Brainteaser

Eye Colors

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Brainteaser

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Question

Eye Colors

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 passage.

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."

Solution

The people who leave are the \(100\) blue-eyed islanders, and they all leave together on the \(100^\text{th}\) night after the Guru speaks.

The key idea is that the Guru does not give anyone new private information. Every blue-eyed person already sees plenty of blue-eyed people. What the Guru gives them is new public information: it is now common knowledge that there is at least one blue-eyed person on the island. That shared fact lets them begin an induction.

Suppose there were only \(1\) blue-eyed person on the island. That person would look around and see no one with blue eyes. Before the Guru speaks, they cannot rule out the possibility that there are zero blue-eyed people. But once the Guru says

“I can see someone who has blue eyes,”

that single blue-eyed person realizes the Guru must be referring to them. So they would know immediately that they have blue eyes, and they would leave on the first night.

Now suppose there were exactly \(2\) blue-eyed people. Call them \(A\) and \(B\). Each sees exactly one blue-eyed person. Each reasons like this:

“If my own eyes are not blue, then there is only \(1\) blue-eyed person on the island, namely the one I see. In that case, that person would leave on the first night.”

So \(A\) watches to see whether \(B\) leaves on night \(1\), and \(B\) watches to see whether \(A\) leaves on night \(1\). When neither leaves on the first night, each concludes:

“There must not be only \(1\) blue-eyed person. Therefore there must be \(2\). So I must also have blue eyes.”

So with exactly \(2\) blue-eyed people, they both leave on the second night.

Now suppose there were exactly \(3\) blue-eyed people. Each one sees \(2\) blue-eyed people. Each reasons:

“If I do not have blue eyes, then there are only \(2\) blue-eyed people on the island, and those two would leave on the second night.”

When no one leaves on night \(2\), each of the three concludes that there must actually be \(3\) blue-eyed people, so they all leave on night \(3\).

This continues in general. If there are exactly \(n\) blue-eyed people, then each blue-eyed person sees \(n-1\) blue-eyed people and reasons:

“If I do not have blue eyes, then there are only \(n-1\) blue-eyed people on the island, and they would all leave on night \(n-1\).”

If nobody leaves on night \(n-1\), each blue-eyed person concludes that their assumption was wrong, so they must also have blue eyes. Therefore all \(n\) blue-eyed people leave on night \(n\).

Here, \(n=100\). So each blue-eyed islander sees \(99\) blue-eyed people and thinks:

“If I do not have blue eyes, then there are only \(99\) blue-eyed people, and they will all leave on the \(99^\text{th}\) night.”

When the \(99^\text{th}\) night passes and nobody leaves, every blue-eyed islander deduces:

“Then there must be \(100\) blue-eyed people, so I must be one of them.”

So all \(100\) blue-eyed people leave together on the \(100^\text{th}\) night.

The brown-eyed people do not leave then, because the Guru’s statement was only about blue eyes. It starts the common-knowledge chain only for the blue-eyed group. Final answer:

\[ \text{All }100\text{ blue-eyed islanders leave on the }100^\text{th}\text{ night.} \]

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Eye Colors

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