Three Hats and a Strange Probability
Three players enter a room and a red or blue hat is placed on each person’s head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players’ hats but not his own.
No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass. The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly.
The same game can be played with any number of players. The general problem is to find a strategy for the group that maximizes its chances of winning the prize.
Communicated by Jim Vere.
no. think of it this way. you see the hats and then you write on a piece of paper ‘blue’, ‘red’ or ‘nothing’. that’s it, no way to pass info along. if
you wish you’re in 3 different rooms and you see on a video screen so no way to pass information.
Okay new strategy, when you see someone with a blue hat you immediately stand in front of them, and if you see someone with a red hat you immediately stand to the right of them.
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I don’t think you can do that well //s (sjm1 At williams.edu)
Can you do better than 7/8 (i.e. random guessing)?
John: correct (feel free to email me at sjm1 AT williams.edu, as I don’t want to post solns as that can spoil other people’s fun)
sadly cannot say a number, can only say a color (sjm1 AT williams.edu)
in the strategic time they will decide that whoever is asked first should shout out the number of red colored hats and the other two can easily guess the colored hat they are wearing as they can see the other two hats and if the person shouts 2 then both of the other two person are wearing red hat. if he shouts 1, the person asked second can see the other person’s hat color and tell which hat is he wearing 🙂
people can’t move — you open your eyes and a seocnd later must speak.
A strategy I thought of that gives the players certainty that they will be able to guess correctly is as follows:
Name the three players A, B, and C. Choose one to be the anchor, i.e., the player that will help the other players choose how to position themselves (any one will do – I’ll choose A). Based on player A’s hat color (which players B and C can see), players B and C will position each other (i.e., player B positions player C because player B can see the color of player C’s hat, and vice versa), such that if the other player has the same color hat as the anchor (i.e., player A), he will be positioned one yard behind player A, if not, he will be positioned 2 yards behind the anchor (player A). Based on how B positions C and C positions B, player A will know the color of his own hat because he can see the color of the other players’ hats and he will take note of their positioning.
By using the above strategy, all players will be certain of the color of their own hat, and thus, will guess correctly.
essentially correct, error in one of the calculations. email me (sjm1 AT williams.edu)
not that simple — they can choose whatever strategy they want, but the hats are randomly placed on them.
Do the players get to pick heads or tails? If so, the strategy could be to have 1 key player pick heads and the rest pick tails (or vice versa). I’m guessing it’s not that simple…
I’ll send a hint //s
Can I have the soln?
sure, I’ll send a hint.
can you post a hint? I feel as if I’m missing a clue as to how to get above 50%
[email protected]: correct, well done (not posting as it’s the correct soln — your email seems not to take messages from me)
For all the guys, tell them if they see 2 matching colors to always say that matching color
For one guy tell him if he sees two different colors to guess red
For the other two guys tell him if he sees two different colors to say blue.
There are only 4 possible groupings, all red and all blue (which are taken care of)
RBR which one of the guys who sees two different will get
BRB by the same argument
What would that yield? That does less than 50%…..
Would it be to all decide on the same color to guess beforehand?
Bren: no, can’t get a higher chance; well done!
not quite — email me at [email protected] to chat more on this problem
glad to hear — working on a student / teacher’s corner for the spring — let me know if you’d like to help, or if you’d like to know when it goes online.
glad to hear — working on a student / teacher’s corner for the spring — let me know if you’d like to help, or if you’d like to know when it goes online.
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They pick one to say the color in advance at random. Then he says the majority color or randomly if no majority. This is the same as first randomly assigning colors, and than randomly picking one to say the majority color.
Noah: correct
no: can’t do blinking
not allowed to convey any information by how speak / what do
can only say a hat color, all are either speaking or staying silent at the same time
Blinking? After making eye contact 1 eye shut is blue 2 eye shut is red. With predetermined partners.
Aldo: correct!
not quite sure what you mean here. they all have to speak or stay silent at a pre-specified time.
Hint: They can agree on certain waiting time periods before they guess their hat colors 🙂
Hard to follow. Doesn’t look like the soln. Problem is everyone can only speak once, and if you are going to speak you must speak at the same time as everyone else.
can do better than 50%
Since the hat determination is random and independent and no communication is possible that determination, wouldn’t the best strategy be to designate one guesser while the others pass. The guess would bot matter also as the outcomes are equally likely for an individuals hat color. This gives a 50% chance if winning.
I would say that ansrew is this:
(this works only if player can deside who to pass his turn, so like 1. player pass his turn to player nro.3, but if they have to pass in an order 1,2,3 this wont work)
if nro 1 pass to 2 it means that 2’s hat is red
if nro 1 pass to 3 it means that 2’s hat is blue
if nro 2 pass to 1 it means that 3’s hat is red
if nro 2 pass to 3 it means that 3’s hat is blue
if nro 3 pass to 1 it means that 1’s hat is red
if nro 3 pass to 2 it means that 1’s hat is blue
if you already know what is ur hat color (with this solution it means that 2 knows their hat alredy), it chages like this:
if it is player 1’s turn and hi knows his hat color, hi will pass to 3 if remaining players hat is red
and to 2 is it is blue.
if it is player 2’s turn and hi knows his hat color, hi will pass to 3 if remaining players hat is red
and to 1 is it is blue.
if it is player 3’s turn and hi knows his hat color, hi will pass to 1 if remaining players hat is red
and to 2 is it is blue.
This solution maybe little bit of confusing… ask if u need clearer ansrew.
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Not quite. Everyone must speak at the same time.
The player who sees that the other two have matching hats should say pass. The next player is going to know their color is the same as the player who didn’t say pass?
but how would that work with more than 3 players?