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 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.

That solution is:

-> loses 1 of 1
-> wins 3, loses 1
-> wins 2 of 2
-> wins 4 of 4
-> wins 3, loses 1
-> loses 1 of 1

...and gets 12 of 16.

Proving 75% is optimal sounds like quite an intersting problem. I have no idea how one could go about it other than brute force and ignorance.

The set of possible deterministic solutions is presumably finite, which makes brute force and ignorance plausible. However, random strategies could be used. I don't see off-hand how they can improve things, but they do have the annoying feature of making the solution space infinite, So some general nonsense result may be needed to cope with this.

Brute force and ignorance can get you an awfully long way in maths...

Any individual player, conditional on that player making a guess, can be correct at most 50% of the time. So, intuitively, how is it that we can do better than 50%?

Simple: imagine we've already decided on player 1's strategy and we're now trying to choose player 2's strategy. If we can arrange it so that player 2's wrong guesses coincide with player 1's wrong guesses, whereas player 2's correct guesses happen at times that player 1 wouldn't have attempted to guess, we can increase the number of outcomes where at least one correct guess happens, without increasing at all the number of outcomes where an incorrect guess happens. As long as we can do this efficiently enough with each subsequent player, it's not hard to see why in theory we might be able to beat 50%.

There are several other classic puzzles with the same concept as this one. In all of them, the key is to choose strategies for the players that cause wrong guesses to be as densely concentrated as possible and correct guesses to be as disjoint as possible. The "bridge" strategy is very effective at this - in the cases that anyone guesses wrongly, 3 people all do so, whereas in the cases that anyone guesses correctly, there is only 1 person doing so. This results in there being 3 hat combinations guessed correctly for every 1 hat combination guessed wrongly, and since this strategy also always has someone guess, the winning percentage is precisely 3:1 = 75%.

A little bit of thought along these lines reveals that since at most 4 players can guess wrongly at once, an immediate upper bound for the best possible winning percentage for any strategy is 4:1 = 80%.

For this specific problem, there is also an interesting geometric visualization. Take all of the hat combinations to be vertices of a graph where there is an edge between two vertices if they differ in exactly one hat. In the 3-person case, this graph is simply a cube. In the 4-person case, it's a hypercube.

Saying that someone will guess a particular color for their own hat when they see a certain combination of other hats is equivalent to choosing a single edge in the graph and marking one of the vertices for that edge as "wrong" and the other as "correct". The choice of edge corresponds to the the choice of person and combination that they see, while the choice of which vertex to mark wrong/correct corresponds to color of hat we want that person to guess in that case. The goal is to maximize the number of vertices marked "correct" at least once and marked "wrong" zero times.

Visualizing it like this, it's now easy to see that 75% is optimal. In general, we will consign some number of vertices to be marked "wrong", and then by choosing all outgoing edges, we can cause all adjacent vertices to them to be marked "correct". And it's clear for any fixed choice of vertices that we'd like to be the "wrong"-marked ones, doing this is the best we can do. Now, if we choose any 4 vertices to be marked "wrong" in such a way that all other vertices are adjacent to at least one of them, we get one of the 75% solutions, because we are correct on all but 4 out of 16 vertices. We can only potentially do better by initially choosing fewer vertices to be "wrong", but if we choose 3 or fewer vertices, then since each vertex has only 4 outgoing edges, we can make at most 12 vertices "correct", which is again only 75%.

Therefore, 75% is optimal.