Life In 19x19

All players follow the same strategy. Their guess is determined by the 3 colors they see.

-> guess
-> no guess
-> guess
-> guess

They will win a bottle with probability 68.75%. Proof:

-> lose
-> win (4 permutations)
-> win (6 permutations)
-> lose (4 permutations)
-> win

They lose 5/16 of the time and win 11/16 of the time, Q.E.D.



More detailed explanations about the cases where they win:

All hats: . One player sees and guesses correctly . The other 3 players see and guess nothing.

All hats: . Two players see and guess nothing. The other two see and guess correctly .

All hats: . All players see and guess correctly .



I don't know if this is the optimal strategy. I also tried to include randomness in the strategy (guess white with probability p1, black with probability p2, nothing with probability p3) and to use different strategies for each player. However, I couldn't improve from 11/16.

strategy : who sees 3 of same color will guess right away.and say opposite color.
when no one i is guessing it means there are two color of each. so after while you know it is 2:2. so you know what color you have.

only time you will get it wrong it when it is all same color which is 2/16
14/16 you will be correct.

The fact they are bridge players is a (minor) clue.

It is basically an information transfer problem. I recognized that early on, and concluded that since they were not allowed to communicate, the problem was mis-stated.
However, jlaire demonstrated that there is more than meets the eye. The pre-arranged agreement is a code, and the stones that they see are the key. Different patterns of stones mean different keys, which effectively allows them to communicate.

jlaire made one slight oversight. He assumed that the
key = the stones that they see.
Whereas, actually,
key = the stones that they see AND where they see them.
This allows more keys, and, therefore, the effective transfer of more information.

. <> <>

This, BTW, explains what Drmwc meant by

drmwc wrote:...The fact they are bridge players is a (minor) clue.

Bridge players think in terms of position. A bid by the opponent to your left is not the same as the same bid made by your opponent to the right.

I think that Jlaire's 4 possible distributions have to be 5. When it is 2+2, there are two possible positional options: adjacent or opposite.


(4 permutations)
(6 2 permutations)
(6 4 permutations)
(4 permutations)

Instead of merely four instructions, like this:

-> guess
-> no guess
-> guess
-> guess

There should be eight like this:

-> guess
-> no guess
-> no guess
-> guess
-> guess
-> no guess
-> guess
-> guess

( everything to the right of the arrows is just a guess about guessing )

Nominate someone to be dummy.

Ignore one player altogether (he doesn't guess, and no one is interested in his hat

The remaining three will do as follows:
- if you see two hats of different color, don't guess
- if you see two hats of same color, guess the other color

In the spirit of the puzzle, an amendment:

Ignore one player altogether (he doesn't guess, and no one is interested in his hat During deliberation, before hat distribution, shoot one player.

I interpreted it to mean that the 4 players separate into partnerships, each with its own strategy. This gives two 12/16 solutions, which are completely different from tj86430's. In particular, nobody is left out - everyone has a potential to make a guess.

The fact that there are several 12/16 solutions suggests that it could be possible to do even better... I suppose the next step is to prove that that is impossible.