Question: Three people enter a room and have a green or blue hat placed on their head. They cannot see their own hat, but can see the other hats. The colour of each hat is purely random. All hats could be green, or blue, or 1 blue and 2 green, or 2 blue and 1 green.
They need to guess their own hat colour by writing it on a piece of paper, or they can write “pass”. They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.
If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing. What strategy would give the best chance of success?
Answer: A simple strategy could be to elect a person who would always guess randomly. This gives them a 50% chance of winning.
There exists a better strategy than this. If you see two green hats or two blue hats write down the opposite colour, else pass. The following cases can occur with equal probabilities-
|GGB||– – B||Win|
|GBG||– B –||Win|
|GBB||G – –||Win|
|BGG||B – –||Win|
|BGB||– G –||Win|
|BBG||– – G||Win|
This gives them a 75% chance of winning!
How much chance do they have if four people enter the room?