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Thread: Prisoner Hat Riddle

  1. #1
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    Prisoner Hat Riddle

    Not really heavy into the math, but a challenge in logic.
    I found it puzzling, but the answer is surprisingly simple.

    Thought I would share it here.

    The Prisoner Hat Riddle by Alex Gendler on TED Ed.
    http://ed.ted.com/lessons/can-you-solve-the-prisoner-hat-riddle-alex-gendler
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  2. #2
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    Re: Prisoner Hat Riddle

    Quite a well-know riddle, along with others that are similar....
    https://www.google.ca/?gws_rd=ssl#q=...prisoners+hats

    Here's another one you'll enjoy:
    http://puzzles.nigelcoldwell.co.uk/seven.htm
    Last edited by DenisB; Jan 11th 2016 at 07:11 PM.
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  3. #3
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    Re: Prisoner Hat Riddle

    How about this one?
    A king has three prisoners. He decides to give them a chance to earn their freedom. The king says he will blindfold all three prisoners and place a white or black hat on the first prisoner's head, a red or green hat on the second's and a yellow or cyan hat on the third's.
    He will then remove the blindfolds so that each prisoner can see the other two hats but not his own.
    He will then ask them to simultaneously write on a piece of paper the color of the hat that each of them thinks he has on his head. The only choices that are permitted are, either a color, or "I don't know". The prisoners will win their freedom if at least one of them correctly guesses his own color, but with the restriction that no one must give a wrong answer (for example, if 2 of the 3 prisoners reply "I don't know" and the 3rd one guesses correctly, they go free). The king allows the prisoners to discuss and coordinate for a few minutes, before the placement of the hats, in order to decide a strategy. What is the optimum strategy that will give them the biggest chance to go free?
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    Re: Prisoner Hat Riddle

    Quote Originally Posted by Alderamin View Post
    How about this one?
    A king has three prisoners. He decides to give them a chance to earn their freedom. The king says he will blindfold all three prisoners and place a white or black hat on the first prisoner's head, a red or green hat on the second's and a yellow or cyan hat on the third's.
    He will then remove the blindfolds so that each prisoner can see the other two hats but not his own.
    He will then ask them to simultaneously write on a piece of paper the color of the hat that each of them thinks he has on his head. The only choices that are permitted are, either a color, or "I don't know". The prisoners will win their freedom if at least one of them correctly guesses his own color, but with the restriction that no one must give a wrong answer (for example, if 2 of the 3 prisoners reply "I don't know" and the 3rd one guesses correctly, they go free). The king allows the prisoners to discuss and coordinate for a few minutes, before the placement of the hats, in order to decide a strategy. What is the optimum strategy that will give them the biggest chance to go free?
    Answer was done by brute force. Couldn't prove if chances of losing could be pared down further.
    Spoiler:

    1,2,3 are Prisoners. WB, RG, YC, colors of hats. "RY - W" means if the prisoner sees a Red and Yellow Hat, write White.

    The result was generated by enumerating along every possibility of what each prisoner could see. I then took a forced starting point that if 1 saw RY, then W R Y is a win, and B R Y is a loss (a 50/50 split). I then 'stacked' upon the B R Y loss as much as possible (ex. if 2 sees BY, say G, because B R Y is a loss anyway) and sealed the W R Y win (ex. if 2 sees WY, Don't Know). After elimination, make another 50/50 win/lose split and proceed again. I think both losses are forced.

    Prisoner 1
    RY - W
    RC - Don't Know
    GY - Don't Know
    GC - B

    Prisoner 2
    WY - Don't Know
    WC - R
    BY - G
    BC - Don't Know

    Prisoner 3
    WR - Don't Know
    WG - Y
    BR - C
    BG - Don't Know

    Result
    W R Y - win
    W R C - win
    W G Y - win
    W G C - lose
    B R Y - lose
    B R C - win
    B G Y - win
    B G C - win

    Chance of being freed = 6/8 = 75% chance.
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