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.
edit: and i see that someone already solved it
"The more we think we know about
The greater the unknown" Words by neil peart, music by geddy lee and alex lifeson
I think that the delayed guessing plans are no different from a secret signal and should be considered communication. In fact, everyone looking at their watches would become an unintended signal.
I think that I have a theoretical understanding of the problem:
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 means that, as a practical matter...
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About Drmwc's clue:
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.
That means that to improve on Jlaire's solution, you probably have to look here:
drmwc wrote:I believe that something like Joaz's solution can work (although I've not checked that solution in detail).
There is a simpler approach that gets 75%.
Big clue:
Nominate someone to be dummy.
Ok, got it now:
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
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.
Nicely done!
Turns out i misinterpreted the bridge hint after all.
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.
