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?