a puzzle(not difficult)

[alexd231232]

All right, here it is

Lets say you have 100 people(this number doesn't matter, it can be any number) and they are all brought into a room before the "experiment" begins and are told exactly what will happen in the experiment. Afterwards, they are seperated and placed into rooms alone.
Once in the room, each of them get a hat, either white or black. This hat is placed on their heads so they can't ever see what color they have on their heads. Also, each room has televisions which have a live telecast feed of all the other people in the expirement. For examle, if I am person 1, my room will have a TV which shows me every other person and the color hat they are wearing.
Now comes the tricky part. The first person is asked, what color hat do you have on, black or white? He answers and if he is wrong, he dies. The question then moves onto person two and he is asked the same exact question with the same consequences. The question is asked of everyone in the expirement.
The puzzle comes in this, how can this be done so that zero or one person is ever killed.

The people are allowed to work out a strategy before hand in the meeting room but cannot speak at all to each other when they get to their individual rooms. They cannot make any gestures by face or otherwise to indicate anything to the other people.

I'm not very good at explaining so if you have any questions just lemme know.

Since everyone in the experiment has a video camera pointed at them and can discuss the strategy before going into the individual rooms I see it going like this. When they are all together this strategy is discussed:
1) Once they are in individual rooms each person looks the screen and specifically at two people (the ones on each side of them, one to the right and one to the left, in terms of room sequence) If they have the same hat color then you face the camera, if not then turn away.
Then everyone will know whether or not they match an individual two rooms down and can make his/her decision.
This will only work if you know the sequence of rooms.

[DougOzzzz]

<font color="white">
The first person will have the following strategy:

Guess black if he counts an odd number of black hats, otherwise guess white. Suppose he guesses black, since he observes 53 black hats and 46 white hats. He's wrong, he dies, but it doesn't matter. The 2nd person now knows that the 1st person saw an odd number of black hats and even number of white hats. If his hat is the same color as the 1st guys then he will see an odd number of black hats (also odd white hats). Otherwise he should see an even number of black hats, and know that his hat is black. Everyone else will follow the same strategy.
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[alexd231232]

You don't know room sequence, its random, sorry i forgot to include that.

[alexd231232]

Doug is correct. I know this wasn't anythign too difficult, just something i enjoyed when i heard it.