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    (Original post by Bustamove)
    when the warden comes in to the jail, it depends on the order of the people he questions? he also can't see the other's prisoners hat, (he may know their hat colour, but he can't see them and he also wouldn't know the order of the questioning?)

    I'm so baffled by this LOL
    It doesn't matter whether or not the prisoners know what order the warden visits them because it can't offer any help. (No prisoner knows what any of the prisoners before them has said.) The warden knows everyone's hat colour, so doesn't need to see them. Each prisoner knows what colour everybody is apart from himself.
    TL DR: The order is irrelevant.
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    (Original post by theOldBean)
    It doesn't matter whether or not the prisoners know what order the warden visits them because it can't offer any help. (No prisoner knows what any of the prisoners before them has said.) The warden knows everyone's hat colour, so doesn't need to see them. Each prisoner knows what colour everybody is apart from himself.
    yea but say they all stand in a line facing forwards and the person at the back gets questioned first, the person at the back can see 19 other hats? then the person infront will see 18 etc...

    I have a feeling i'm over thinking this
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    (Original post by Bustamove)
    yea but say they all stand in a line facing forwards and the person at the back gets questioned first, the person at the back can see 19 other hats? then the person infront will see 18 etc...

    I have a feeling i'm over thinking this
    But the person in front actually sees 19 because he knows everyone's hat apart from himself. You're confusing it with a different problem where the answer is 'guess red or blue depending upon whether the number or red or blue hats in front is odd or even.'
    But this strategy doesn't work here because if I am in second place, the first person's guess gives me no information because I cannot hear it.

    Try going back a page in the thread - Xiuchen posted a solution which, even though it doesn't work, comes close to hitting the key idea.
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    (Original post by theOldBean)
    But the person in front actually sees 19 because he knows everyone's hat apart from himself. You're confusing it with a different problem where the answer is 'guess red or blue depending upon whether the number or red or blue hats in front is odd or even.'
    But this strategy doesn't work here because if I am in second place, the first person's guess gives me no information because I cannot hear it.

    Try going back a page in the thread - Xiuchen posted a solution which, even though it doesn't work, comes close to hitting the key idea.
    Another big hint:
    Spoiler:
    Show
    Divide the prisoners up into 10 pairs. Can you arrange for exactly one person in each pair to guess correctly?. Although just 'one person guesses red and one blue' is obviously no good.
    Ah, I see, that is a good suggestion by Xiuchen.
    I'm gonna make another complete guess

    Divide the prisoners into 10 pairs. Count the number of colours for red or blue from the prisoner hat colours from the guard. Only half of them have to get it right? So assign the pairs of prisoners to call out a colour depending on the ratio.
    e.g. 15 red, 5 blue, so therefore the 10 pairs will be split up as 7:3?
    e.g. 10 red, 10 blue, so split as 5:5
    e.g. 9 red, 11 blue, so split as 4:5

    This method won't get everyone to get the answer correctly.. but I think it has a chance of getting half...
    just a complete guess.... my head hurts from thinking... This is probably my last guess, I have to revise for my statistics exam ahaaaaa XD
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    (Original post by theOldBean)
    But the person in front actually sees 19 because he knows everyone's hat apart from himself. You're confusing it with a different problem where the answer is 'guess red or blue depending upon whether the number or red or blue hats in front is odd or even.'
    But this strategy doesn't work here because if I am in second place, the first person's guess gives me no information because I cannot hear it.

    Try going back a page in the thread - Xiuchen posted a solution which, even though it doesn't work, comes close to hitting the key idea.
    Oh so you put them in pairs!!! THAT CLUE I GOT IT NOW
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    (Original post by theOldBean)
    But the person in front actually sees 19 because he knows everyone's hat apart from himself. You're confusing it with a different problem where the answer is 'guess red or blue depending upon whether the number or red or blue hats in front is odd or even.'
    But this strategy doesn't work here because if I am in second place, the first person's guess gives me no information because I cannot hear it.

    Try going back a page in the thread - Xiuchen posted a solution which, even though it doesn't work, comes close to hitting the key idea.
    Oh wait I didn't i was confused sorry
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    So the pirsoners get split into pairs. One person in the pair will guess the same colour as their partner, the other will guess the same colour as their partner BOOOM THERE YA GO FINALLY WORKED IT OUT! took me agess good riddle!!!!
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    (Original post by theOldBean)
    But the person in front actually sees 19 because he knows everyone's hat apart from himself. You're confusing it with a different problem where the answer is 'guess red or blue depending upon whether the number or red or blue hats in front is odd or even.'
    But this strategy doesn't work here because if I am in second place, the first person's guess gives me no information because I cannot hear it.

    Try going back a page in the thread - Xiuchen posted a solution which, even though it doesn't work, comes close to hitting the key idea.
    I'm sure I've got it now!! That wasss sooo difficult!!!
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    (Original post by Xiuchen)
    So the pirsoners get split into pairs. One person in the pair will guess the same colour as their partner, the other will guess the same colour as their partner BOOOM THERE YA GO FINALLY WORKED IT OUT! took me agess good riddle!!!!
    Correct! (As long as you replace the second 'same' with 'opposite' )

    Interesting fact: There cannot possibly be a strategy which guarantees that 11 of them will guess correctly. The intuitive explanation for this is that each prisoner has a 50% chance of guessing correctly, so you expect, on average, 10 to get it right. To demand that 11 people guess correctly all the time is like insisting that 'everyone must be better than average', which is impossible. To make this precise, you need something called Markov's Inequality. Ask your probability/statistics teachers or look at the wikipedia article if you are interested.
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    (Original post by Bustamove)
    Ah, I see, that is a good suggestion by Xiuchen.
    I'm gonna make another complete guess

    Divide the prisoners into 10 pairs. Count the number of colours for red or blue from the prisoner hat colours from the guard. Only half of them have to get it right? So assign the pairs of prisoners to call out a colour depending on the ratio.
    e.g. 15 red, 5 blue, so therefore the 10 pairs will be split up as 7:3?
    e.g. 10 red, 10 blue, so split as 5:5
    e.g. 9 red, 11 blue, so split as 4:5

    This method won't get everyone to get the answer correctly.. but I think it has a chance of getting half...
    just a complete guess.... my head hurts from thinking... This is probably my last guess, I have to revise for my statistics exam ahaaaaa XD
    I know this doesn't work, because as I said in another hint, 'any valid strategy guarantees that EXACTLY 10 get it correct, whatever the distribution of hats.' Xiachen has worked out the answer now, so you can either read the thread to find it, or put the puzzle aside until after exams. Alternatively here's a massive hint:
    If there were only two prisoners, how could you guarantee that exactly one of them guessed correctly?
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    (Original post by theOldBean)
    Correct! (As long as you replace the second 'same' with 'opposite' )

    Interesting fact: There cannot possibly be a strategy which guarantees that 11 of them will guess correctly. Indeed, the probability that any individual prisoner guesses correctly is 1/2, so the expected number of people to guess correctly is 10. Hence, the only way guarantee that 10 get it right is for it to be certain that exactly 10 will be correct. More generally, by Markov's inequality, the chance that 11 or more will be correct cannot possibly be more than 10/11, or 91%
    (Original post by Xiuchen)
    So the pirsoners get split into pairs. One person in the pair will guess the same colour as their partner, the other will guess the same colour as their partner BOOOM THERE YA GO FINALLY WORKED IT OUT! took me agess good riddle!!!!
    I'm still baffled by the solution LOL

    So they're split into pairs and in one pair (prisoner 1 and prisoner 2 being in one pair), prisoner 1 guesses the opposite as prisoner 2 and prisoner 2 guesses the opposite as prisoner 1.... If their answers are based on each other, who starts first? and if they're all being questioned individually in the cell, then lets say prisoner 1, won't know prisoner's 2 guess? and prisoner 2 won't know prisoner 1's guess? unless they decide to assign each other colours before the questioning and before information is revealed to them.....

    can someone explain? looool, It's driving me nuts...:facepalm:

    Edit= Sorry, meant to change same to opposite
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    (Original post by theOldBean)
    Correct! (As long as you replace the second 'same' with 'opposite' )

    Interesting fact: There cannot possibly be a strategy which guarantees that 11 of them will guess correctly. The intuitive explanation for this is that each prisoner has a 50% chance of guessing correctly, so you expect, on average, 10 to get it right. To demand that 11 people guess correctly all the time is like insisting that 'everyone must be better than average', which is impossible. To make this precise, you need something called Markov's Inequality. Ask your probability/statistics teachers or look at the wikipedia article if you are interested.
    Hahaha, ooops i accidentally wrote same instead!! Hahaha, omg that really interesting!!!!! And i stalked your profile a bit (SORRY!) you go to cambridge right? Thats how you come up with such amazing riddles! I want to go to Cambridge when I grow up
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    (Original post by Bustamove)
    I'm still baffled by the solution LOL

    So they're split into pairs and in one pair (prisoner 1 and prisoner 2 being in one pair), prisoner 1 guesses the same as prisoner 2 and prisoner 2 guesses the same as prisoner 1.... If their answers are based on each other, who starts first? and if they're all being questioned individually in the cell, then lets say prisoner 1, won't know prisoner's 2 guess? and prisoner 2 won't know prisoner 1's guess? unless they decide to assign each other colours before the questioning and before information is revealed to them.....

    can someone explain? looool, It's driving me nuts...
    Its not based on what each other guess. Its about each other's colour. So if one guesses the same as the other and the other guess the opposite ONE IS BOUND TO BE CORRECT. Prisoner 1 guesses the same colour as prisoner 2 prisoner 2 guesses the oppsite of prisoner 1! Think about the combinations.
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    (Original post by Bustamove)
    I'm still baffled by the solution LOL
    etc etc.
    (Original post by Xiuchen)
    Hahaha, ooops i accidentally wrote same instead!! Hahaha, omg that really interesting!!!!! And i stalked your profile a bit (SORRY!) you go to cambridge right? Thats how you come up with such amazing riddles! I want to go to Cambridge when I grow up
    Bustamove: Let me rephrase what Xiuchen said:

    Consider the easier problem where there are only two prisoners, A and B, and at least one has to get it right. The strategy is:
    i) A guesses whatever colour B is.
    ii) B guesses the OPPOSITE colour to what A is.
    This way, if they are the same colour, then A guesses correctly, and if they are the opposite colour then B guesses correctly.
    To translate this to the case where there are 20 prisoners, divide them into ten pairs (A1,B1), (A2,B2),...,(A10,B10) and have each pair (A,B) carry out the same strategy given above. This guarantees that exactly 10 of them will be correct.

    Xiuchen: Glad you enjoyed it. If you do apply to Cambridge in the future, then I wish you the best of luck.
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    (Original post by Xiuchen)
    Its not based on what each other guess. Its about each other's colour. So if one guesses the same as the other and the other guess the opposite ONE IS BOUND TO BE CORRECT. Prisoner 1 guesses the same colour as prisoner 2 prisoner 2 guesses the oppsite of prisoner 1! Think about the combinations.

    (Original post by theOldBean)
    Bustamove: Let me rephrase what Xiuchen said:

    Consider the easier problem where there are only two prisoners, A and B, and at least one has to get it right. The strategy is:
    i) A guesses whatever colour B is.
    ii) B guesses the OPPOSITE colour to what A is.
    This way, if they are the same colour, then A guesses correctly, and if they are the opposite colour then B guesses correctly.
    To translate this to the case where there are 20 prisoners, divide them into ten pairs (A1,B1), (A2,B2),...,(A10,B10) and have each pair (A,B) carry out the same strategy given above. This guarantees that exactly 10 of them will be correct.

    Xiachen: Glad you enjoyed it. If you do apply to Cambridge in the future, then I wish you the best of luck.
    Ahhhhh I see, thank you!
    Damn that was a tough riddle LOL
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    (Original post by theOldBean)
    Bustamove: Let me rephrase what Xiuchen said:

    Consider the easier problem where there are only two prisoners, A and B, and at least one has to get it right. The strategy is:
    i) A guesses whatever colour B is.
    ii) B guesses the OPPOSITE colour to what A is.
    This way, if they are the same colour, then A guesses correctly, and if they are the opposite colour then B guesses correctly.
    To translate this to the case where there are 20 prisoners, divide them into ten pairs (A1,B1), (A2,B2),...,(A10,B10) and have each pair (A,B) carry out the same strategy given above. This guarantees that exactly 10 of them will be correct.

    Xiachen: Glad you enjoyed it. If you do apply to Cambridge in the future, then I wish you the best of luck.
    Thank youuuuu <3
    Did you come up with the riddle yourself btw? Its really good!!!
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    (Original post by Xiuchen)
    Thank youuuuu <3
    Did you come up with the riddle yourself btw? Its really good!!!
    Alas, no. I got it from this:
    http://www.math.vt.edu/people/brown/doc/dozen_hats.pdf

    There is an extremely famous hat puzzle (PLEASE don't google it if you haven't already seen it - the problem is too nice to spoil by looking up the solution) which I shall give below:

    Four prisoners; four hats. They are buried in the sand as shown below:

    A-->B-->C-->WALL<--D
    R---B---R----------------B

    A is wearing a red hat and can see B and C. He can't see D because there is a wall in the way.
    B is wearing a blue hat and can see C. He can't see A or D.
    C is wearing a red hat, and can't see anyone else.
    D is wearing a blue hat, and can't see anyone else.

    They are told that each prisoner is wearing either a red or blue hat, and that there are two of each. They are not allowed to communicate in any way. They are told that if, within 10 minutes, one of them correctly guesses the colour of their hat, then they will be released. If more than 10 minutes go by, or anybody guesses incorrectly, then they will all be executed.

    After 9 minutes, one of them correctly states the colour of his hat. Who was it, and how did he know?
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    (Original post by theOldBean)
    Alas, no. I got it from this:
    http://www.math.vt.edu/people/brown/doc/dozen_hats.pdf

    There is an extremely famous hat puzzle (PLEASE don't google it if you haven't already seen it - the problem is too nice to spoil by looking up the solution) which I shall give below:

    Four prisoners; four hats. They are buried in the sand as shown below:

    A-->B-->C-->WALL<--D
    R---B---R----------------B

    A is wearing a red hat and can see B and C. He can't see D because there is a wall in the way.
    B is wearing a blue hat and can see C. He can't see A or D.
    C is wearing a red hat, and can't see anyone else.
    D is wearing a blue hat, and can't see anyone else.

    They are told that each prisoner is wearing either a red or blue hat, and that there are two of each. They are not allowed to communicate in any way. They are told that if, within 10 minutes, one of them correctly guesses the colour of their hat, then they will be released. If more than 10 minutes go by, or anybody guesses incorrectly, then they will all be executed.

    After 9 minutes, one of them correctly states the colour of his hat. Who was it, and how did he know?
    OK, I'm not going to try and figure it now haha, I need to revise for GCSEs, I'll do it tomorrow, outch my brain hurts after your first puzzle!!
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    (Original post by theOldBean)
    Alas, no. I got it from this:
    http://www.math.vt.edu/people/brown/doc/dozen_hats.pdf

    There is an extremely famous hat puzzle (PLEASE don't google it if you haven't already seen it - the problem is too nice to spoil by looking up the solution) which I shall give below:

    Four prisoners; four hats. They are buried in the sand as shown below:

    A-->B-->C-->WALL<--D
    R---B---R----------------B

    A is wearing a red hat and can see B and C. He can't see D because there is a wall in the way.
    B is wearing a blue hat and can see C. He can't see A or D.
    C is wearing a red hat, and can't see anyone else.
    D is wearing a blue hat, and can't see anyone else.

    They are told that each prisoner is wearing either a red or blue hat, and that there are two of each. They are not allowed to communicate in any way. They are told that if, within 10 minutes, one of them correctly guesses the colour of their hat, then they will be released. If more than 10 minutes go by, or anybody guesses incorrectly, then they will all be executed.

    After 9 minutes, one of them correctly states the colour of his hat. Who was it, and how did he know?
    Ok so A can see two people in front of him. He doesn't say anything, which means B and C are different colours. B therefore knows he is different to C, and as C is red B therefore knows he is blue?

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    (Original post by theOldBean)
    Alas, no. I got it from this:
    http://www.math.vt.edu/people/brown/doc/dozen_hats.pdf

    There is an extremely famous hat puzzle (PLEASE don't google it if you haven't already seen it - the problem is too nice to spoil by looking up the solution) which I shall give below:

    Four prisoners; four hats. They are buried in the sand as shown below:

    A-->B-->C-->WALL<--D
    R---B---R----------------B

    A is wearing a red hat and can see B and C. He can't see D because there is a wall in the way.
    B is wearing a blue hat and can see C. He can't see A or D.
    C is wearing a red hat, and can't see anyone else.
    D is wearing a blue hat, and can't see anyone else.

    They are told that each prisoner is wearing either a red or blue hat, and that there are two of each. They are not allowed to communicate in any way. They are told that if, within 10 minutes, one of them correctly guesses the colour of their hat, then they will be released. If more than 10 minutes go by, or anybody guesses incorrectly, then they will all be executed.

    After 9 minutes, one of them correctly states the colour of his hat. Who was it, and how did he know?
    B, bc the other's are quiet, a would've said if he knew, meaning b and c are different, therefor, b said his colour

    ok riddle:
    a man went to a party and drank some punch. he then left early. after, everyone else who drank from the same punch died from poisoning. why didn't the man die?
 
 
 
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