Famously the hardest logic puzzle in the world: Blue Eyes

https://xkcd.com/blue_eyes.html

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 answer

The answer is >!the 100 people with blue eyes leave on the 100th night.!< Double spoiler tagged so it hopefully works on every Lemmy client.

For the logic, imagine if it were actually just three people: the guru, one brown eyes, and one blue eyes. For the sake of clarity, I’ll speak through the perspective of a blue eyes in these examples. Guru says “I see someone with blue eyes.” Brown eyes also sees someone with blue eyes. But I see no one with blue eyes. I deduce that I must have blue eyes, and leave that same night.

Now >!imagine there are two blue eyes, two brown eyes, and the guru. Guru says “I see someone with blue eyes.” Brown eyes both see two sets of blue eyes, but I only see one set of blue eyes. I figure if the other person sees no other blue eyes, they’ll leave the first night. They don’t, which can only mean that they also saw someone with blue eyes, which must be me. We both leave the second night.!<

You >!can expand this logic all the way out to 100, so on the 100th night, all the blue eyes leave.!<