This puzzle is one of the hardest puzzle in the world. If you are able to solve this, believe me you are genius.
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 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 or anyone lying or guessing, 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.”
The knowledge of each islander consists of:
- the color of the eyes of every other islander.
- any past pronouncement from the guru.
- the history of who left the island on previous days (including their eye color), which provides knowledge about other’s knowledge (that either they did or did not know their own eye color on previous days).
Blue eyed people leave on the 100th night.
If you (the person) have blue eyes then you can see 99 blue eyed and 100 brown eyed people (and one green eyed, the Guru). If 99 blue eyed people don’t leave on the 99th night then you know you have blue eyes and you will leave on the 100th night knowing so.
Imagine a simpler version of the puzzle in which, on day #1 the guru announces that she can see at least 1 blue-eyed person, on day #2 she announces that she can see at least 2 blue eyed people, and so on until the blue-eyed people leave.
So long as the guru’s count of blue-eyed people doesn’t exceed your own, then her announcement won’t prompt you to leave. But as soon as the guru announces having seen more blue-eyed people than you’ve seen yourself, then you’ll know your eyes must be blue too, so you’ll leave that night, as will all the other blue-eyed people. Hence our theorem obviously holds in this simpler puzzle.
But this “simpler” puzzle is actually perfectly equivalent to the original puzzle. If there were just one blue-eyed person, she would leave on the first night, so if nobody leaves on the first night, then everybody will know there are at least two blue-eyed people, so there’s no need for the guru to announce this on the second day. Similarly, if there were just two blue-eyed people, they’d then recognize this and leave on the second night, so if nobody leaves on the second night, then there must be a third blue-eyed person inspiring them to stay, so there’s no need for the guru to announce this on the third day. And so on… The guru’s announcements on the later days just tell people things they already could have figured out on their own.