Why is it optimal to always defect in the iterated prisoner's dilemma with a known finite number of iterations? From the Prisoner's dilemma Wikipedia page: "If the game is played exactly N times and both players know this, then it is optimal to defect in all rounds." and "For cooperation to emerge between game theoretic rational players, the total number of rounds N must be unknown to the players."
From The Evolution of Cooperation by Robert Axelrod: "If the game is played a known finite number of times, the players still have no incentive to cooperate."
I am having trouble understanding the claim that always defecting is optimal. I understand that the best strategy against 'always defect' is to always defect as well and that this is a Nash equilibrium. I have also seen and understand the backward inductive proof. I can see this proof being valid for proving the Nash equilibrium part, but not for the optimal part. I can give a simple example.
Player x applies the tit for tat strategy.Player y applies the tit for tat strategy as well. Both do not care about N. Player z applies the claimed optimal strategy of always defecting.
Now if we set N equal to 1, then obviously player z's strategy is optimal. No matter what, their strategy is always preferred.
But if we set N equal to, for example, 100 this is no longer the case as player x will perform better against player y than player z would. Doesn't this show that always defecting is suboptimal?
 A: I think the claims you cited are making the implicit assumption that all agents involved in the game are implementing Causal Decision Theory. Under this assumption, the inductive argument proves that the other agent will defect in all cases, and then it becomes optimal for the first agent to defect as well.
Once you drop this assumption, the claim is no longer true. In fact, non-CDT agents may cooperate even when $N=1$. You may be interested in this paper. Tit for Tat is an example of a strategy incompatible with Causal Decision Theory.
A: This theorem assumes both players play optimal (i.e. get the best possible result for themselves). Both tit for tat players are not optimal, since they can improve their own score by always defecting on the last round. This theorem can therefore not be applied to this situation.
It is true if you have this population, tit for tat players might get a higher score on average, as can be seen in your example, if there exist players which also sometimes cooperate, if the strategy is the same for a player throughout its whole life. However if one tit for tat player decides to change its strategy and defect on every last round instead, this player will get an even higher score. The other tit for tat player will know this (since he knows the opponent will play optimal) and will therefore also change its strategy and defect in the last round. This will lead to iteratively changing the strategies to defect on one more round. In the end we reach the same strategy: always defecting
