Puzzle: How can you save the 100 prisoners from execution?

We can arrive at this by considering some of the simpler cases. Let's assume that the warden placed only 1 white hat on my head and 99 black hats on everyone else's head. So at the time when the warden said start, I would look around and see 99 black hats. Since there has to be at least one 1 hat, I would immediately announce that I had a white hat. Everyone else would then know that they had black hats and would say so.

Now consider the case where I have a white hat and someone else also has a white hat. I would look and see 1 white hat. If I had a black hat, that person would immediately announce that they had a white hat. But that doesn't happen. Instead they wait. Now since neither of us was able to announce right away, we know that there must be 2 white hats. So we both announce that we have white hats.

You can follow this pattern with 3 white hats, etc. but it requires that you have a plan in place for timing. Let's say that everyone will wait ten seconds for every hat they see less of. So if I see no white hats, I will immediately announce "WHITE" after 0 seconds. Everyone else will follow with "BLACK" right after. Similarly, if I saw 1 white hat, I would plan to say WHITE after 10 seconds. If no one announced after 0 seconds, I would be correct. Then everyone else would say "BLACK" after that.

It doesn't matter the number of hats or whether there are more or less of one color as long as we wait a discrete period of time to announce. I chose periods of 10 seconds so that there wouldn't be a chance of mistakes.

If there are W white hats and B black hats, then:

Everyone wearing a white hat will see W-1 white hats and B black hats.

Everyone wearing a black hat will see B-1 black hats and W white hats.

There are 3 cases:

If W < B then all white hats will say 'White' at the same time (10(W-1) seconds), and everyone else knows they are wearing black.

If B < W then all black hats will say 'Black' at the same time (10(B-1) seconds), and everyone else knows they are wearing white.

If W = B then all prisoners will know their color at the same time. (10(B-1) seconds, or 10(W-1) seconds, they are equivalent).

Let's take an example. If there were 51 white hats and 49 black hats, and I was wearing a white hat. Then I would plan to say "black" after 490 seconds. However, the people in black hats would beat me to the punch saying "black" after 480 seconds. Then I would say "white" along with all the rest of the prisoners wearing white hats.

The "hat puzzle" is described in detail at Wikipedia and I've included the link below. I think the first answerer quoted from there but didn't include the reference.

1

Each prisoner counts the number of white hats and black hats they see, and waits for 10N seconds, where N is the lower of the two numbers. After that time, (providing the other prisoners are all doing the same thing), he must be wearing the hat of the colour which he is seeing fewer of.

Everyone wearing a white hat will see W-1 white hats and B black hats.

Everyone wearing a black hat will see B-1 black hats and W white hats.

* If W < B then all white hats will say 'White' at the same time (10(W-1) seconds), and everyone else knows they are wearing black.

* If B < W then all black hats will say 'Black' at the same time (10(B-1) seconds), and everyone else knows they are wearing white.

* If W = B then all prisoners will know their colour at the same time. (10(B-1) seconds, or 10(W-1) seconds, they are equivalent).

prisoner #1 must count the number of black hats he sees subtract this from 100 to get a potential number of white hats. He must then count the number of white hats he sees. If this count is 1 less than the difference he obtained by subtraction then he must answer "white". If the two answers are the same then he must answer "black". Assuming no prisoners are removed from sight then each prisoner must use this locution when answering the warden's question.

Black And White Caps

All have the same covered what expert one person

all one hundred persons line up in a line. One tap on the butt for white; Two taps on the shoulder for black. That way everyone cooperates and everyone enjoys from the death sentence.

Who said maths were essential?

It is agreed in the 20 mins that everyone will put a white hat on the next prisoner. Since there is an unlimited number of hats of either color....everyone grabs or asks for a white hat and places it on the person infront of them.

This is the same as the "brown eyes, blue eyes" question. Each person will see the other person, and then repeat. They will simulate "days". As soon as someone figures out he/she has a white cap, he/she will leave to a part of the room known as "whiteland". Same thing as the "brown eyes, blue eyes" problem, except that the number of white caps is the number of "days" it takes to figure out if one has a white cap or not.

Or a simpler answer would to be that all prisoners take off their hats...

First of all, prisoners aren't smart - or else they wouldn't be in prison

and second, they all say the next guys color and then when asked they say the color they were told