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Let p_n be the probability that I say n. Then the probability I escape on exactly the nth round is at most p_n/2 since the coin has to come up on the correct side, and then I have to say n. In fact the probability is normally less than that since there is a possibility that I have already escaped. So the probability I escape is at most the sum over n of p_n/2. Since p_n is a probability distribution it sums to 1, so this if at most 1/2. I'll escape with probability less than this is I have any two p_n nonzero. So the optimal strategies are precisely to always say the same number, and this can be any number.
I got the same answer, with essentially the same reasoning.
Assuming that each guess is a draw from the same probability distribution over positive integers, the expected number of correct guesses is 0.5 if I keep guessing forever (rather than leaving after 1 correct guess), regardless of what distribution I choose.
So the probability of getting at least one correct guess (which is the win condition) is capped at 0.5. And the only way to hit that maximum is by removing all the scenarios where I guess correctly more than once, so that all of the expected value comes from the scenarios where I guess correctly exactly once.