What strategy would give the best chance of success? 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.
bjahnavi:
i didnt understand could u explain better plz.
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Elect one person to be the guesser, the other two pass. The guesser chooses randomly "green" or "blue". This gives them a 50% chance of winning.
The better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down "pass".
It works like this ("-" means "pass"):
Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose
Result: 75% chance of winning!
The better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down "pass".
It works like this ("-" means "pass"):
Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose
Result: 75% chance of winning!
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