I've created this quiz, but I'm not sure if the answer that I've found is correct or not.
Three people meet at a pub, each of them has a blue or a red hat on his\her head. Nobody knows the colour of the hat on his head but he\she can see the other people's hat. Simultaneously each one of them guess the colour of the hat on his\her head. They win if and only if everybody answer correctly, and obviously they can't speak to each other. They know that the hats where assigned by doing a random permutation of 3 red hats and 3 blue hats (i-th hat to the i-th guy). Question: consider the case where everybody have a red hat, and provided that they will use the best strategy to win (and that they can count on the fact that everybody will use the best strategy), what will be their answer?
***** SOLUTION ALERT *****
This is my solution:
At a first glace you are probably tempted to say that they should all say "blue" since from the point of view of each of them the probability to have a blue hat is $3/4$. But what if you really have a blue hat on your head? If this is the case the others will see a red hat and a blue hat and, since they don't have any other clue, the best thing that they could do is to pick at random, and so they guess with probability $1/2$. So the probability to win if you say "blue" is just $\frac{1}{2}*\frac{1}{2}*\frac{3}{4} = \frac{3}{16}$ whether if you say "red" is $\frac{1}{4}$ because you can count that, if you have a red hat, the others will reason in the same way.
Suppose everybody plays your strategy: If you see two colours the same, you guess that colour; otherwise you guess at random. Then the probability of success is just the probability that all colours are the same, which is $\frac1{10}$.
Now suppose that everybody plays my strategy: If I see two colours the same, I guess the other colour; otherwise I guess at random. Then the probability of success is $\frac14 \times$ (the probability that all colours are not the same), which is $\frac14 \times \frac9{10} = \frac{9}{40}$. This is more than twice as good as $\frac1{10}$.