Question

Heya all !

THE 100 HAT RIDDLE PROBLEM


Hello there Brainliacs! Today, I've got an interesting puzzle. It's a little bit tricky and can be confusing, so make sure you're thinking it through :)


THE 100 HAT RIDDLE 

100 prisoners are lined up by an executioner, who places a red or blue hat upon each of their heads.

The prisoners can see the hats of the people lined up in from of them, but they cannot look at the hats behind them, or at their own.

Starting at the back of the line, the executioner asks the last prisoner to state the colour of his hat.

In order to live, the prisoner must answer correctly. If he doesn't, he is killed 'instantly and silently.'

This means that the other prisoners will hear the answer, but will not know whether or not it was correct. 

The night before the line-up, the prisoners can discuss a strategy to help them survive. What should they do?

Enjoy guys :)

Answer 1

Heya!
here's your answer!
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This is a really tricky question indeed.
I had to do a lot to reasearch work and hear researchers like Terrence Gaffney who's also a mathematics professor. Credit goes to him.

the easy yet tricky and understandable solution is here as per Gaffney :-


• Since 100 prisoners is a really big number to work with. Terrence Reduced it to just two people

• There was a little complication in the solution that there's 99% chances of survival for 99 prisoners , but the one who's standing in the last will have only 50% chances of being alive.


• The last person can save the prisoner who's in front of him by shouting out the color of his hat.


• If there are 3 people , then the third one in line can see either zero , one , or two blue/red hats as per the people in front of him are wearing.


• The two possibilities out of the three have something in common . Guess what? Let's see :- if we suppose the last prisoner shouts “blue” then he can tell no.1 and no.2 that he sees an even number of blue hats. (Everything's dependent on even or odd number of hats)


• the second prisoner, looks in front and by counting the number of blue hats, knows his must be blue ( if he sees one blue hat, and red if he sees no blue hats.) The last prisoner agrees to shout “red” if the number of blue hats seen is odd.


• Then if number2 sees a blue hat on Number1, his must be red, and then if Number1 has a red hat, his hat must be the blue one. By shouting out the color of his hat, Number1 also knows his hat color.


• 2 blue hats or two red ones in a row must mean he wears blue, while one blue and one red means he wears red.


• Seems like this always works, because there are only two possibilities as far as the number of blue hats worn–they are either even or odd as told before.


• So, check as in the three person case that using this strategy (“blue” for an even number of blue hats “red” for an odd number)


• Now, the last prisoner tells number99 the color of his hat, and then each prisoner in turn can learn the color of his hat by taking into account the number of blue hats he can see, the number of blue hats number100 saw and the number of prisoners behind him wearing blue hats and red ones and so on...


Source of My understanding : Internet.

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Hope this helps you!