A group of people 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 his own eyes. Every night at midnight, a ferry stops at the island. If anyone has figured out the color of his own eyes, he [must] leave the island that midnight.
On this island live 100 blue-eyed people, 100 brown-eyed people, and the Guru. The Guru has green eyes, and does not know her own eye color either. Everyone on the island knows the rules and the properties stated above (except that they are not given the total numbers of each eye color) and is constantly aware of everyone else's eye color. Everyone keeps a constant count of the total number he sees of each (excluding themselves). However, they cannot otherwise communicate. So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes, but that does not tell him his own eye color; it could be 101 brown and 99 blue. Or 100 brown, 99 blue, and the one could have red eyes.
The Guru speaks only 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 with blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb, It is not a trick question, and the answer is logical. It doesn't depend on tricky wording, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
