So here's the original question:
$n$ players enter a room and a red or blue hat is placed on each person’s head. The color of each hat is determined by [an independent] coin toss. No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats [but not their own], the players must simultaneously guess the color of their own hats or pass. The puzzle is to find a group strategy that maximizes the probability that at least one person guesses correctly and no-one guesses incorrectly.
Now the solution for $n=3$, I could somehow solve (winning probability = 0.75):
Say Red/Blue if you see 2 Blue/Red (in that order), pass otherwise
Although I do not know how to generalize this result to $n$, my issue with any arbitrary $n$ is this.
Looking at it from the perspective of say any one player Bob. Bob has 2 jobs, decide whether he will pass or not based on what the others' hats are, and the guess (again based on the hats of the rest).
Both of these steps require basing his decision on the hats of the rest. So why does any decision based on the information from independent variables affect the chances of his being right when he does guess?
Irrespective of what the other hats are, from Bob's point of view, whenever he does choose to guess (based on information which is in no way related to the color of his own hat), it will always be a 50% chance that he is right. And this reasoning can be extended to every other player, so that from their own perspective, all the guesses are in fact only 50% valid. How does the winning probability be any higher than 0.5?
Isn't basing your prediction about the next outcome on a sequence of IID previous outcomes very similar to the Gambler's fallacy? I am significantly confused, since I did solve the $n=3$ case somehow.
PS: I hear that this is somewhat a famous puzzle. Any link to a solution for the general case?