Puzzle: How can you save the 100 prisoners from execution?: HOW CAN WE CALL THIS ANSWERED????

Question

First, look up that "answered" riddle. So first, there is no method for giving a 100% chance of survival for the first prisoner to guess. His chances are always 50/50. What he can do, is pass information forward. But what information would be useful to ALL 99 other prisoners? None of the answers work yet.

It would be agreed that the last man in line would count the number of hats. If he saw and odd number of black hats, he would say his was black. If he saw an even number of black hats, he would answer white. They agree to this. He gives his answer and 50/50 is freed or not. Working forward, the next prisoner in line can then count black hats. If he sees a number that is opposite (as in odd vs even) to what the rear-most man reported, then he knows he has a black hat. Each person would then keep track of each time a person reported a black hat and would switch what their guess would be when it gets to them back and forth. Try it with 5 prisoners to see it work.

Answer 1

§ Note first that the prisoners are not arranged on a line.... there is no first and last person... they will have to announce the color of their hats correctly at the appropriate time.

All they know is that there is at least one hat of one color. There is a black and a white color.

I will still follow the plan of cg... the answerer from the first question posed... the plan works... as this is mathematically and logically sound.

The general rule is:
If there are fewer colors of one type that you see, call the color of the fewer type after (10n) seconds. n is the number of fewer colors you see. If the other group calls first... then you know that the color of your hat is the other one.

Say on 5 prisoners...
For simplification purposes, let us assume that the number of white hats is less than or equal to black hats. If this is not the case, simply reverse the color... and the rule still works.

Supposing we have WBBBB
since prisoner1 sees 0 white hats, he has to call the color "white". Then the other prisoners must know they are black.
Let us observe prisoner2, he sees 1 white and 3 blacks... he intends to call "white" after 10 seconds... but prisoner1 called out first... so his color now must be black. The same goes with the other prisoners.


now suppose we have WWBBB
then prisoners 1 & 2 will call the color "white" after 10 seconds. this time prisoners 3,4&5 will have a problem since they see two white and two black... they dont know which color they will call... but now since the first two prisoners had beaten them in the call, there is no problem, the rest of the prisoners must be black.

Again, if there are more white hats really, reverse the color from the earlier arguments. The plan still works.



This trick will work for all number of prisoners, even 100, even greater. As long as the prisoners know how to count correctly and quickly, hahaha.

Say in 100 prisoners. 40 wear white and 60 wear black.
Those wearing white will see 39 white hats and 60 black hats.They will call "WHITE" after 390 seconds. Those wearing black will see 40 white hats and 59 black hats. They intend to call "WHITE" after 400 seconds... but they were beaten by 10 seconds by the first group so they are the ones wearing black.

Amazing!

Edit: Ummm... as to your plan...
what if the order is
WWBBB ... the last man sees even number of whites and blacks and he will call "WHITE" ... they will all die.

Answer 2

Okay, so do all of the prisoners have watches? How can we be certain they all have a perfect and consistent idea of how long a second is. I mean, the 1 mississippi thing might work, but..um..well I have my doubts about that. Second of all, you may want to read the method listed before using an example counter to it to try to say it doesn't work. The white hats don't apparent;y matter. Only the black. So running your scenario, it would still work properly. That was a very long post considering how little it actually said.

Answer 3

each and each prisoner counts the kind of white hats and black hats they see, and waits for 10N seconds, the place N is the decrease of the two numbers. After that component, (offering the different prisoners are all doing a similar element), he might desire to be wearing the hat of the colour which he's seeing fewer of. all and sundry wearing a white hat will see W-a million white hats and B black hats. all and sundry wearing a black hat will see B-a million black hats and W white hats. * If W < B then all white hats will say 'White' on a similar time (10(W-a million) seconds), and all and sundry else knows they are wearing black. * If B < W then all black hats will say 'Black' on a similar time (10(B-a million) seconds), and all and sundry else knows they are wearing white. * If W = B then all prisoners will understand their shade on a similar time. (10(B-a million) seconds, or 10(W-a million) seconds, they are equivalent).

Answer 4

I don't have five prisoners to try it on. Now I'll have to go out and kidnap three more girls.
I say execute them all. They must have done something to deserve the death penalty.Why mess around with hats?