Me too. I was thinking that 51 answers that agree with the majority is enough, but we need 51 answers that are accurate to their own hat. In a 51-49 split, the 51 folks who see 49-50 are those who would get the right answer under your strategy, but not necessarily the modified. The 49 folks who see the 51-49 split as 51-48 are *still* getting the wrong answer under the simple or modified strategy (Hence Dan's idea about switching). Since 51-49 or 49-51 is about twice as likely as 50-50, it seems tricky to make a rule that gives 50-50 a chance of success without dropping the success of 49-51 below 50%. On Mon, Jun 19, 2017 at 10:56 AM, James Propp <jamespropp@gmail.com> wrote:
Why strike it? It convinced me! What's the fallacy?
Jim
On Monday, June 19, 2017, James Davis <lorentztrans@gmail.com> wrote:
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 <javascript:;>> 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.
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