Philosophy questions - Morality & Moral Theories

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I haven't read the other threads on this, but I'll go further than both of you and state that cooperating in the prisoner's dilemma does require either altruism or irrationality.

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I think it's rational self-interest.

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If it's rational self-interest, then both players defect (in the one-shot game), because they both have a dominant strategy.

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I am talking about an iterated prisoner's dilemma...

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There are some other assumptions that have to be made in the finitely repeated prisoner's dilemma (most importantly, common knowledge of rationality), but including all the assumptions leads to a unique equilibrium where players defect on all rounds.

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How do you figure? If both players are rational, a tit-for-tat strategy (or slightly modified), ensures cooperation on all rounds. That is the rational strategy.

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Well, if we make the assumptions:
1) the game is finite
2) both players know that the other player is rational (CKR)
3) both players have perfect recall
4) both players are capable of calculating as many steps as the game has iterations
5) both players have perfect information, i.e., they know how many rounds the game will last, etc.

Then the unique equilibrium is the backward induction one where both players defect on all rounds. E.g., since both players will defect on the last round, they know that the other player will defect on the next to last round, so they'll defect on the next to last round, but the other player knows that ... and so on.

Now, I think this result is fairly absurd, but then one of the assumptions has to go. So which one is it?

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I think #5 is incorrect, they don't know when the game is going to end. However, even if they did, if they were both rational, then they would cooperate every round. This is especially true if they will be playing multiple games with multiple other people, all of whom know how each other have played in previous games. The best strategy is a tit-for-tat, and knowing this, it is best to cooperate. A constant defector may fair a smidgeon better than the tit-for-tat player in one game (since on the first round, the defector will gain more than the cooperator), but when playing multiple games, the defector will end up costing himself a lot -- as everyone else will soon be defecting against him, but cooperating with those that are cooperating -- thus, the cooperators will be gaining more than the defector.

Anyway, even in a single game, the defector will know that the tit-for-tat player will defect on every round after the first if the defector defects on the first round. However, if he cooperates on the first round, then he will gain more on subsequent rounds, and thus will maximize his gain over the course of the game. This is the most rational play that both can make.