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by weinzierl
3186 days ago
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A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph. On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes. The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following: "I can see someone who has blue eyes." Who leaves the island, and on what night? There are many versions of this puzzle, this is the xkcd version (https://www.xkcd.com/blue_eyes.html) |
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Imagine three girls sitting in a circle, each wearing either a red hat or a white hat. Suppose that all the hats are red. When the teacher asks if any student can identify the color of her own hat, the answer is always negative, since nobody can see her own hat. But if the teacher happens to remark that there is at least one red hat in the room, a fact which is well-known to every child (who can see two red hats in the room) then the answers change. The first student who is asked cannot tell, nor can the second. But the third will be able to answer with confidence that she is indeed wearing a red hat.