4 bridge players are presented with the following problem:

They are all randomly assigned black or white hats. They cannot see their own hat colour, but they can see the other 3 hats.

At a given signal, they simultaneously must make one of the following two actions:
1) remain silent, or
2) guess their hat colour

If anyone on their table guesses correctly, with no incorrect guesses, they win a bottle of champagne. If anyone guesses incorrectly,or no-one guesses, they don't win the champagne but there are no nasty side effects.

If anyone tries to cheat by finding their hat colour before the guesses, they will all be shot. If there is signalling, they will all be shot etc.

The odds of being given a black hat is 50%, and the odds do not change dependant on the other players' hat colours. (This is to avoid Baye's theory in possible solutions.)

They are highly intelligent, and can agree on a strategy before the hats are given out.

What strategy maximises their chances of champagne?

(Zeno: this is is maths problem. But feel free to move it if there is a better home for it elsewhere.)

Well, this tells me that the chance of guessing correctly are exactly 50%. If there were exactly two black and two white hats, then they would win the champagne 100% of the time.

I take it that, if someone doesn't guess, that it isn't counted against the group unless no one guesses, i.e. if one person guess correctly and no one else guesses, then they get the champagne. Then I think the right strategy is clearly to have only one person guess.

If it's random and it's possible for them to all have the same colour hat, then the odds of them all guessing correctly is (1/2)^4. Knowing other people's hats doesn't matter. Therefore, the optimal strategy is for ONE person to guess and just guess...they maximize at 50% probability of winning.

I think OP made it too easy for them. They should all have to guess correctly to win. I this case





rather than



I'll let others elaborate.

The 50% solution is along the right lines, but it's not optimal.

The optimal solution has a 75% chance of champagne. Knowing this figure won't help you, though....

This problem is surprisingly non-trivial.

Okay I changed the problem. So in my version where every player must guess black or white

In your version where every player must guess black or white or no-guess, and they win as long as there is no incorrect guesses, and at least one correct guess

I didn't prove this is best possible. I'll let someone else do that.

If the hat distribution is random from all black to all white, why does anyone do better by knowing what hats other people wear?

For each of them to sit at a table by themselves.

Their guesses aren't independent. They plan a strategy together in advance.

Alright.



This gives one players a 50% chance of success, and all the others a 100% chance of success.

Edit:

Edit - Oh ****, simultaneous. Never mind, reading comprehension fail.

Do you see why that's only a 50% chance though?

You haven't solved for the 75% solution. Your strategy need only stop w/ the 1st person. If they guess right, why would any of the others guess? They win if the first person is right...if the first is wrong, then it's game over.

Okay, I think I got it:

You are doing a different problem. OP specified guesses are simultaneous, not sequential. Oh haha I just noticed you editted.

No, because too often some will guess wrong and they lose.

Everyone wins in a 2/2 case. One person wins and everyone else loses in a 1/3 case. Everyone loses in a 0/4 case.

So in 3/8 of all cases, everyone wins. In 1/8 of cases, nobody wins. And in half of cases, 25% win. That only adds up to 50% again, though. So yeah, no help.

It seems like the only information a player can have is that of which colors the hats of the other players are and the knowledge of the strategy formulated previously. Given ONLY this knowledge, they have to get a 75% correct rate?

Even mixed strategies are coming up on 50% for me, but it seems like the formulated strategy must provide some special information. Maybe if each player has a different strategy, and the strategies of some of the players are based on the strategies of other players?

Oh, wait, they can remain silent. I thought they had to guess every time. Okay, I'm just going to shut up now. At least until I'm not so tired.

I don't know if this question is phrased "correctly".

Answer to question if players are allowed to remain silent only for some time and then guess:

This looks correct and "trivial", no?

thyalcine's solution is correct for 4 palyers. The non-triviality comes about (with 4 players) if you do not force someone to pass - without this you can (with difficultly) construct a strategy with 9/16 success rate.

I have no idea if the solution is optimal with N players (N>4). It probably is, but proving it seems too much like hard work.

The most common call for bridge players is "pass" so having the word "bridge" can be considered a hint.

If everyone has to guess black or white, then they have no strategy to win more than 1/2 the time.

So what are you saying with this 9/16? Is it yet another version of the problem?

With the original problem (where every player must guess black or white or no-guess, and they win as long as there is no incorrect guesses, and at least one correct guess) with n>0 players, they can do no better than to win n/(n+1) of the time, though they can attain that upper bound if and only if n+1 is a power of two. So FWIW I guess they can always do at least

1-2^{-\lfloor\log_2(n+1)\rfloor}

Grunch:

http://en.wikipedia.org/wiki/Hamming_code

How do we generalise the solution using hamming codes to cases where the number of players is not 2^k-1 ? (as in the OP for instance).

It seems like there is a 3/4 success solution according to this webpage:

http://narandus.blogspot.com/2009/02/hat-problems.html

Thylacine: I think what river_tilt meant with regard to your original 3/4 solution is that you are not permitted to nominate any one person who will pass in advance. All players have to make their decision to pass, or choose a color solely based on the colors of hats that they observe, rather than their 'ordering'. You can phrase the problem to make it impossible to choose any 'ordering' anyway but it would probably be a bit wordy !

LOL, I actually figured it out for myself, during this thread, but I figured every conceivable version of this problem was thoroughly researched. After all the OP obviously heard the problem from somewhere.

That's called making up the rules as you go along. There are obviously many possible variations of this problem so it needs to be made clear which one is intended.

I still want to know which variation gives 9/16.

Given that the action is simultaneous and each person's hat outcome is independent, how can the answer be anything other than one person just guess, thus 50% probability of champagne?