BrainTeaser 3

[Grymtongue Vazaelle]

Hey all,

Thinking this through there may be a way to save 9 people for sure with a 50-50 chance for the 10th person. From what is said the people can talk to one another before they get lined up to talk about a strategy. Talking to each other before they get put in line they can agree that the first person (the one who can see all the other hats) will use the word "Black" to confer that he/she sees an odd number of black hats. Conversely he/she would say "White" if they saw an even amount of black hats.

This way the next person to call can look and see how man hats of black color they can see and say the correct color of their hat.

A little illustration:

B W W W B W B W B ?

The ? is the person who goes first. This person can see all the other nine hats in front of them. So this person would say "White!" (They see an even amount of black hats) The next person in line would then know that ? had seen an even amount of black hats. Since this person can only see 3 black hats (they are wearing the fourth) they can assume that they are wearing a black hat and respond accordingly.

The person after that needs to then assume (since the person behind them is still alive) that there is NOW an odd amount of black hats since an even number minus 1 is an odd number. Looking ahead they see 3 black hats. Since 3 is an odd number they can then say that they are wearing a white hat.

So this is a way (If everyone understands) to save 9 people. The first person will only have a 50-50 chance though of saying the correct color.

[papyscott]

<bogus answer removed to protect the stupid /bonk>

ARRGG Rocsalt pointed out the bogusness of my answer before I could remove it. To make rocsalt's post make sense, the answer I gave was to have each person say the color of the hat in front of them. As roc points out this is bogus. You do know your hat color, but you can't both say it and your neighbors hat color.


Papy

[Rocsalt]