It isn't making a move that I suggested was equivalent to making a binding commitment. (In this case, it's working out one's best strategy, in the presence of a perfect predictor.) It's equivalent in the sense that both have the effect of narrowing the options the other player thinks you might take. That's not a good notion of equivalence in all contexts, but I think it is here; the impact on the game is the same.
Yes, there are situations in which UDT-as-understood-by-the-OP produces predictable results. That doesn't mean that UDT (as understood etc.) Is consistently predictable, and it remains the case that the OP explicitly characterized the UDT-using agent as a superhumanly effective predictor.
I don't know enough math and I don't know if this is important, but in the hopes that it helps someone figure something out that they otherwise might not, I'm posting it.
In Soares & Fallenstein (2015), the authors describe the following problem:
More precisely: two agents A and B must choose integers m and n with 0 ≤ m, n ≤ 10, and if m + n ≤ 10, then A receives a payoff of m dollars and B receives a payoff of n dollars, and if m + n > 10, then each agent receives a payoff of zero dollars. B has perfect predictive accuracy and A knows that B has perfect predictive accuracy.
Consider a variant of the aforementioned decision problem in which the same two agents A and B must choose integers m and n with 0 ≤ m, n ≤ 3; if m + n ≤ 3, then {A, B} receives a payoff of {m, n} dollars; if m + n > 3, then {A, B} receives a payoff of zero dollars. This variant is similar to a variant of the Prisoner's Dilemma with a slightly modified payoff matrix:
Likewise, A reasons as follows:
And B:
I figure it's good to have multiple takes on a problem if possible, and that this particular take might be especially valuable, what with all of the attention that seems to get put on the Prisoner's Dilemma and its variants.