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.
My solution would be that person number 1 then the 4th,8th,12th and so on would analise the next 3 people and when asked for the colour of their hat would give the majority colour of these three hats as their answer. In this way the next three would have at least a 66% success rate and he wfould have a 50% chance. It would be a responsibility of each perdson to keep count of people as they are called. I do not beleive that in keeping to the spirit of the puzzle that you could say anything other than one word red or blue so that no hidden coded messages could be related. Not sure of the overall formula to calculate the end success rate for the above sinerio, I'm sure someone will supply it.