Why is a raven like a writing-desk? Part II
Yesterday, I wrote about the Hat Problem. I figured out some more late last night, but haven't had a chance to post until now. I cancelled my brute-force program because I figured out the solution to the 4-person version: have 1 person pass all the time, and have the other 3 people pretend it's the 3-person problem. I know this because I got an upper bound on the best strategy: with N people, you can never do better than winning N out of N+1 times (so the solutions based on Hamming codes are always optimal). If you add up all the non-pass responses over all possible hat configurations, you need an equal number of correct and incorrect answers (otherwise, the chances that your own hat is red are not independent of the colors of everyone else's hat, and probability is broken).
The best you can bunch up the wrong answers is to have everyone guess wrong at the same time, and the best you can spread out the right answers is to have each right answer in its own hat configuration. So, for every time the group loses, you can have at most N wins, and the best chances you can ever get are N in N+1.
How does this help in the N=4 situation? We now know that best you can do is win 80% of the time. Moreover, there are 16 distinct hat configurations. Winning 13 of them would be more than 80%, so we must win fewer than that many. However, we can win 12 of them by reducing ourselves to the N=3 version, so we know that's the optimal solution. For N=5, there are 32 distinct hat combinations, and the optimal strategy will win at least 24 and at most 26 of them (but I don't know the exact answer). For N=6, the optimal strategy will win at least 48 and at most 54 of the 64 possible configurations. The brute-force method won't finish within our lifetimes for N>4, so I can't try that again. Any other ideas on how to progress? FWIW, I've found a second way to get 75% success when N=6.
The best you can bunch up the wrong answers is to have everyone guess wrong at the same time, and the best you can spread out the right answers is to have each right answer in its own hat configuration. So, for every time the group loses, you can have at most N wins, and the best chances you can ever get are N in N+1.
How does this help in the N=4 situation? We now know that best you can do is win 80% of the time. Moreover, there are 16 distinct hat configurations. Winning 13 of them would be more than 80%, so we must win fewer than that many. However, we can win 12 of them by reducing ourselves to the N=3 version, so we know that's the optimal solution. For N=5, there are 32 distinct hat combinations, and the optimal strategy will win at least 24 and at most 26 of them (but I don't know the exact answer). For N=6, the optimal strategy will win at least 48 and at most 54 of the 64 possible configurations. The brute-force method won't finish within our lifetimes for N>4, so I can't try that again. Any other ideas on how to progress? FWIW, I've found a second way to get 75% success when N=6.