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).