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What color is my hat problem

  • Thread starter mamamia
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here is a question that i am having trouble with for my math class:

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 independent coin tosses. Each person can see the other player’s hats but not his/her 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 prize if at least one player guesses correctly and no other player guesses incorrectly. Find a strategy for the group that maximizes its chances of winning the prize.

i have realised that since only one person needs to guess and with the greater number of people who guess comes the greater the chances of someone guessing incorectly, that only one person should guess. also the color of the others hats will not help the person who is guessing determine their own hat color. therefore the best strategy that i can come up with is to have one predetermined person randomly guess red or blue and the other two pass (50% of being right and winning). but winning only half the time is not a good startegy.

if you can think of a better, mor effective way to do the proble or can figure out a better way if the problem was slightly changed in any way (i.e. the three people would not guess simultaneously), please help me out
 

NateTG

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Classic problem. It's possible to win the price 3 times out of 4.
 
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would it work when whoever sees two hats of the same color guesses the opposite color?
 

OlderDan

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mamamia said:
would it work when whoever sees two hats of the same color guesses the opposite color?
Sounds like a plan. What is the probability of all hats being the same color? If one is different, by your plan will they always win the prize?
 
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2^3 possibilities or eight chances
2 chances with all the same color
6 chances with 2 hats of one color and one of the other color
does that make it 3:1 odds?
 

NateTG

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mamamia said:
2^3 possibilities or eight chances
2 chances with all the same color
6 chances with 2 hats of one color and one of the other color
does that make it 3:1 odds?
You got it. Whether you think of it as 3:1 or 3:4 depends on what the ratio is suppsed to be.
 
OK sooo if you see 2 people wearing the same hat you wait a minute, if no one else says anything you say the opposite color.

But if you see two people wearing the same hat and wait a minute you let someone else answer and say that their hat is the color that they see. eh?
 

OlderDan

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sandstorm03 said:
OK sooo if you see 2 people wearing the same hat you wait a minute, if no one else says anything you say the opposite color.

But if you see two people wearing the same hat and wait a minute you let someone else answer and say that their hat is the color that they see. eh?
A condition of the problem is that each person has to simultaneously state their guess or pass. There are other variations on this problem where one person could respond first, but this is not one of them.
 
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You could, before you start, tell your team mates that the person that see two hats of the same color say the opposite; everyone else should pass. That way 1:3 times everyone would answer the question wrong; however, 2:3 times only one person would answer and you would would win the prize.
Is it worth the chance?
 
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what if they didn't guess similtaneosly but they had a predetermined order. i did the math with the same strategy and i think i found that it was still 3:4 odds of winning.
1:1 odds of winning if the first person guesses (right 50%)
1:0 odos if the second person guesses (right 100%)
1:0 odds if the thrid person guesses (right 100%)
does that mean that it doesn't matter whether or not they guess at the same time? or do i need a new strategy for that new condition?
 
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mamamia said:
would it work when whoever sees two hats of the same color guesses the opposite color?
Hmm, maybe I'm wrong, but this doesn't sound quite right. It's like saying if a woman gives birth to a girl, then another girl, the probability of her giving birth to a boy is .75, not .5. So if a coin is flipped and the first person gets a red hat, then another coin is flipped and the second person gets a red hat, there is still a 50/50 chance that the third person gets a blue hat.

On the other hand I can't can't seem to fault the logic when you consider the system as a whole. Perhaps someone can clear this up for me.
 
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Never mind, just worked out my problem, odds are 3:4.
 

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