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View Full Version : Need help on a riddle!

k0n_
11-02-2006, 06:11 PM
So theres this prison gaurd and he lines up 10 prisoners. Each facing the person in front of thems back.

The gaurd has 10 hats. In 2 colors black and white. a random color combination (i.e. 6 black 4 white)

He places them on each one.

Now heres the riddle.

he tells them that if you want to live, you have to guess what color hat you have. If your wrong, you die. The only person that can garuntee your survival is the person behind you because he can tell the person in front of him.

They are allowed to say only ONE color and no other words. It has to be what they have on their head or they die.

They can choose to guess or help the guy in front of them,

the highest possible survival is 9 prisoners. How?



I can only figure out for 5 because the last person tells them, and the person in front of him lives, he dies. The third last person does the same. he dies. and front of him lives...

HALP!
Shay
11-02-2006, 06:44 PM
It is possible all could live.
k0n_
11-02-2006, 06:47 PM
tell me. how
HoppyPotty
11-02-2006, 06:50 PM
no, cuz the guy in the back dusnt have ne1 to tell him. everyone else does.

he got fucked over
k0n_
11-02-2006, 06:57 PM
but "everyone else" can only say one color

If they dont say theirs, they die
Shay
11-02-2006, 06:58 PM
Hmm, I think you're right. I'm just an idiot.

I think I got 9.

Example:

10 tells 9 black, 9 tells 8 black, 8 tells 7 black, 7 tells 6 black, 6 tells 5 black, 5 tells 4 white (5 dies), 4 tells 3 white, 3 tells 2 white, 2 tells 1 white. 10 hopes he has black.

.......:squint:

I should go to bed.
k0n_
11-02-2006, 07:04 PM
i think i said the riddle wrong somewhere. If any body heard of this before please correct me.
ThePope#15
11-02-2006, 08:50 PM
So if someone chooses to help the guy in front of them, they die?
sNiPe_
11-02-2006, 09:12 PM
no they can help.. but as they are helping, they have to say the color of their own hat as well...

so if im behind u.. and u have a black hat. i can help you by saying "black" but i might die cuz i might have a white hat.
Mattjam
11-02-2006, 10:49 PM
no they can help.. but as they are helping, they have to say the color of their own hat as well...

so if im behind u.. and u have a black hat. i can help you by saying "black" but i might die cuz i might have a white hat.

So in that case, Shay's got it right, right?
MooCowzRock
11-02-2006, 11:10 PM
no, its impossible...you got the riddle wrong, somewhere...
acdcrulsthecoop
11-04-2006, 03:02 AM
its possible they all live...if they all guess right
DarcSystems
11-05-2006, 06:12 AM
pretty simple. 10 is the only one who can't guarantee his survival. If "Black" is a code for an even number of black hats infront of him, and "White" is a code for an odd number of black hats infront of him, then the other 9 prisoners can guess what they are based on the first thing that 10 says.

Since the hats are random, and 4 white and 6 black, assuming 10 is wearing a black hat, he will be able to see 4 white, and 5 black. Thinking that there are 5 and 5, he shouts "White" because he see's 5 black hats infront of him (an odd number)

now number 9 only sees 4 black hats infront of him, and based on 10's reply, he knows he's wearing one. He then shoults "black" and lives.

Now 8 knows the guy behind him was wearing a black hat, there were originally 5 black hats visible from prisoner 10, then 4 black hats visible from 9, if prisoner 8 still see's 4 black hats, then he in turn is wearing a white hat. He shouts white and lives.

etc...

So the prisoners base their response off of the prisoner behind them, and factor out how many black or white hats are still in play.
sNiPe_
11-05-2006, 08:48 PM
pretty simple. 10 is the only one who can't guarantee his survival. If "Black" is a code for an even number of black hats infront of him, and "White" is a code for an odd number of black hats infront of him, then the other 9 prisoners can guess what they are based on the first thing that 10 says.

Since the hats are random, and 4 white and 6 black, assuming 10 is wearing a black hat, he will be able to see 4 white, and 5 black. Thinking that there are 5 and 5, he shouts "White" because he see's 5 black hats infront of him (an odd number)

now number 9 only sees 4 black hats infront of him, and based on 10's reply, he knows he's wearing one. He then shoults "black" and lives.

Now 8 knows the guy behind him was wearing a black hat, there were originally 5 black hats visible from prisoner 10, then 4 black hats visible from 9, if prisoner 8 still see's 4 black hats, then he in turn is wearing a white hat. He shouts white and lives.

etc...

So the prisoners base their response off of the prisoner behind them, and factor out how many black or white hats are still in play.

makes sense.. but how would they be able to communicate to each other that "black" and "white" are the codes for EVEN NUMBER OF BLACKS and ODD NUMBER OF BLACKS?
canadianbacon58
11-05-2006, 10:57 PM
I think you guys are reading into the riddle too much, I see a scenario where 9 live. If I completely misunderstood the riddle, my bad.

But say, counting from the back of the line, the last four prisoners are wearing white hats, and the next 6 are wearing black. If they all decide to help the person in front of them, then the person at the very back would say "white". Not only would he be helping the person in front of him, but he'd also save his own ass inadvertantly. Follow this scenario all the way down the line, and the only person who dies is the 4th person from the back of the line, because he would say "black", which would help the person in front of him, but he would die because he was wearing white.
MooCowzRock
11-05-2006, 11:16 PM
I think you guys are reading into the riddle too much, I see a scenario where 9 live. If I completely misunderstood the riddle, my bad.

But say, counting from the back of the line, the last four prisoners are wearing white hats, and the next 6 are wearing black. If they all decide to help the person in front of them, then the person at the very back would say "white". Not only would he be helping the person in front of him, but he'd also save his own ass inadvertantly. Follow this scenario all the way down the line, and the only person who dies is the 4th person from the back of the line, because he would say "black", which would help the person in front of him, but he would die because he was wearing white.

Oh, I thought it was random, like it could be b-b-w-w-b-w-b-b-b-w-w or something like that...

nevermind, then, you have the answer...
DarcSystems
11-07-2006, 07:53 AM
makes sense.. but how would they be able to communicate to each other that "black" and "white" are the codes for EVEN NUMBER OF BLACKS and ODD NUMBER OF BLACKS?

the original riddle gives them time to conspire about how they are going to do it. Like, they do the hat thing the next day or something, and all figure out a way to help each other out.