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Tue 9 Dec, 2003 03:44 am
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.
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 percent chance of winning the prize.
Question: Can the group do better than 50%?
The ratio of blue hats to red hats tends towards 1:1, so the guesser should say that her hat is the colour that she can see less of.
surely this is the same as coin tossing, and the odds for that were worked out ages ago.
If the strategy is "honest" the chance of guessing correctly if one names a color and two say "pass" is 50% no matter what.
If two guess and one passes the chance of both being correct is 25%
If all three guess the chance of all three being correct is 12.5%.
This will be true no matter what the strategy.
If the strategy is a "Cheat", e.g. they agree that only 'C' makes a guess while 'A' and 'B' both say "Pass" and at the same time both 'A' and 'B' look at 'C' if he has a red hat and at one another if he has a blue hat, the chance of them winning is 100%.
Unless they are detected.
Quote:The ratio of blue hats to red hats tends towards 1:1, so the guesser should say that her hat is the colour that she can see less of.
Tends towards...isn't exact.
Quote:If the strategy is "honest" the chance of guessing correctly if one names a color and two say "pass" is 50% no matter what.
If two guess and one passes the chance of both being correct is 25%
If all three guess the chance of all three being correct is 12.5%.
This will be true no matter what the strategy.
If the strategy is a "Cheat", e.g. they agree that only 'C' makes a guess while 'A' and 'B' both say "Pass" and at the same time both 'A' and 'B' look at 'C' if he has a red hat and at one another if he has a blue hat, the chance of them winning is 100%.
Unless they are detected.
No cheating, and there is a strategy to get over 50.
If you say there is a solution then of course I accept that there is a solution.
But it is hard to see where such a solution is possible if the answer is derived by logic and mathematics.
If however it is a matter of the 'hidden sneky' and/or some word-juggling, then I freely admit to being remarkably dumb about those.
I think they can do much better than 50%. I think they can get the odds up to 66.6% in their favor..
At the strategy session, they should decide that if any of them sees two different hats on the tohere two, that person should pass. If any of them see two hats of the same color on the other two -- that person should guess the opposite color.
Possible scenarios
Player A Red
Player B Red
Player C Red They lose because all pass.
Player A Blue
Player B Blue
Player C Blue They lose because all pass.
There are four other combinations -- and all four will provide a winner for the group using that strategy.
Franki
First the '28 coins and blindfold' and now this one.
Do you have some kind of mission in life to show me that I am dumb? (Kidding. A clever answer)
Your two 'wrong' answers should have read that they fail because all say the same wrong color, but that is a detail; you are still smarter than I!
P.S. Franki
I make six right out of eight = 75%, don't you?
(I gotta take every opportunity to save face here!)
Mungo wrote:P.S. Franki
I make six right out of eight = 75%, don't you?
(I gotta take every opportunity to save face here!)
Gotta think about a bit.
I originally came up with 8 possible scenarios -- but later it appeared as though only 6 were available. But my head just cannot seem to get it straightened out.
I'll work it out with paper and pencil -- and get back to you.
EXCELLENT RIDDLES! I love 'em.
If I try to answer any others you post, I'll PM them so that the maximim number of people get a chance to answer.
You were right, Mungo.
Here is how it works out with all the possible combinations (permutations?):
Person A - R R R R B B B B
Person B - R B R B R B R B
Person C - R R B B R R B B
In the case of the first and last combination (all red or all blue) all three will simultaneously pass -- and they will lose.
In all the other cases -- one person will see two of one color -- and will guess the opposite color -- and they will win.
So the odds are 6 in 8 -- or 3 in 4. 75% -- not 66.6%.
Ain't pencil and paper wonderful.
Sorry Tet, I was thinking that Mungo posted the riddle instead of you.
EXCELLENT RIDDLE, TET!
I'll PM you if I solve any others you offer.
Re: More hats
TetsuoDowntime wrote:Three players enter a room and a red or blue hat is placed on each person's head.
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.
For more players, the chances of one player seeing everyone else in the same colour hat gets less, so I think that my answer is still OK.
Frank - can your pencil and paper work out the odds for 15 people? (my brain can't
)
K8
Do you realise just what you are asking? The number of possible outcomes is the number of choices - in this case two; red or blue - to the power of the number of people - in this case three. So two to the power of three gives Franki's eight. Two to the power of fifteen is around thirty-two and threequarter thousand. Not only difficult to do in your head but you'd need a really big piece of paper and a lot of time.
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