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Iterated prisoner's dilemma with $n$ rounds, where both players know $n$, has a sub-game perfect equilibrium where both players always defect.

What if the number of rounds is not known though. Suppose that there is a random variable $N$, and we iterate prisoner's dilemma for $N$ rounds (both player's know $N$'s probability distribution, but not its value). How can we calculate the sub-game perfect equilibrium.

In particular, given a probability distribution on $\mathbb N$, how do we calculate the sub-game perfect equilibrium for iterated prisoner's dilemma, where the number of rounds is sampled from that distribution?


The obvious case is if a give number has a 100% chance of occurring, in which case the strategy is always defect.

As noted by mlc, both players always defecting will always be a subgame perfect equilibrium (since neither player can improve their outcomes by changing their strategy). So how do we calculate the other equilibrium?

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  • $\begingroup$ "Always defect" will always be an equilibrium. Perhaps you meant to ask whether there may be other subgame-perfect equilibria. $\endgroup$ – mlc Nov 9 '17 at 21:33
  • $\begingroup$ @mlc That is correct. $\endgroup$ – PyRulez Nov 9 '17 at 21:35
  • $\begingroup$ In one case, where there is a fixed probability of playing another round (conversely, some probability of the game ending) you can think of it like an infinite horizon game, where the probability the game ending is the discount factor (or added to the discount factor). Then you have trigger strategies of various types (cooperate until the other defects, for example). $\endgroup$ – Trurl Nov 9 '17 at 22:09
  • $\begingroup$ Axelrod showed that Tit For Tat is an equilibrium in this game provided that the probability of playing again is sufficiently high, in The Evolution of Cooperation $\endgroup$ – dbx Nov 10 '17 at 0:47

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