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.
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.