The optimal strategy to maximize the odds is this:
-Spoiler-
There has to be at least one guy that sees 2 hats of the same colour. This guy has to speak and say that his hat is a different color. This maximizes the odds of winning over 50%.
Yes, and when each player says a colour, they're still only right 50% of the time. But the strategy means the cases where you'd be wrong are the same cases that others would also vote incorrectly, but the cases where you're right, the others pass, so you win.
MMM O O O Lose
MMO Pass Pass O Win
MOM Pass O Pass Win
MOO M Pass Pass Win
OMM O Pass Pass Win
OMO Pass M Pass Win
OOM Pass Pass M Win
OOO M M M Lose
I had read the solution but it still didn't make sense in my head how seeing the colour of unrelated hats could somehow improve the chance of you predicting your own hat colour. But as you explain, no one individual guess is better than 50%, it's just that the wrong guesses get lumped together.
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[ 2.7 ms ] story [ 21.1 ms ] threadThere has to be at least one guy that sees 2 hats of the same colour. This guy has to speak and say that his hat is a different color. This maximizes the odds of winning over 50%.
See below chart -- The probability that my hat is M/O is the same (50%) regardless of what the others are wearing.
Others Me
MM M
MM O
MO M
MO O
OM M
OM O
OO M
OO O
I had read the solution but it still didn't make sense in my head how seeing the colour of unrelated hats could somehow improve the chance of you predicting your own hat colour. But as you explain, no one individual guess is better than 50%, it's just that the wrong guesses get lumped together.