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The two gas stations from the dilemma face each other again tomorrow, and the day after, with no end date they can point to. Suddenly holding prices is not obviously foolish. A cut today buys one good morning and a long, cold price war after.

Repeated games and discounting

Played once, the Prisoner’s Dilemma ends in mutual defection: betrayal is dominant. Played over and over, the calculation changes. A defection can be punished tomorrow, and if tomorrow matters enough, the threat of that punishment holds cooperation in place. The turning point is a single number.

From earlier lessons

Two quick ones before we start.

In the one-shot Prisoner's Dilemma, defecting is...

Backward induction in a finite game starts from...

Predict, then play

Each round is the Prisoner’s Dilemma, the same game and the same payoffs as the gas stations: $T=5$ (defect on a cooperator), $R=3$ (both cooperate), $P=1$ (both defect), $S=0$ (cooperate against a defector). Only the ranking $T>R>P>S$ matters, not the units. The slider is $\delta$ , the chance the relationship continues to another round, which is also how much you value next round versus this one. Pick your strategy against a grim-trigger bot and run it.

Why did that happen?

Take a guess, then reveal the reason

Defecting still pays more this round: 5 beats 3. What changed is the shadow of the future. Against a grim-trigger opponent, one defection ends cooperation for good, trading a stream of 3s for a stream of 1s. When the future is heavy enough, that lost stream costs more than the one-time gain of 2, so you cooperate. When the future is light, the quick 2 wins and cooperation collapses. The whole thing turns on how much tomorrow is worth.

The math

Definition, discounted repeated game

The stage game is repeated, and a player values a payoff $t$ rounds from now at $\delta^t$ times its face value, where $\delta \in (0,1)$ is the discount factor. Equivalently, after each round the game continues with probability $\delta$ . A strategy is a rule that maps the whole history so far to a move.

Consider grim trigger: cooperate until the opponent defects once, then defect forever. Suppose both play it. Cooperating forever is worth

R + \delta R + \delta^2 R + \cdots = \frac{R}{1-\delta}.

Deviating means defecting once for $T$ , after which the opponent punishes and you both get $P$ forever:

T + \delta P + \delta^2 P + \cdots = T + \frac{\delta P}{1-\delta}.

Cooperation is a best response, so grim trigger is a subgame-perfect equilibrium, exactly when the first is at least the second:

\frac{R}{1-\delta} \ \ge\ T + \frac{\delta P}{1-\delta} \quad\Longleftrightarrow\quad \delta \ \ge\ \frac{T-R}{T-P}.

With $T=5, R=3, P=1$ the threshold is $\tfrac{5-3}{5-1} = \tfrac12$ . Value the future at least as much as the present-by-half, and cooperation stands; value it less, and it falls.

This checks the tempting one-shot deviation on the cooperative path. After any defection, grim trigger prescribes Defect forever; because $P>S$ , deviating back to Cooperate during punishment is not profitable. The strategy is therefore optimal after every history, which is the requirement for subgame perfection.

Work it out

Derive the threshold, then move the numbers

Set cooperating forever, $R/(1-\delta)$ , against defecting once and taking the punishment, $T + \delta P/(1-\delta)$ . Multiply through by $1-\delta$ :

R \ \ge\ T(1-\delta) + \delta P.

Rearranging, $R - T \ge -\delta(T - P)$ , so $\delta \ge (T-R)/(T-P)$ . For $T=5, R=3, P=1$ this is $\tfrac12$ .

Faded. Suppose mutual defection were less bad, $P=2$ . The threshold becomes $(5-3)/(5-2) = \tfrac23$ : a softer punishment means you need more patience to deter cheating.

Solo. At $\delta = 0.6$ , does tit-for-tat sustain cooperation? No, and the reason is worth seeing. Tit-for-tat punishes for one round, not forever, so a defector suffers $S$ once and then returns to $R$ . Comparing $R/(1-\delta)$ against $T + \delta S + \delta^2 R/(1-\delta)$ reduces to $R(1+\delta) \ge T + \delta S$ , giving $\delta \ge (T-R)/(R-S) = \tfrac23$ here. A milder punishment needs more patience, so $0.6$ is not enough.

Key idea

Repetition does not make cooperation automatic. It makes it possible, and only when players value the future enough: here, when the game continues with probability at least one-half.

Check yourself

Check yourselfapply

With T=5, R=3, P=1, the grim-trigger threshold is delta = 1/2. At delta = 0.4, what happens?

Check yourselfanalyse

If mutual defection were less painful (P rises from 1 to 2), what happens to the patience needed for cooperation?

Check yourselfbrings back: backward induction

Why can cooperation be sustained in the infinitely repeated PD but not in one repeated a known, finite number of times?

Sources

Friedman, J. W. (1971). A non-cooperative equilibrium for supergames. Review of Economic Studies, 38(1), 1-12.
Axelrod, R. (1984). The Evolution of Cooperation. Basic Books.