Oops, strike that first paragraph! I'm inclined to agree with Propp. On Mon, Jun 19, 2017 at 4:12 AM, James Davis <lorentztrans@gmail.com> wrote:
I believe that this is best possible, but I don't have even the beginnings of a proof.
Adding that a prisoner who sees 49-50 pick randomly gives ~46% chance of success on a 50-50 split, and only decreases the failure on a 49-51 overall split from 100% to ~100%, so I think the overall success is around 96% as the problem suggests.
It's interesting that the solution smacks of a (reverse) gambler's fallacy. Even more, if the problem slightly favored one color, you often put the prisoners in the paradoxical position of having to play an individually losing/irrational strategy that somehow works en masse.