Guys, consider this solution:

Let the players be A, B and C. They make the following strategy.

           2R      1R1B      2B
A       Red       Pass      Blue
B       Pass      Blue      Red
C       Blue      Red       Pass

That is, if A sees two red hats (on B and C), he says red. If C sees a red and a blue hat, he says red, and so on.

Now consider the following cases:

1) 3 red hats (A-red, B-red, C-red)
Here, according to the table, A will say red, B will pass and C will say blue. Since A’s answer is correct, they will win.

2) 2 red hats, one blue hat
(a) A-blue, B-red, C-red
A will say red, B will say blue and C will say red. Since C’s answer is correct, they will win.
(b) A-red, B-blue, C-red
A and B will pass, while C will say red. Since C’s answer is correct, they will win once again.
(c) A-red, B-red, C-blue
A will pass, while both B and C will say blue. Since C’s answer is correct, they will win the game.

3) 1 red hat, 2 blue hats
(a) A-red, B-blue, C-blue
A and B will both say blue, while C will say red. Since B’s answer is correct, they will win.
(b) A-blue, B-red, C-blue
A passes, B says red and C also says red. Since B gave the correct answer, they win.
(c) A-blue, B-blue, C-red
A passes, B says blue and C passes. Since B is correct, they win.

4) 3 blue hats (A-blue, B-blue, C-blue)
In this case, A will say blue, B will say red while C will pass. Since A gave the correct answer, they win.

Thus this strategy provides for a 100% hit rate, since they win in every possible case. But every answer to this question, here as well as elsewhere, gives a 75% accuracy. Could someone please point out an error in this strategy?