The Puzzle of 100 Hats
100 persons are standing in line, each facing the same way. Each person is wearing a hat, either red or blue, but the hat color is not known to the person wearing the hat. In fact, a person knows the hat color only of those persons standing ahead of him in line.
Starting from the back of the line (that is, with the person who can see the hat colors of all of other 99 persons), in order, and ending with the person at the head of the line (that is, with the person who can see the hat color of no one), each person exclaims either “red” or “blue”. These exclamations can be heard by all. Once everyone has spoken, a score is calculated, equal to the number of persons whose exclamation accurately describes their own hat color.
What strategy should the 100 persons use in order to get as high a score as possible, regardless of how the hat colors are assigned? (That is, what strategy achieves the best worst-case score?)
For example, if everyone exclaims “red”, the worst-case score is 0. If the first 99 persons exclaim the color of the hat of the person at the head of the line and the person at the head of the line then exclaims the color he has heard, the worst-case score is 1. If every other person exclaims the hat color of the person immediate in front and that person then repeats the color he has just heard, then the worst-case score is 50. Can you do better?
Hint: Instead of using just red and blue as the possible hat colors and exclamations, use N different colors.
The person standing at the back of the line can see all the people. I think it is important to consider that this problem has specifically 100 people in it, instead of some variable amount. This means the person in the back can see 99 people in blue and red hats in front of him. Since there are an odd number of hats the last person can see, it means that:
1. The number of red and blue hats the last person can see is not equal.
2. Either the red or the blue hats contain an odd number of hats, but not both.
With this in mind, we will be able to consistently communicate hat colors to all the participants. The person in the back of the line will begin by counting all the hats, and saying whatever color has an odd number of hats. The second person will then be able to count the hats in front of him (meaning all hats the last person saw minus the one he is wearing) if he sees an even number of the same color the first person said, he must be wearing that colored hat, otherwise he is wearing the other colored hat.
For example, if the person in the back sees an odd number of red hats, he will declare “red”. The second to last person will then count the amount of red hats he sees, if he sees an odd number, it means he is wearing a blue hat, if he sees an even number, it means he is wearing a red hat.
This process will work if every person in line keeps track of which colors have an odd or even count. Thus the worst case for this solution is 99, since the person at the front will answer solely based on what is in front of him, he will have a 50% chance of guessing correctly.
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