## 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 :-) |

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

### #451

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

### #452

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?

### #453

Posted 26 September 2011 - 08:38 PM

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

### #454

Posted 26 September 2011 - 11:34 PM

### #455

Posted 29 September 2011 - 08:10 PM

### #456

Posted 29 September 2011 - 08:54 PM

### #457

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.

### #458

Posted 11 October 2011 - 07:55 AM

### #459

Posted 03 November 2011 - 08:57 PM

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

### #460

Posted 03 November 2011 - 08:58 PM

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users