Talk:Iterated prisoners dilemma

I know the thing: if there is 1 round left, there is no reason to cooperate. If there are 2 rounds left, ditto. If 3 rounds, ditto, etc. So why should it matter whether you know the number of rounds or not? You know that, whatever the number is, cooperation does not pay, so why does it matter whether or not you know the number? THAT is the paradox!

It's not really a paradox. If I know this is the last round, then I ought to defect. If I know this is the second-to-last round, then I don't necessarily know I ought to defect. It depends on how smart I think the other guy is. Will he be smart enough to figure out that he ought to defect on the last move no matter what? If he's smart, then I should defect now. On the other hand, if he's planning to cooperate on the last round, then maybe I shouldn't defect now and make him mad. It might be better for me to cooperate now and defect on the next turn.

The Nash equilibrium assumes both players are smart, so yes, I should defect now. But if we don't know when the game will end, then my opponent is no longer "smart" in this sense, and I know he isn't, so I probably should cooperate.

There are other games that have even stranger results. For example, you can set up a game of Chicken where a powerful computer is playing against a simple finite state machine. If you set it up right, then the Nash equilibrium is for the less intelligent machine to win every time.