Puzzle time - prisoners and hats
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I think I now remember where I saw the puzzle.
There's a mathematician who publishes such puzzles on a regular basis, Peter Winkler. He has a monthly column in the "Communications of the ACM", of which I'm a subscriber. I believe I saw that puzzle there. If I think about the puzzle more, I may remember parts of the solution, so I guess I better just keep my mouth shut.
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I guess the set of rules would attempt to clump wrong guesses together while right guesses would tend to be alone. The rules would also want to ensure that at least one person guessed.
@Horace said in Puzzle time - prisoners and hats:
I guess the set of rules would attempt to clump wrong guesses together while right guesses would tend to be alone. The rules would also want to ensure that at least one person guessed.
So, is the answer that simple? Those who see two different colors would not vote. If the other two people are same, guess opposite? This will cause any configuration of colors other than all-same to be a win (one correct vote), while all-same would be a loss (3 wrong votes).
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Yes that is the % of my proposed solution. I got there by figuring that each guess would be 50/50 so the rules would have to clump wrong guesses together and single out right ones. One such set of rules leaves 75% of color combinations yielding one right guess.
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It gets interesting if you generalize the number of prisoners and/or the number of colors.
