Each player has additional information about how the other player has played in the past. I wasn't trying to say that iterated PD for N=100 rounds becomes N=0, I was saying it becomes N=99, followed by one straight game.
And that is inaccurate, because your decision in round 99 may affect Clippy's decision in round 100. There's no rule anywhere that says Clippy isn't, for example, assuming that your decision making processes are similar, and that if it decided to cooperate the last round after 99 identical turns, there'd be a good chance that you'd cooperate as well because of that. Sure, that's not a very likely scenario, and obviously you should always defect the last round—but this shows why N=100 does not ever become N=99.
(I'm not very familiar with EDT, but it seems like a decision theory that would be prone to not defecting the last round following identical 99 rounds).
And my decision in round 99 is also part if Iterated Prisoners Dilemma. My decision in round 100 is no longer iterated PD, but normal PD with additional information about how my partner played a IPD.
Key feature of PD as opposed to IPD: In Prisoner's Dilemma, you will never, ever, interact with the other player again. If that's a possibility, then you are playing a game with many similarities but a different premise.
Today's post, The Truly Iterated Prisoner's Dilemma was originally published on 04 September 2008. A summary (taken from the LW wiki):
Discuss the post here (rather than in the comments to the original post).
This post is part of the Rerunning the Sequences series, where we'll be going through Eliezer Yudkowsky's old posts in order so that people who are interested can (re-)read and discuss them. The previous post was The True Prisoner's Dilemma, and you can use the sequence_reruns tag or rss feed to follow the rest of the series.
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