Hat problem

[Joaz Banbeck]

Nope. Here's a simple counterexample...

I was being dense too.


However, the counterexample still does not get above 50%, so I still think that Drmwc forgot something.

[jlaire]

However, the counterexample still does not get above 50%

It wasn't supposed to.

I have found a strategy that gives 10/16 = 62.5%. Edit: make that 11/16 = 68.75%.

I think it's crazy how much effort people have spent on looking for loopholes in the wording. I think the problem statement was very clear on first reading.

[Boidhre]

Yeah, my problem was I read it and my brain went, "so he didn't rule out an infinite pool of hats and I got all confused.

[tj86430]

I think it's crazy how much effort people have spent on looking for loopholes in the wording. I think the problem statement was very clear on first reading.

I think the stament was clear except for how the guessing is implemented: whether a time element can be utilized or not. I believe not, but if that is the case then it would be clearer to say e.g. that each player writes down their guess so that no one can see when and what others write, and the guesses are revealed simultaneously; not writing is permitted and equals not guessing.

If a time element could be utilized, a simple strategy would be e.g. that the players decide who will go first. That player will look at the player on the right, and if he sees a white hat, he will immediately guess something. If he sees a black hat he will stay silent, and after a while the next player will guess black. This strategy has 75% chance of success (whenever the second player has a black hat and 50% of the time when he has a white hat)

[HermanHiddema]

If a player sees three hats of the same color, he immediately guesses the other color.

If all hats are the same color, this means all players immediately guess wrong.

If the colors are divided 3-1, only one player guesses immediately, and his guess is correct.

If, after 10 seconds, nobody has made a guess, then the hats must be divided 2-2. One player is preassigned to guess after 10 seconds the color that makes it 2-2 (i.e. if he sees WWB, he guesses B, if he sees WBB, he guesses W).

This strategy fails if all hats are the same color (1/8 chance), succeeds in all other cases (7/8 chance).

[jlaire]

My solution is here, don't read if you want to solve it yourself. No silly tricks with time delays or anything like that.

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.

[Magicwand]

my percentage 87.5%

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

[Splatted]

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.

[drmwc]

Time delay solutions will result in everyone being shot. The guesses all have to happen at exactly the same time, pre-assigned time.

jlaire is on the right lines. However, it's possible to do better than his solution.

For LocoRon, the wine is 1995 Dom Perignon, and so well worth taking a punt at...

Clue:

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

[Joaz Banbeck]

I think that I am beginning to understand this problem...

jlaire is on the right lines.

..

...All players follow the same strategy. ...

However, it's possible to do better than his solution.

[palapiku]

I wrote a brute force program using drmwc's hint and got a 12/16 solution. However I don't really understand it, so I won't post it for now.

[Joaz Banbeck]

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...

. <> <>

About Drmwc's clue:

This, BTW, explains what Drmwc meant by

...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:

...
-> lose (4 permutations)...

[palapiku]

That's not how I interpreted the hint, by the way.

[Joaz Banbeck]

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 )

[drmwc]

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.