Three players enter a room and a red or blue hat is placed on each person's head. The color of each hat is

Question

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 will win if at least one player guesses correctly and no players quess incorrectly. One obvious strategy for the players, for instance, would be for one player to always guess "red" while the other players pass. This would give the group a 50% chance of winning the game. Can you device a better strategy?

To Jim Brain: Read the problem again! It says: "The color of each hat is deetermined by a coin toss, ..."

To Shelbola: Then why bother to answer?

To Kevin!:
You should read the question again! They all lose if anyone guesses incorrectly!
Pavi

To Celtic 263: Read the question again! No communication is allowed after the initial session!
Pavi

To SmW: There seems to be two requirements to be in this site: 1) To have a brain. 2) To enjoy using it. If you are lacking on either or both, you shouldn't be here. Pavi

Answer 1

Here is a better strategy:

When each person walks in, if he sees two hats of the same colour, he says the other colour. If he sees two hats of different colours, he says nothing,

Let's analyze when this works:
RRR - lose, each player says blue
RRB - win, player three says blue and is correct, other two say nothing
RBR - win, as above, but with player 2
BRR - win, as above but with player 1
RBB - win, as above but with red instead of blue
BRB - win, as above
BBR - win, as above
BBB - lose, all say red.

This gives a 75% chance of winning.



Further, you should ignore what tbolling is saying. His statement that "An independent event is inherintly independent and there is no information to be gained by looking at the other two hats." shows a lack of understanding.

The strategy has nothing to do with there being a better chance that your hat is the blue if the other two are red. In fact, whenever you guess anything, there is exactly a 50% chance you will be right. However, what this strategy does is makes it so that when there is 1 wrong guess there are actually three wrong guesses, but when there is one right guess, the other people aren't saying anything.

So if 50% of your guesses are wrong, but the wrong ones are made in groups of three and the correct ones are made by themsevles, your ratio of correct instanec to incorrect is 3:1, which is 75% chance of guessing correctly.

Answer 2

Perhaps I'm misunderstanding something. If the announcement of 'Pass' has to be made at the same time that the others are announcing their guesses, then there is no guessing strategy.

The key here is in the coin toss and it being independent of the others. An independent event is inherintly independent and there is no information to be gained by looking at the other two hats.

Now, if someone is allowed to say something before the others announce, my strategy is to have the person going first to guess his own hat Red if he sees 2 Red hats, guess his own hat Blue if he sees 2 Blue hats, and pass if he sees 1 Red and 1 Blue. But, that isn't really want you're looking for. I just wanted to throw that in as it occurred to me as a good strategy :)

Answer 3

Sure:

You have 4 distinct possibilities with an associated pobability.

3R = 1/8; 2R,B = 3/8; R,2B =3/8; 3B = 1/8

The chance that there will be 2 of one color and one of the other is 3/4, so base your strategy on this fact.
The stategy is that if a player sees two R(or B) hats he calls out B(or R) and if a player sees two hats of different colors the player passes. In this strategy, only if all the hats are of the same color will they be wrong (they'll all state the wrong color simultaneously) But they have a 75% chance of winning the game.

Answer 4

The players agree that:
1. If a player sees both red and blue hats, he will pass.
2. If a player sees two red hats, he will guess blue.
3. If a player sees two blue hats, he will guess red.

This strategy will win 3/4 of the time, failing only when there are three red hats (1/8 of the time) or three blue hats (1/8 of the time).

Answer 5

Any player (A) who sees the other 2 players with hats of the same color to say pass.

The 2 other players can safely guess the color of his hat by looking at each others hat.

Answer 6

alright bob and bobbette, you guys both pass on your chance and i'll be the one to guess the color of my hat. bob if my hat is red then cough and if not then dont do anything. ok? good.

actually the coin toss does count if for example on heads they get a blue hat and on tails they get red. then you would know what color you got by knowing which side of the coin you got.

Answer 7

well there is a chance that two of the hats are the same color and the other one is the other color...if one person looks at the two peoples hats and he sees that they are the same color he will say the other color (such as the two are red and the other is blue) he will say blue. if they are different colors he passes.

Answer 8

They can all guess the hat red. This will give a 7/8 = 87.50% chance of winning. (The only loss is when all hats are blue).

^_^

Answer 9

Blink twice to signal blue, once for red

Answer 10

School's out! My brain's on vacation. Ask again in Sept.