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On September 10 2010 06:27 The_Pacifist wrote:AcrossFiveJulys believes my answer is invalid and uses communication, so I came up with option two: + Show Spoiler +Prisoners meet and introduce themselves before getting the hats and sitting at the table. One person introduces himself as "Bob," and all the prisoners agree to write down the color of Bob's hat, whatever it happens to be, once they all sit down and get hats. Cookie nao? :3
what if bob has color A and all the 6 other prisoners got color B? then this method doesnt work...
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So why not assume that all the people are completely logical and rational.
Then have one person say a color that he sees. Then the remaining 6 prisoners all say that same color.
Eventually they'll get it right.
*edit* nevermind. They're not saying the colors out loud.
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Are you sure that you have not left out any rules? Because as it stands right now I can not see a single way that this can be solved since there is NO 100% way of knowing what color your hat has.
Doesn't matter what the other guys are wearing, you could be alone with your hat and still have the exact same chance of calculating the right answer.
Although, if they know the colors beforehand they can simply designate one person to each color so that all seven colors are covered and thereby auto-win. That would be really easy but you said this was a math-problem so...
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On September 10 2010 06:27 AcrossFiveJulys wrote:
I think this is the solution. If all the hats are a different color the chosen guy gets it right, if there are repeats then the other guy wearing the same color as the chosen guy gets it. Well done.
unless the repeat does not include the chosen guy ;-)
Bob => red
everyone else => blue
1/6 chance for Bob to save the team.
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On September 10 2010 06:28 Black Gun wrote:Show nested quote +On September 10 2010 06:27 The_Pacifist wrote:AcrossFiveJulys believes my answer is invalid and uses communication, so I came up with option two: + Show Spoiler +Prisoners meet and introduce themselves before getting the hats and sitting at the table. One person introduces himself as "Bob," and all the prisoners agree to write down the color of Bob's hat, whatever it happens to be, once they all sit down and get hats. Cookie nao? :3 what if bob has color A and all the 6 other prisoners got color B? then this method doesnt work...
It doesn't matter. The conditions state that only one person has to get the correct answer. Whether Bob is right or wrong, there is still a 100% chance that all the prisoners will go free, according to the game's rules.
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On September 10 2010 06:27 AcrossFiveJulys wrote:Show nested quote +On September 10 2010 06:07 0nega wrote:Hello! lurked at this site for some time and couldn't resist to post the solution(hopefully) + Show Spoiler +Before they get the hats, they agree, that they will all take the colour of the hat of one guy, this guy instead takes the colour which no hat he sees has I think this is the solution. If all the hats are a different color the chosen guy gets it right, if there are repeats then the other guy wearing the same color as the chosen guy gets it. Well done.
What if his hat is red and every other hat is yellow?
They are all going to pick yellow and fail, then he has a 1 in 6 chance of getting it right.
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Could the OP clarify that no prisoner has access to any information whatsoever other than the color of the hats, and whatever they decided beforehand?
This includes: timing of writing, what other people are writing, etc
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On September 10 2010 06:30 Seth_ wrote:Show nested quote +On September 10 2010 06:27 AcrossFiveJulys wrote:
I think this is the solution. If all the hats are a different color the chosen guy gets it right, if there are repeats then the other guy wearing the same color as the chosen guy gets it. Well done. unless the repeat does not include the chosen guy ;-) Bob => red everyone else => blue 1/7 chance for Bob to save the team.
Ah, that's true. I think we are getting closer though.
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On September 10 2010 06:28 Stenstyren wrote:
Are you sure that you have not left out any rules? Because as it stands right now I can not see a single way that this can be solved since there is NO 100% way of knowing what color your hat has.
Doesn't matter what the other guys are wearing, you could be alone with your hat and still have the exact same chance of calculating the right answer.
Although, if they know the colors beforehand they can simply designate one person to each color so that all seven colors are covered and thereby auto-win. That would be really easy but you said this was a math-problem so...
if everyone tells one color they can still fail. lets denote A the color which prisoner 1 decides to write down, B the color of prisoner 2 and so on. if now the hats are put to the prisoners in the order "1-B,2-C,...,7-A", everyone guesses wrong with this method and their heads are gonna roll...
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Communication is not allowed, but in what form? Direct or indirect?
Is this allowed?
They assign themselves as follow.
Guy1 sacrifices himself if he sees 6 people with the same color.
Guy2 sacrifices himself if he sees 5 people with the same color.
Guy3 sacrifices himself if he sees 4 people with the same color.
...
I know this is not likely to be the correct step, but is this form of communication allowed?
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How about where they sit? Can they choose where they sit AFTER they have seen each others hats?
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If any of the prisoners guesses their hat color correctly, they will all go free. Otherwise, they are all executed on the spot.
Please do not forget this rule. 100% survival rate does not mean that everyone has to be right. It only means that at least 1 person has to get it right each time it's played.
0nega's solution still works.
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On September 10 2010 06:31 tissue wrote:
Could the OP clarify that no prisoner has access to any information whatsoever other than the color of the hats, and whatever they decided beforehand?
This includes: timing of writing, what other people are writing, etc
Correct, there is no communication. No timing, no looking at other people's writing, or anything else of this gimmicky nature.
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On September 10 2010 06:32 LunarDestiny wrote:
Is this allowed?
They assign themselves as follow.
Guy1 sacrifices himself if he sees 6 people with the same color.
Guy2 sacrifices himself if he sees 5 people with the same color.
Guy3 sacrifices himself if he sees 4 people with the same color.
don't know what you mean by sacrificing, but there's nothing wrong with that, since the prisoners only follow the rules they discussed beforehand and don't act on anything other than the color of the hats.
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Since we're given that the naive strategy (completely random guesses) will achieve a success rate of ~66%, I'll assume that the hat distribution is also completely random.
Since there is no communication between the prisoners, for each prisoner, the other prisoners' answers cannot affect his own. Therefore, he ONLY has the knowledge of the other prisoners' hats. Based on this information, he must somehow improve his chances of guessing his color correctly to over 1/7.
Well, it's impossible without cheating. You have no additional information. No matter what the other prisoners' hats are, your own color is equally likely from among the 7 colors in the pool.
That is, if x is the event that your hat is color x, then
P(x) = 1/7
If y1 is the event that the 6 other prisoners' hats are of a particular ordered sequence, then
P(x | y) = 1/7, still. In fact, for any y, P(x | y) = 1/7.
In short, there is no way to improve on the completely random guess.
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On September 10 2010 06:33 The_Pacifist wrote:Show nested quote +If any of the prisoners guesses their hat color correctly, they will all go free. Otherwise, they are all executed on the spot. Please do not forget this rule. 100% survival rate does not mean that everyone has to be right. It only means that at least 1 person has to get it right each time it's played. 0nega's solution still works.
0nega's solution does not work, as multiple people have demonstrated with the simple case.
Guy has red hat.
Others have blue hat.
Guy guesses yellow, or purple, or whatever other colors are in the pool besides blue.
Others guess red.
They all get it wrong.
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Do they all have to hand their answer into the king at the same time, or can they go one by one?
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On September 10 2010 06:34 Slithe wrote:Show nested quote +On September 10 2010 06:31 tissue wrote:
Could the OP clarify that no prisoner has access to any information whatsoever other than the color of the hats, and whatever they decided beforehand?
This includes: timing of writing, what other people are writing, etc Correct, there is no communication. No timing, no looking at other people's writing, or anything else of this gimmicky nature.
Yes there is, tell us the gimmick already.
Is it that they can see the others hats as they sit at the table?
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On September 10 2010 06:27 AcrossFiveJulys wrote:Show nested quote +On September 10 2010 06:07 0nega wrote:Hello! lurked at this site for some time and couldn't resist to post the solution(hopefully) + Show Spoiler +Before they get the hats, they agree, that they will all take the colour of the hat of one guy, this guy instead takes the colour which no hat he sees has I think this is the solution. If all the hats are a different color the chosen guy gets it right, if there are repeats then the other guy wearing the same color as the chosen guy gets it. Well done.
I don't think this is right.
If the colors are ROYGBBV (2 blues) and the guy they decide on beforehand is wearing say a red hat, then 6 of the prisoners would pick red. They would all be wrong. The last prisoner would then see OYGBBV as the remaining colors and would still have to guess between R (red) and I (indigo).
Same thing if there were 6 red hats and 1 blue hat. If they agreed to choose the color of the blue-hatted guy, what color is the blue-hatted guy going to choose?
This wouldn't work 100% of the time if there were any multiples of any color.
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On September 10 2010 06:36 Slithe wrote:Show nested quote +On September 10 2010 06:33 The_Pacifist wrote:If any of the prisoners guesses their hat color correctly, they will all go free. Otherwise, they are all executed on the spot. Please do not forget this rule. 100% survival rate does not mean that everyone has to be right. It only means that at least 1 person has to get it right each time it's played. 0nega's solution still works. 0nega's solution does not work, as multiple people have demonstrated with the simple case. Guy has red hat. Others have blue hat. Guy guesses yellow, or purple, or whatever other colors are in the pool besides blue. Others guess red. They all get it wrong.
if there is no form of communication, even indirect communication there is no way to get 100% probability. unless you mean 99.999999% rounded up, perhaps some people can find a way
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EPT
