Prisoner Hat PUZZLE 10 Prisoners RED & BLUE Hats
video description
Guy 10 says the 9 hats he sees in front of him, it was never specified how many answers he could give, just that they had to be red/blue
So guy 10 says B R R B B B R B R and the other 9 just listen and figure out where they are in that list.
Another solution is saying the answer loudly means the person in front of you is red, but saying the answer not loudly means blue
So guy 10 would say blue quietly, guy 9 would say blue loudly because 8 is next and its red.
Another solution involves various other methods of sending a signal like in police movies when people need to answer yes or no on a phone by saying different words of coughing. These could include tapping the feet, coughing, sneezing.
Also a flaw is the riddle doesn't state they need to stay silent, so guy 10 could just tell each guy the correct answer,
Date: 2023-11-15
Related videos
Comments and reviews: 29
Shanmukha
Here's how the logic works: -
If there are an even number of red hats in front of the counter, the guessers can deduce the color of their hat by counting the number of red hats they see. If the count of red hats among the first 9 prisoners (excluding the counter) is even, then the color of their own hat must be blue. If it's odd, their own hat must be red. -
If there are an odd number of red hats in front of the counter, the guessers can similarly deduce the color of their hat. -
This strategy ensures that the counter can make a correct guess based on the parity (odd or even) of the red hats in front of him, and the guessers can correctly infer their hat color based on the counter's response. This way, at least 9 prisoners will survive, and they will all be released.
reply
Here's how the logic works: -
If there are an even number of red hats in front of the counter, the guessers can deduce the color of their hat by counting the number of red hats they see. If the count of red hats among the first 9 prisoners (excluding the counter) is even, then the color of their own hat must be blue. If it's odd, their own hat must be red. -
If there are an odd number of red hats in front of the counter, the guessers can similarly deduce the color of their hat. -
This strategy ensures that the counter can make a correct guess based on the parity (odd or even) of the red hats in front of him, and the guessers can correctly infer their hat color based on the counter's response. This way, at least 9 prisoners will survive, and they will all be released.
reply
fomori2
Took about 3 minutes to figure out that the last guy in line will say -blue- if he sees an even number of blue hats, and -red- if he see an odd number of blue hats. This alerts the prisoners what the starting target count of hats is, even or odd.
Each prisoner will count the number of blue hats and will know if his hat is blue, if the number of blue hats he sees is the opposite of the odd/even target count, or red, if the number of blue hats matches the odd/even target count.
Whenever someone calls out blue the odd/even count target switches to the other. Zero is considered even.
reply
Took about 3 minutes to figure out that the last guy in line will say -blue- if he sees an even number of blue hats, and -red- if he see an odd number of blue hats. This alerts the prisoners what the starting target count of hats is, even or odd.
Each prisoner will count the number of blue hats and will know if his hat is blue, if the number of blue hats he sees is the opposite of the odd/even target count, or red, if the number of blue hats matches the odd/even target count.
Whenever someone calls out blue the odd/even count target switches to the other. Zero is considered even.
reply
Uzumaki
There is another way, since the last prisoner has the leverage to answer wrongly, so the prisoners can come up with the strategy that the 10th prisoner will call out the colour of the hat of 9th prisoner, if the hat of the 9th prisoner has the same colour as that of the 8th prisoner(since he can see his hat) he will boldly and confidently call out the colour of his hat, but if the colours are different, then he will call out his hat colour in a feeble and soft voice indicating that the next prisoner has the opposite colour hat, and thats how the chain will continue
reply
There is another way, since the last prisoner has the leverage to answer wrongly, so the prisoners can come up with the strategy that the 10th prisoner will call out the colour of the hat of 9th prisoner, if the hat of the 9th prisoner has the same colour as that of the 8th prisoner(since he can see his hat) he will boldly and confidently call out the colour of his hat, but if the colours are different, then he will call out his hat colour in a feeble and soft voice indicating that the next prisoner has the opposite colour hat, and thats how the chain will continue
reply
Alabama
The puzzle initially stated that the red/blue hats were placed randomly on the ten prisoners. However, it did not state that there were five red hats and five blue hats. (Indeed, the first illustration showed the prisoners wearing six blue and four red hats) Conceivably, there could be eight blue and only two red hats, or any other combination of red and blue hats, including all the hats being either red or blue. Not knowing the number of red and blue hats in the total of hats would make logically solving the puzzle impossible.
reply
The puzzle initially stated that the red/blue hats were placed randomly on the ten prisoners. However, it did not state that there were five red hats and five blue hats. (Indeed, the first illustration showed the prisoners wearing six blue and four red hats) Conceivably, there could be eight blue and only two red hats, or any other combination of red and blue hats, including all the hats being either red or blue. Not knowing the number of red and blue hats in the total of hats would make logically solving the puzzle impossible.
reply
sorsocksfake
Prisoner 10 will count whether the red hats are even or odd. If even, he says blue; if odd, he says red.
Thus the other prisoners know whether the first 9 -including them- are an odd or even number.
They can also see the prisoners in front of them, and have heard the ones behind them. So they know whether the first 9 -except themselves- is an odd or even number.
If the two match, then their own hat must be blue. If they do not match, it's because their own hat is red. They speak accordingly.
reply
Prisoner 10 will count whether the red hats are even or odd. If even, he says blue; if odd, he says red.
Thus the other prisoners know whether the first 9 -including them- are an odd or even number.
They can also see the prisoners in front of them, and have heard the ones behind them. So they know whether the first 9 -except themselves- is an odd or even number.
If the two match, then their own hat must be blue. If they do not match, it's because their own hat is red. They speak accordingly.
reply
Gilberth
What about if they say the color of the hat in front of them in a louder or lower voice. Louder = Blue, Lower = Red. For example, the last prisoner sees that prisoner number 9 have a blue hat so he says any color with a louder answer so prisoner number 9 already know that he has a blue hat so before answer, he sees the hat of the number 8 and it's red so he says Blue in a lower voice. And so on that way, they only have to listen to the guy before them and see the hat of the one in front of them.
reply
What about if they say the color of the hat in front of them in a louder or lower voice. Louder = Blue, Lower = Red. For example, the last prisoner sees that prisoner number 9 have a blue hat so he says any color with a louder answer so prisoner number 9 already know that he has a blue hat so before answer, he sees the hat of the number 8 and it's red so he says Blue in a lower voice. And so on that way, they only have to listen to the guy before them and see the hat of the one in front of them.
reply
Laughter
There is another simplest way, prisonor 10 will say colour of 9th hat. Now 9th know the color of his hat, so if the color of his next is same as him he will answer his hat color within 1 second or if the color of his next hat is different he will answer his hat colour in 5 seconds. So this 1 or 5 second time interwal will give hint to the next prisnor and so on. As we know that 10th prisonor will be release even if he is incorrect so in this way all of them will be released.
reply
There is another simplest way, prisonor 10 will say colour of 9th hat. Now 9th know the color of his hat, so if the color of his next is same as him he will answer his hat color within 1 second or if the color of his next hat is different he will answer his hat colour in 5 seconds. So this 1 or 5 second time interwal will give hint to the next prisnor and so on. As we know that 10th prisonor will be release even if he is incorrect so in this way all of them will be released.
reply
Stan
This logic assumes that there is an even number of red & blue hats, which there is no indication of that being the case. It just says he (the guard) then randomly places either red hat or blue hat on each prisoners head. So there could be 6 blue and 4 reds for instance, in which case the logic for this falls apart, when the first (tallest guy) sees 4 reds, and yells out red, only to realize he has a blue hat as well, and gets killed. Thus this is an idiotic puzzle.
reply
This logic assumes that there is an even number of red & blue hats, which there is no indication of that being the case. It just says he (the guard) then randomly places either red hat or blue hat on each prisoners head. So there could be 6 blue and 4 reds for instance, in which case the logic for this falls apart, when the first (tallest guy) sees 4 reds, and yells out red, only to realize he has a blue hat as well, and gets killed. Thus this is an idiotic puzzle.
reply
What
Actually my solution is in morse code but for every hat red would be represented as -. (Dot)- And blue represented as -_ (Dash)- but the last prisoner would do 1 long click for blue and two short clicks for red starting from prisoner 1-9 therefore even if the last prisoner gets his hat colour wrong the others would know exactly which hat they were wearing
But then I realized its an indication and it wouldn't work
reply
Actually my solution is in morse code but for every hat red would be represented as -. (Dot)- And blue represented as -_ (Dash)- but the last prisoner would do 1 long click for blue and two short clicks for red starting from prisoner 1-9 therefore even if the last prisoner gets his hat colour wrong the others would know exactly which hat they were wearing
But then I realized its an indication and it wouldn't work
reply
Prasun
Answer: The jailor was colour blind. So the answer is -Grey-
Counter: Only Red and Blue are acceptable answers.
Answer: Grey
Counter: not the possible answers
All live until the conversation goes on. Jailor will give up first. 1 against 10.
Hence when the Jailor's shift ends all live. they would have lived anyway. The jailor does not execute anyone. The executioner does. Lol-
reply
Answer: The jailor was colour blind. So the answer is -Grey-
Counter: Only Red and Blue are acceptable answers.
Answer: Grey
Counter: not the possible answers
All live until the conversation goes on. Jailor will give up first. 1 against 10.
Hence when the Jailor's shift ends all live. they would have lived anyway. The jailor does not execute anyone. The executioner does. Lol-
reply
canale2303
I'd have answered differently. Prisoner 10 always says the hat's color of the prisoner 9. Prisoner 9 knows his hat's color. The agreed strategy is that each prisoner looks at the color of the prisoner in front of him and in case it is blu he SHOUTS his hat's color, otherwise it says it normally. So each prisoner knows his own hat's color just because of his partner's voice volume.
reply
I'd have answered differently. Prisoner 10 always says the hat's color of the prisoner 9. Prisoner 9 knows his hat's color. The agreed strategy is that each prisoner looks at the color of the prisoner in front of him and in case it is blu he SHOUTS his hat's color, otherwise it says it normally. So each prisoner knows his own hat's color just because of his partner's voice volume.
reply
vineet
If 9 prisoner give answers right then all the 10 Prisoner are released Hence there is Simple formula that 10th prisoner look at what 9th prisoner hat colour is and Say that. and 9th prisoner will look at 8th prisoner hat colour and say. and so far and so on. here if 10th Prisoner give wrong answer then remaining 9 are giving correct answer.
Please correct me if i am wrong.
reply
If 9 prisoner give answers right then all the 10 Prisoner are released Hence there is Simple formula that 10th prisoner look at what 9th prisoner hat colour is and Say that. and 9th prisoner will look at 8th prisoner hat colour and say. and so far and so on. here if 10th Prisoner give wrong answer then remaining 9 are giving correct answer.
Please correct me if i am wrong.
reply
Sagar
I have easier approach to this:
Last prisoner will tell second last prisoners hat color, if third last prisoner's hat color is equal to the second last prisoner's hat color then second last prisoner will tell his own hat color in louder voice otherwise in lower voice.
From this third last prisoner will conclude his own hat color
reply
I have easier approach to this:
Last prisoner will tell second last prisoners hat color, if third last prisoner's hat color is equal to the second last prisoner's hat color then second last prisoner will tell his own hat color in louder voice otherwise in lower voice.
From this third last prisoner will conclude his own hat color
reply
Rupesh
Why taking so much calculations they can simply say the hat colour infront of them so the other person says the same colour as his colour.
AND ALSO there is logic missing in your process who carries the 10th person's message to 8th, &7th, & and. 1st person? while they shouldn'tnt shout as per rules.
reply
Why taking so much calculations they can simply say the hat colour infront of them so the other person says the same colour as his colour.
AND ALSO there is logic missing in your process who carries the 10th person's message to 8th, &7th, & and. 1st person? while they shouldn'tnt shout as per rules.
reply
education
I have a much simpler solution:
Prisoner 10: Says whatever color in front of him.
Prisoner 9: Says the color said by prisoner 10, in a low pitched voice if the next one has the same color, and in a high pitched voice if the next one has a different color.
And so on.
reply
I have a much simpler solution:
Prisoner 10: Says whatever color in front of him.
Prisoner 9: Says the color said by prisoner 10, in a low pitched voice if the next one has the same color, and in a high pitched voice if the next one has a different color.
And so on.
reply
Baba
Looks like all those prisoners were mathematicians because a much simpler approach can be that Each prisoner (when asked, starting from the 10th) can call out the colour of the hat of the prisoner standing in front of him. This way, no need to do complex parity check calculation. -
reply
Looks like all those prisoners were mathematicians because a much simpler approach can be that Each prisoner (when asked, starting from the 10th) can call out the colour of the hat of the prisoner standing in front of him. This way, no need to do complex parity check calculation. -
reply
vadermike777
Easier solution, that doesn't require math at all. The rules dont state that you cant touch the shoulder of the person in front of you. You simply agree beforehand that the person behind you touches your left shoulder if your hat is red, or your right shoulder if blue.
reply
Easier solution, that doesn't require math at all. The rules dont state that you cant touch the shoulder of the person in front of you. You simply agree beforehand that the person behind you touches your left shoulder if your hat is red, or your right shoulder if blue.
reply
uberfu
There's a flaw to the puzzle -- the puzzle assumes that the jailor will go in order of prisoners from back to front. Which is the only way this puzzle works. If the jailor calls out random prisoners or starts fromt eh front the entire puzzle fails.
reply
There's a flaw to the puzzle -- the puzzle assumes that the jailor will go in order of prisoners from back to front. Which is the only way this puzzle works. If the jailor calls out random prisoners or starts fromt eh front the entire puzzle fails.
reply
Kids
Nice. This logic will
Work only if the total number of prisoners are even. If it is odd, difficult to solve. Because last prisoner can saw both Color ok odd number or even numbers. So difficult to give the clue to others.
reply
Nice. This logic will
Work only if the total number of prisoners are even. If it is odd, difficult to solve. Because last prisoner can saw both Color ok odd number or even numbers. So difficult to give the clue to others.
reply
KausCl
Too complex solution for a simple answer: just say the color of the prisioner in front of you. Even if the prisioner #10 have 50% of prob to match color, all others can answer with 100% prob, so everyone will be released.
reply
Too complex solution for a simple answer: just say the color of the prisioner in front of you. Even if the prisioner #10 have 50% of prob to match color, all others can answer with 100% prob, so everyone will be released.
reply
Rajan
What if there are 0 red hats? Will it be seen as an even count? I think that calling out the color with an odd number of appearances is better. Cos even if red is 0, there will be 9 blue hats and vice versa.
reply
What if there are 0 red hats? Will it be seen as an even count? I think that calling out the color with an odd number of appearances is better. Cos even if red is 0, there will be 9 blue hats and vice versa.
reply
Abcdefg
Why should they even do that. Except the last one, everyone can know their colour from the previous person. Only the last person has to apply the given strategy to guess his own colour.
reply
Why should they even do that. Except the last one, everyone can know their colour from the previous person. Only the last person has to apply the given strategy to guess his own colour.
reply
Time
I have ez way, if the last prisoner sees the guy ahead of him wearing a blue hat he says-bblue-(stressing b) and if it's red he says blue normally all others can do the same. Ez
reply
I have ez way, if the last prisoner sees the guy ahead of him wearing a blue hat he says-bblue-(stressing b) and if it's red he says blue normally all others can do the same. Ez
reply
Mridul
Instead of so much confusion. if they can call a colour. Then just call the colour of the hat which u see in front. all will answer correct and the last one will also survive
reply
Instead of so much confusion. if they can call a colour. Then just call the colour of the hat which u see in front. all will answer correct and the last one will also survive
reply
Jamm
If those guys were so smart, how did they get banged up in the first place? I guess they must have been living under a Biden-like feminist Marxist administration.
reply
If those guys were so smart, how did they get banged up in the first place? I guess they must have been living under a Biden-like feminist Marxist administration.
reply
mayank
Prisoner behind can see the color of the prisoner next to him. So he can tell thr color of hat to prisoner next to him and this way 9 of them will be right.
reply
Prisoner behind can see the color of the prisoner next to him. So he can tell thr color of hat to prisoner next to him and this way 9 of them will be right.
reply
alex
It can be solved by another method of shouting, like means if prisoner behind me answers loudly then my hat is red and if answer calmly then my hat is blue.
reply
It can be solved by another method of shouting, like means if prisoner behind me answers loudly then my hat is red and if answer calmly then my hat is blue.
reply
Benjyi
Your solution would make sense only if the quantity of hats of at least 1 color would be defined which isnt given so the whole video is absolete.
reply
Your solution would make sense only if the quantity of hats of at least 1 color would be defined which isnt given so the whole video is absolete.
reply
zoran
Can't just every prisoner say the color of the next guy? The first one (guy 10) might be wrong but each of then next guys will know their color.
reply
Can't just every prisoner say the color of the next guy? The first one (guy 10) might be wrong but each of then next guys will know their color.
reply
Add a review, comment
Other channel videos
