In a one-shot prisoner's dilemma, cheating is a dominant strategy. Yet firms in a cartel, countries in a trade agreement and neighbors often cooperate for years. Sequential Games and Backward Induction and Infinite Geometric Series explain when.
Key idea
In a game repeated with no known last round, cooperation can be an equilibrium if players value the future enough. With stage payoffs $T > R > P$ (temptation, reward for cooperating, punishment) and the grim trigger strategy (cooperate until the other cheats, then cheat forever), cooperation is sustainable when $$\frac{R}{1 - \delta} \ge T + \frac{\delta P}{1 - \delta}, \quad\text{that is,}\quad \delta \ge \frac{T - R}{T - P}.$$
Why it works
Cooperating forever pays $R + R\delta + R\delta^2 + \cdots = \dfrac{R}{1 - \delta}$. Cheating pays $T$ once, then the punishment $P$ in every later round: $T + \dfrac{\delta P}{1 - \delta}$. Multiply both by $1 - \delta$ and rearrange: $R \ge (1 - \delta)T + \delta P$, which gives the threshold. A high $\delta$ (patient players, or a high chance the game continues) makes future losses loom large. If the game has a known last round, cooperation unravels: in the last round cheating is best, so in the round before nothing can be gained by cooperating, and so on back to the start.
Worked example: Finding the threshold
$T = 10$, $R = 6$, $P = 2$. When can grim trigger sustain cooperation?
- $\delta^* = \dfrac{T - R}{T - P} = \dfrac{10 - 6}{10 - 2} = \dfrac{1}{2}$.
- At $\delta = 0.8$: cooperating is worth $\dfrac{6}{0.2} = 30$; cheating is worth $10 + \dfrac{2(0.8)}{0.2} = 18$. Cooperate.
- At $\delta = 0.4$: cooperating is worth 10; cheating is worth $10 + \dfrac{0.8}{0.6} \approx 11.33$. Cheat.
Answer: $\delta \ge 1/2$.
Worked example: A known last round
The same game is played exactly 5 times, and both players know it. Can cooperation last?
- Round 5 is a one-shot game: both cheat.
- In round 4, round 5's outcome is fixed whatever happens, so round 4 is effectively one-shot too: both cheat.
- The argument repeats back to round 1. With a known end, the subgame-perfect outcome is cheating every round.
Answer: No: it unravels from the end.
Common mistake
Thinking a big payoff from cooperating is enough. What matters is the comparison: the one-time gain from cheating, $T - R$, against the future loss from punishment, $R - P$ in every later round, weighted by $\delta$. Check: compute $\delta^*$ and compare it with the discount factor, not just $R$ with $P$.
The discount factor combines patience with the chance that the game continues: if each round is followed by another with probability 0.9 and money is discounted at 0.95 per round, $\delta = 0.9 \times 0.95 = 0.855$. Robert Axelrod's computer tournaments (1980s) found that simple reciprocating strategies such as tit-for-tat did well in repeated prisoner's dilemmas.
Experiment with strategies in Repeated Games and Cooperation.