Instead choose a score S=(X-Y)/N, where X+Y=N, after N bits and X matching. This is just a linear scaling, but moving the target to zero seems to make analysis easier. For example, it allows us to write: S=N/M*(X-Y)/N + (Z1+Z2+Z3+...)/M, with additional choices Zm=+/-1 after N up to M. The first term is always zero in the large M limit, so in some sense, initial values never really matter. — Brad
On Oct 4, 2019, at 1:18 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Oct 4, 2019 at 1:39 PM James Propp <jamespropp@gmail.com> wrote:
I don’t see why this will work. If there’s a tie for success-rate-furthest-from-0.5, moving one of them toward .5 might move the other farther away.
Yes. 'and if the first one moves A closer and B further away, then B will be further, and the next one will move B closer. Suppose it moves A further away, which feels like the worst case. Then we're back in a tie where we started. Repeat this for two million more guesses, and each of A and B has a million more right and a million more wrong, and both of their averages are now way closer to .5 than they used to be. On the other hand, if sometimes we make a guess that moves both A and B closer to .5, that seems even better, and like it will make both ratios converge to .5 faster.
This isn't a formal proof, but do you see the intuition here?