Comment by furyofantares
12 hours ago
Failing to cooperate with literal clones of yourself in a prisoner's dilemma would be a spectacular failure. There's only two things that can happen with identical decision makers: they both cooperate or they both defect. So identical decision makers who know they're identical can cross off the asymmetrical entries in the payoff matrix and the decision to cooperate becomes trivial.
Ah, but what if one of your "clones" is actually the wicked and persuasive "All-Defector" in disguise? (No, really, I agree with your analysis but if you haven't read "The Quantum Thief" you might like it.)
no? you can choose a mixed strategy.
Sure, you can break symmetry (in this case making the decision makers not identical because they have different random number generators available), but the remaining symmetry means identical mixes must be chosen, and so a mixed strategy would only be chosen if it maximize his value for both people cooperatively.
Maybe it's a bit subtle that they said clones and I said identical decision makers; I'm letting you fill in the gap for how much clones may diverge and how much that matters.
Even if mixed strategies are allowed, I'm getting that it's still optimal to always cooperate as long as 2R>=S+T, which is usually assumed to be true (this condition also appears in iterated prisoner's dilemma, where it prevents alternating cooperation and defection giving a greater reward than mutual cooperation).