the same game repeated indefinitely : after each round, the game continues with probability $\delta \in (0,1)$. Equivalently, players discount future payoffs by the factor $\delta$ — a payoff expected next round is worth $\delta$ times a payoff today, either because next round only happens with probability $\delta$ or because later gains are simply valued less; the algebra is identical. Since there is never a round known to be last, backward induction finds no foothold, and the strategic landscape changes completely. Consider the grim trigger strategy: cooperate until the opponent defects once, then defect forever. Suppose your opponent plays it, and weigh your two futures. Cooperating forever earns $R$ every round. Summing with discounting requires one first-year tool, the geometric series: if $V = R + \delta R + \delta^2 R + \dots$, then $\delta V$ is the same sum shifted one term, and subtracting gives $V(1 - \delta) = R$, so $$V_{\text{cooperate}} = \frac{R}{1-\delta}.$$ Now defect instead. Today you collect the temptation payoff $T$. But from tomorrow on, the grim opponent defects unconditionally forever, and against permanent defection your own best reply is to defect too (each round, $P > S$). So the deviation is worth $T$ today plus $P$ in every later round: $$V_{\text{defect}} = T + \delta P + \delta^2 P + \dots = T + \frac{\delta P}{1-\delta}.$$ (A standard result of dynamic programming, the one-deviation principle, guarantees that checking this single deviation is enough: if no one-round deviation pays, no elaborate multi-round scheme pays either, because any such scheme would have to contain a first profitable step.) Cooperation is a best response whenever $V_{\text{cooperate}} \ge V_{\text{defect}}$. Multiply both sides by the positive quantity $(1-\delta)$: $$R \;\ge\; T(1-\delta) + \delta P \quad\Longleftrightarrow\quad \delta (T - P) \;\ge\; T - R \quad\Longleftrightarrow\quad \delta \;\ge\; \frac{T-R}{T-P}.$$ The threshold has a readable anatomy: the numerator $T - R$ is what defection gains today, the denominator $T - P$ is what the defector loses per round once cooperation collapses from the exploiting payoff to the punishment payoff, and $\delta$ says how much that future loss weighs. With Axelrod's payoffs the threshold is $\delta \ge (5-3)/(5-1) = 2/4 = 1/2$: if the odds of meeting again are better than even, mutual grim trigger is a Nash equilibrium and permanent cooperation sustains itself. A modest shadow of the future suffices. The folk theorem and its honest caveat Grim trigger is one instance of a sweeping result, called the folk theorem because game theorists knew it informally long before anyone published a proof (Friedman, 1971, gave an early formal version; Fudenberg and Maskin, 1986, the definitive one). Two terms make it stateable. A payoff level is feasible if some pattern of play achieves it on average. A player's minmax payoff is the worst the opponents can force on her when she responds as well as she can — in the prisoner's dilemma this is $P$,