Common Knowledge and the Blue-Eyed Islanders

An island holds 100 people, all perfect logicians. Every one of them has blue eyes, but nobody knows their own eye colour: there are no mirrors, no reflections, and discussing eye colour is forbidden. Each islander can see the other 99. The rule is that if you ever deduce your own eye colour, you must leave on the ferry at midnight that night.

One day a visitor stands up and says, out loud, to everyone: "I can see at least one person with blue eyes." Then she leaves.

Everybody on that island already knew there were blue-eyed people — each of them can see 99 of them. The visitor said nothing new. And yet on the hundredth midnight, all 100 islanders board the ferry together. Why?

Stop here and try it. The way in is to shrink the island.

The worked solution, one island at a time

One blue-eyed islander. She looks around and sees nobody with blue eyes. The visitor said there is at least one. It can only be her. She leaves on night 1.

Two blue-eyed islanders, A and B. A sees exactly one blue-eyed person, B. A reasons: if my eyes are not blue, then B is the only one, and by the argument above B leaves tonight. Midnight comes; nobody leaves. That silence is data. A now knows B saw a blue-eyed person too — and the only candidate is A. B runs the identical argument. Both leave on night 2.

Three blue-eyed islanders. Each sees two. Each reasons: if I am not blue, those two are in the two-person world and leave on night 2. Night 2 passes in silence, so each concludes they are blue. All three leave on night 3.

The induction. Suppose that on an island with exactly $n$ blue-eyed people they all leave on night $n$. Now take $n+1$ of them. Each one sees $n$ blue-eyed neighbours and reasons: if I am not blue, they are the whole set and go on night $n$. Nobody goes on night $n$, so each of them knows they are blue, and all $n+1$ leave on night $n+1$. With 100, they leave on night 100.

Notice what each islander is really doing: converting a count they can see into a prediction about a date, then watching the calendar to test it. If you can see 99 blue-eyed people, there are either 99 or 100 of them, and those two worlds differ only in when the ferry fills up. Night 99 passing in silence settles it.

So what did the visitor actually add?

For 100 blue-eyed islanders, everyone knows at least one person has blue eyes. Everyone also knows that everyone knows it, and so on — but only to a finite depth. Islander A knows it; A knows B knows it; A knows B knows C knows it; the chain runs out after about 99 links, because at the bottom sits an imagined islander who imagines an island with nobody blue-eyed.

The visitor's sentence was heard by everyone, in front of everyone. That converts the fact into common knowledge: everyone knows it, everyone knows everyone knows it, forever, with no bottom to the chain. The induction climbs that chain one night per rung. Without a public announcement there is no rung 1 to start from, and the whole ladder collapses.

The reusable technique

Three moves come out of this puzzle and they transfer everywhere.

1. Solve the smallest island, then induct. When a puzzle has a parameter $n$, do $n = 1$ and $n = 2$ honestly before guessing.
2. Treat silence as a measurement. If a rational agent would have acted by now and didn't, cross off the world in which they would have.
3. Ask what is public. Track not just who knows what, but who knows that others know.

Those three moves crack the muddy children puzzle (a parent announces "at least one of you is dirty" and the dirty children identify themselves on round $k$), the sum-and-product puzzles where each "I don't know" narrows the field, the two generals problem where no finite sequence of messengers ever reaches common knowledge, and level-k games such as guessing two-thirds of the average.

The market version is worth carrying around: a fact that every trader privately knows is not the same as a fact the market has publicly digested. Prices move on the announcement, not on the knowledge — which is exactly the islanders' hundredth midnight.