Jump to content


Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.

Of course, you can also enjoy our collection of amazing optical illusions and cool math games.

If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.

Thanks and enjoy the Den :-)
Guest Message by DevFuse
 

Photo
- - - - -

Hats on a death row!! One of my favorites puzzles!


Best Answer bonanova, 08 January 2008 - 06:25 PM

Spoiler for Here's my idea
Go to the full post


  • Please log in to reply
470 replies to this topic

#451 OmniQ

OmniQ

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 23 September 2011 - 06:13 PM

cant the first person tell the person in front what colour hat they are wearing and so on... then so long as they are all truthfull all should be saved exept the first one who only has a 50/50 chance.


Thank you! I was wondering if I was the only one who thought of that.

The questions are "...find a way to guarantee the freedom of some prisoners tomorrow? How many?"

So I agree 100%, the best answer is:
Each person simply states the color of the person's hat in front of them.
If they are honest, that guarantees 19 live and the first person has a 50% chance!

Pretty good odds, but I still wouldn't want to be the first person in line, or in front of someone who had it out for me. 8^)

-Ken
  • 0

#452 Molly Mae

Molly Mae

    Moderator

  • Members
  • PipPipPipPip
  • 3188 posts
  • Gender:Male
  • Location:La Salle, Illinois, USA

Posted 23 September 2011 - 08:29 PM


Thank you! I was wondering if I was the only one who thought of that.

The questions are "...find a way to guarantee the freedom of some prisoners tomorrow? How many?"

So I agree 100%, the best answer is:
Each person simply states the color of the person's hat in front of them.
If they are honest, that guarantees 19 live and the first person has a 50% chance!

Pretty good odds, but I still wouldn't want to be the first person in line, or in front of someone who had it out for me. 8^)

-Ken


What if the guy behind you says black (indicating that your hat colour is black) but the person in front of you is red?
  • 0

A recipe for honey-pickled apples


Awards:

Bonanova Gold Star

Spoiler for Molly's Rules to Live By

#453 Kia

Kia

    Newbie

  • Members
  • Pip
  • 3 posts

Posted 26 September 2011 - 08:38 PM

Some of you guys seem to be missing the point. You cannot guarantee to be saved if you say the color of the person in front of you. In the worst case scenario, every other person is wearing a different color (black, red, black, red, ...). If each person says the color of the person in front of him, then everyone is dead. Although, using this method you can guarantee that 50% will survive (the prisoners standing in the even positions) and the ones in the odd positions have a 50/50 chance, so that's 75%. But the real answer guarantees all will survive except the first one who has a 50% chance. Here how it works:

Before hands the prisoners need to agree that the person at the end of the line counts the red hats in front of him and if even he calls red, and if odd, he calls black. From that point on, everyone needs to keep track of two things:The number of red hats in front, and how many people (aside from the first person) have called red. Then follow this basic logic

If the first person called red and these two numbers are both even or both odd, then call black, otherwise red
If the first person called black and these two numbers are both even or both odd, then call red, otherwise black

Let's go over a simple example to clear this out. Let's say there are 10 prisoners instead of 20 and this is the hat order:
r, b, r, r, b, r, b, b, b, r

The first person sees 4 red hats and calls red (luckily he survives)
Based on the logic above (first line)
The second person sees 4 red hats and 0 other prisoner has called red. Both numbers are even so he calls black
The third person sees 3 red hats and 0 other prisoner has called red. One number is odd and one even, so he calls red
The fourth person sees 2 red hats and 1 other prisoner has called red. One number is odd and one even, so he calls red
The 5th person sees 2 red hats and 2 other prisoners have called red. Both numbers are even so he calls black
The 6th person sees 1 red hat and 2 other prisoners have called red. One number is odd and one even, so he calls red
The 7th person sees 1 red hat and 3 other prisoners have called red. Both numbers are odd so he calls black
The 8th person sees 1 red hat and 3 other prisoners have called red. Both numbers are odd so he calls black
The 9th person sees 1 red hat and 3 other prisoners have called red. Both numbers are odd so he calls black
The 10th person sees 0 red hats and 3 other prisoners have called red. One number is odd and one even, so he calls red

In this case everyone was saved, but the first one was lucky. I'll do another example with odd number of red hats in fron of the first person

r, r, r, b, b, r, b, r, r, b

The first person sees 5 red hats and calls black (dead)
Based on the logic above (second line)
The second person sees 4 red hats and 0 other prisoner has called red. Both numbers are even so he calls red
The third person sees 3 red hats and 1 other prisoner has called red. Both numbers are odd so he calls red
The fourth person sees 3 red hats and 2 other prisoners have called red. One number is odd and one even, so he calls black
The 5th person sees 3 red hats and 2 other prisoners have called red. One number is odd and one even, so he calls black
The 6th person sees 2 red hats and 2 other prisoners have called red. Both numbers are even so he calls red
The 7th person sees 2 red hats and 3 other prisoners have called red. One number is odd and one even, so he calls black
The 8th person sees 1 red hat and 3 other prisoner has called red. Both numbers are odd so he calls red
The 9th person sees 0 red hats and 4 other prisoner has called red. Both numbers are even so he calls red
The 10th person sees 0 red hats and 5 other prisoner has called red. One number is odd and one even, so he calls black

I hope the was clear enough
  • 0

#454 omnifrog

omnifrog

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 26 September 2011 - 11:34 PM

I'm thinking about this as an information theory problem. It will take 4 bits of information to encode the next 16 possibilities. This means that at best, 16 are saved and 4 have a 50% chance.
  • 0

#455 Argon

Argon

    Newbie

  • Members
  • Pip
  • 11 posts

Posted 29 September 2011 - 08:10 PM

like flipping a coin, the answer is a 50/50 guess considering we do not know how many black and red colors are used..
  • 0

#456 Molly Mae

Molly Mae

    Moderator

  • Members
  • PipPipPipPip
  • 3188 posts
  • Gender:Male
  • Location:La Salle, Illinois, USA

Posted 29 September 2011 - 08:54 PM

Curious, but what's the best solution for 3 hat colours and 20 prisoners?
  • 0

A recipe for honey-pickled apples


Awards:

Bonanova Gold Star

Spoiler for Molly's Rules to Live By

#457 storygirl101

storygirl101

    Junior Member

  • Members
  • PipPip
  • 47 posts

Posted 10 October 2011 - 07:40 AM

YOU COULD JUST ASK YOUR OTHER CELLMATES, "WHAT COLOR IS MY HAT???" IF YOUR CELLMATES AREN'T SICK AND CRUEL, THEY WOULDN'T LIE, GUARANTEEING YOUR SURVIVAL.



Yeah, but, you have to guess BEFORE that night. And you can't change your answer after that night.
  • 0

#458 seshan

seshan

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 11 October 2011 - 07:55 AM

According to me, The tone of the word plays a major role to save 19 peoples!!! last one can see the one who is standing before him.. Black or red in low tone means the hat is red in color and Black or red in brisk tone the its blacky!!! by this 19 can be saved but worrying for the last one
  • 0

#459 Laws

Laws

    Newbie

  • Members
  • Pip
  • 2 posts

Posted 03 November 2011 - 08:57 PM

The people are standing in a single file line, all facing one direction. The person in back is seen by noone. I'd say it's up to a test of memory. The guy in back has to spoit off as many colors, in order, as he possible can and be sent to death. While the people in front of him have to remember the order. When memory fails. It's your turn to spoit off the order of hats. Hopefully saving the maximum amount of people. There's really no solution to save the poor basterd in back. Unless he takes his chances with a 50\50 guess. Thus leaving the person in front of him with the same chances...

Edited by Laws, 03 November 2011 - 09:00 PM.

  • 0

#460 Laws

Laws

    Newbie

  • Members
  • Pip
  • 2 posts

Posted 03 November 2011 - 08:58 PM

Ah... I love the tone idea. good one!
  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users