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.(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
TL DR: The order is irrelevant.
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theOldBean
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 21052015 19:34
Last edited by theOldBean; 21052015 at 19:36. 
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 21052015 19:38
(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.
I have a feeling i'm over thinking this 
theOldBean
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 21052015 19:41
(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 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.Last edited by theOldBean; 21052015 at 19:45. 
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 21052015 20:02
(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:ShowDivide 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.
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 XDLast edited by Bustamove; 21052015 at 20:10. 
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 21052015 20:03
(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. 
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 21052015 20:06
(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. 
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 21052015 20:15
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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 21052015 20:16
(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. 
theOldBean
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 21052015 20:25
(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!!!!
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.Last edited by theOldBean; 21052015 at 20:38. 
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 21052015 20:42
(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
If there were only two prisoners, how could you guarantee that exactly one of them guessed correctly? 
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 21052015 20:42
(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!!!!
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...
Edit= Sorry, meant to change same to oppositeLast edited by Bustamove; 21052015 at 20:50. 
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 21052015 20:44
(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. 
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 21052015 20:50
(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... 
theOldBean
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 21052015 20:51
(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
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.Last edited by theOldBean; 21052015 at 20:56. 
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 21052015 20:54
(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.
Damn that was a tough riddle LOL 
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 21052015 20:57
(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.
Did you come up with the riddle yourself btw? Its really good!!! 
theOldBean
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 21052015 21:07
(Original post by Xiuchen)
Thank youuuuu <3
Did you come up with the riddle yourself btw? Its really good!!!
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
RBRB
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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 21052015 21:10
(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
RBRB
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? 
boods8897
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 21052015 21:42
(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
RBRB
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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username1259045
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 21052015 22:02
(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
RBRB
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 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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