Well theoretically couldn’t all of them be saved except for the first?
each person simply tells the person in front of them what they’re wearing, and then, since no one can do this to the first guy, he has to take a shot. (unless there are 50 of each color and then you could just ask each prisoner to yell out their color and tally them) am i right?

oh no shit! that was a quick answer! I will take it back!

all can get free
thats an easy Game theory question

We can arrive at this by considering some of the simpler cases. Let’s assume that the warden placed only 1 Blue 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 Blue hat. Everyone else would then know that they had black hats and would say so.

Now consider the case where I have a Blue hat and someone else also has a Blue hat. I would look and see 1 Blue hat. If I had a black hat, that person would immediately announce that they had a Blue 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 Blue hats. So we both announce that we have Blue hats.

You can follow this pattern with 3 Blue 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 Blue hats, I will immediately announce “Blue” after 0 seconds. Everyone else will follow with “BLACK” right after. Similarly, if I saw 1 Blue hat, I would plan to say Blue 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 Blue hats and B black hats, then:
Everyone wearing a Blue hat will see W-1 Blue hats and B black hats.
Everyone wearing a black hat will see B-1 black hats and W Blue hats.

There are 3 cases:
If W < B then all Blue hats will say ‘Blue’ at the same time (10(W-1) seconds), and everyone else knows they are wearing red
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 Blue.
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 Blue hats and 49 black hats, and I was wearing a Blue 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 “Blue” along with all the rest of the prisoners wearing Blue 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.

Source(s):
http://en.wikipedia.org/wiki/Hat_Puzzle

and for those who took a communication course
this puzzle is like the Hamming code solution
see
http://en.wikipedia.org/wiki/Hamming_code

i love copy and paste, and the internet ofc
with them, i dont need to understand a thing ,

severus! what u post is a solution for a different type of questions! in ur case the guy can see every ones hat, but not his. in s3d examplem he gets to see only guys infront of him!

2nd, this was an interview question, unless they do not want that guy to join them, they will not ask him this question! it is a thesis man, and not a solution

I was thinking in the 10 secs thingy! but if u were the gardues u can figure a way to excute them all using this method!

i cant try, am simply stupid
i really am

and btw, the answer to my solution is u are that u are gaurds, u can kill them with or without a reason =) kinda like bohr answers x)

so, supreme ki, do u have an answer? u mentioned before u had an answer then u said u were too quick to judge, so what about now?

yea man! it is the forth comment here man!

u did not see it?

or u want another answer?

tell me!

50 is not enough, u can save much more

here is the biggest hint

u can SURELY save 99 prisoner

we have been through this, u r not allowed to convey any kind of message through saying blue or red except the fact that u said blue or red

u have to say in a little bit more details what do u meant by even and odd, because this can be the right answer and it can be not

i solved it….x-ray vision goggles, every prisoner except for the first one (he is poor) wears one and see the hat by himself. this is how u save 99 prisoners.

that would be a good one! or u know, the gaurds can color the rare half with a color and the front half with another and then excute them all heheheh!

ok, yuri officially got it. and he gave a new hint, it get to do with evens and odds

the answer spreme ki gave me is not posted and your method only saves 50 for sure, i am saying you can save 99 for sure.

each prisoner has one shot, so u can’t say the color of your own hat AND tell the one in front of you his color, only one of these two can be accomplished.

and there is no distribution for the colors at all, so not necessary 50/50

So you have a bullet to you head and 99 prisoners infront of you. How the heck can you count 99 people’s hats and then decide the odd and even in 5 seconds (or even 10 seconds)? And when did the prisoners decide on this scheme? Before they communed to be executed! Because usually, the 100 people will not be together… even if they were, chances are someone will miss up…