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!)
Your idea, a great one, only works as stated if there are an odd number
of prisoners. If there are an even number of prisoners. The
last prisoner will always say "BLUE", and no information is transmitted.
Here's how to fix it. The last prisoner calls out "RED" if he
sees an odd number of red hats and "BLUE" if he sees an even
number of red hats. Now all remaining prisoners can figure out
their hat color, no matter how many prisoners there are.