I think an algorithm for N outcomes is: spin twice, gain 1 every time you get the answer right but lose 1 if both guesses are the same.
One can “see intuitively” why it works: when we increase the spinner-probability of outcome i by a small delta (imagining that all other probabilities stay fixed, and not worrying about the fact that our sum of probabilities is now 1 + delta) then the spinner-probability of getting the same outcome twice goes up by 2 x delta x p[i]. However, on each spin we get the right answer delta x q[i] more of the time, where q[i] is the true probability of outcome i. Since we’re spinning twice we get the right answer 2 x delta x q[i] more often. These cancel out if and only if p[i] = q[i]. [Obviously some work would need to be done to turn that into a proof...]
Just to be clear: if you spin twice and both come up right, you’re gaining 2 and then losing 1? (I.e., this is equivalent to what you wrote in an earlier version of the comment?)
I think an algorithm for N outcomes is: spin twice, gain 1 every time you get the answer right but lose 1 if both guesses are the same.
One can “see intuitively” why it works: when we increase the spinner-probability of outcome i by a small delta (imagining that all other probabilities stay fixed, and not worrying about the fact that our sum of probabilities is now 1 + delta) then the spinner-probability of getting the same outcome twice goes up by 2 x delta x p[i]. However, on each spin we get the right answer delta x q[i] more of the time, where q[i] is the true probability of outcome i. Since we’re spinning twice we get the right answer 2 x delta x q[i] more often. These cancel out if and only if p[i] = q[i]. [Obviously some work would need to be done to turn that into a proof...]
Just to be clear: if you spin twice and both come up right, you’re gaining 2 and then losing 1? (I.e., this is equivalent to what you wrote in an earlier version of the comment?)
That’s right.