The Prisoner's Dilemma, and Axelrod's Tournament

Set the four payoffs. The tool finds the best reply to each of the other player's moves, then runs a round robin of eight strategies and shows what each earns per round as the match gets longer.

Worked example — The Prisoner's Dilemma, and Axelrod's Tournament

The game	Prisoner's dilemma
If they cooperate	Defect
If they defect	Defect
Better against both of their moves	Defect
What the equilibrium pays each	1
The condition for repeating	Yes
Strategy	Tit for Two Tats
Per round	2.564

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — The Prisoner's Dilemma, and Axelrod's Tournament

A game is a prisoner's dilemma when its four payoffs satisfy two inequalities. T > R > P > S makes defecting the better reply to either move, so both players defect and both do worse than if both had cooperated. And 2R > T + S makes mutual cooperation better for the pair than taking turns exploiting each other — which is what makes the repeated game worth playing rather than a race to alternate.

What each symbol means

T: the temptation: what you get for defecting on a cooperator
R: the reward: what you both get for cooperating
P: the punishment: what you both get for defecting
S: the sucker: what you get for cooperating with a defector

The winner of the tournament never won a match 🖖

Look at the last column. Always defect wins six of its eight matches at every round count from 1 to 200, and from the seventh round onwards it finishes last. Tit for Two Tats wins none of its eight at any round count, and over the 200 rounds Axelrod used it finishes first. Watch how that happens. Tit for Two Tats waits for a second defection before it retaliates, so always defect takes five points from it twice, and then spends 198 rounds collecting one point from a player who has stopped cooperating: 5 + 5 + 198 = 208, against 0 + 0 + 198 = 198. It wins that match by ten and it does not matter, because the table pays a total and not a margin. While always defect was earning 1.04 points a round there, the five strategies that open by cooperating were banking three a round off each other. Robert Axelrod, whose 1980 tournament gave this its design and its payoffs, wrote the conclusion as advice. Don't be envious. You are not playing the other player, you are playing the field.

Four rounds is the whole difference 🖖

Drag the rounds slider down to one. Always defect tops the table on 4.000 points a round and every strategy that opens by cooperating sits beneath it. Two rounds: still first, on 3.250. Three: 2.833, still first. At four it drops to third and it never leads again. Put the slider back at two hundred and the dashed vertical line in the chart marks the first round at which the lead changed hands. It changes hands eleven more times after round twenty, the last at round 48, and Tit for Two Tats holds it from there. Sweep every set of whole-number payoffs from 0 to 10 that makes a proper dilemma, 250 of them, and a nice strategy takes the lead in 237: usually by round 4, at the latest by round 20, and in 13 of them never within 200 rounds. The game itself has not changed. Every round of it still has the same dominant move and the same equilibrium. What the fourth round buys is a bill, because a defection now costs you something later, and four rounds is long enough for it to arrive.

Two slips in a hundred do almost all the damage that twenty would 🖖

With the noise off, 62.2% of all the moves played are cooperation. Move the slider to 2%, so two moves in a hundred come out as the opposite of what the strategy intended, and cooperation falls to 50.2%. Then keep going: 50.6% at 4%, 50.2% at 10%, 50.0% at 20%. Ten times the error rate, and nothing further is lost. Whatever two percent breaks stays broken. Grudger shows the breaking at its most brutal, because it never forgives: one defection by its opponent, anywhere in 200 rounds, ends cooperation in that match for good, and it falls from second place to fifth. Generous Tit for Tat, which forgives one defection in three, climbs to second. With the noise off it is only third, which is what forgiveness costs when nobody needs it.

Problem solved in full

How long does the relationship have to be? The tool plays a fixed number of rounds, and none of its eight strategies plays any differently as the end approaches. A real repeated game ends at a moment nobody can name, so the question is not how many rounds there are but how likely another one is.

Open this problem in the tool
1. Call w the chance that the match continues after this round. Against Grudger, which punishes one defection for ever, cooperating for ever pays R each round, and that geometric series sums to R over 1 minus w.
  
  $R + w R + w^{2} R + \dots = \frac{R}{1 - w}$
2. Defecting once pays T this round and P in every round after it, because Grudger never comes back.
  
  $T + w P + w^{2} P + \dots = T + \frac{w P}{1 - w}$
3. Cooperation holds when the first is at least the second. Multiply through by 1 minus w and the series disappears, leaving one inequality in w.
  
  $\frac{R}{1 - w} \geq T + \frac{w P}{1 - w} ⟺ R \geq T - w (T - P)$
4. With Axelrod's payoffs that is two over four. The match needs an even chance of continuing or better, which is an expected length of two rounds.
  
  $w \geq \frac{T - R}{T - P} = \frac{5 - 3}{5 - 1} = \frac{2}{4} = \frac{1}{2}, \frac{1}{1 - w} = 2$
5. Now the same deviation against Tit for Tat, which punishes for one round and then forgives: the deviator gets T, then S, then R again for ever. The threshold rises to two thirds, an expected length of three rounds.
  
  $R (1 + w) \geq T + w S ⟺ w \geq \frac{T - R}{R - S} = \frac{5 - 3}{3 - 0} = \frac{2}{3}, \frac{1}{1 - w} = 3$
6. Here the more forgiving punisher needs the longer relationship, and the reason is on the line above: T − P is the larger gap. It is not a general rule. Over the 250 whole-number dilemmas these four inputs allow it holds in 50 and reverses in 130 — set T to 4 and P to 2 and Tit for Tat needs only a third.
  
  $T - P = 4 > R - S = 3 ⟹ \frac{T - R}{T - P} < \frac{T - R}{R - S}$
Answer

Against a punisher that never forgives, cooperation is stable once w reaches (T − R)/(T − P) = 1/2, an expected two rounds. Against one that forgives after a single round it takes (T − R)/(R − S) = 2/3, an expected three. The tool's own crossover is 4, and the gap between two and four is the field: the threshold asks when cooperating beats defecting between two players, while the table asks when a cooperator's average across eight opponents beats a defector's, and three of those eight defect first. Two rounds buys a stable pair. Four buys a league.

$w^{*} = \frac{T - R}{T - P} = \frac{1}{2} vs \frac{T - R}{R - S} = \frac{2}{3}$

Learning path

The best reply, and what happens when nobody computes one

References (4)

The tournament this tool reproduces, reported by the man who ran it — the source for the 1980 date, for the fourteen entries plus RANDOM, and for the nice/not-nice split: R. Axelrod, "Effective Choice in the Prisoner's Dilemma." Journal of Conflict Resolution 24(1), 1980, pp. 3–25.
"Don't be envious", and the argument that a strategy losing every match can still win the tournament: R. Axelrod, The Evolution of Cooperation. Basic Books, New York, 1984.
The 1950 RAND experiment the game came out of, written up by one of the two men who invented it. Albert Tucker gave it the prisoners and the name; this is the experiment: M. M. Flood, "Some Experimental Games." Management Science 5(1), 1958, pp. 5–26.
Why the winner changes when moves come out wrong, and where the one-in-three forgiveness rate this tool gives Generous Tit for Tat comes from: P. Molander, "The Optimal Level of Generosity in a Selfish, Uncertain Environment." Journal of Conflict Resolution 29(4), 1985, pp. 611–618.

By the 42Math team · Reviewed against published sources. Corrections welcome.