A group of prisoners is under sentence of death and the warder decides to give them a test to gain their freedom. He tells them, "I will place a red or blue hat on each of your heads and then I'm going to arrange you in random order in a row so that no prisoner will be able to see his own hat but each one will see all the hats in front of him. Starting with the guy at the back each of you in turn must loudly say what color hat you think you have. Correct answers will go free, incorrect ones will be thrown to the alligators in the moat. I will give you time for a brief meeting before we start, so you can plan your optimum strategy."
What strategy can the prisoners  there are N of them  adopt to improve their odds above 50:50?
Hint: They need to agree on a strategy which allows each person to identify his/her own hat while simultaneously providing as much information as possible for all those in front.
(In reply to
Solution by Old Original Oskar!)
Nice solultion O.O.O! Yet, I agree with Steve. The information given by calling out one's own color must be revealed for a single colored hat, which would also reveal the quantity of the other hats seen, and not if only both hats seen are of an odd quantity.
Edited on July 15, 2006, 2:23 pm

Posted by Dej Mar
on 20060715 14:13:58 