Question: There are 100 prisoners lining up to go to jail. Each prisoner is wearing a hat that is either black or white. The prisoners don't know their own hat color, just the hat color of those in front of them in line (the first prisoner in line can't see anyone's hat and the last prisoner can see everyone's hat except their own). Starting from the back, one of the guards asks each prisoner what color their hat is. If they are correct they get to go free but if they are wrong they go to jail.If the prisoners get to discuss a plan, how can at least 99 of them be saved?
Answer: The back prisoner will yell "black" if there is an odd number of black hats in front of him and "white" if there is an even number of black hats in front of him. The next prisoner will then count the number of black hats in front of him and if it was odd and is now even, or vice versa, then that person knows what color their hat is. The next person then knows their hat color based on what the people before them said and how many black hats are in front of them. In this way the front 99 prisoners will know their hat color and will be set free. The prisoner in the back who goes first has a 50 percent chance of being set free.