The Announcement That Told Nobody Anything

On an island, nobody can see their own eye colour and everybody can see everybody else's. A stranger passes through, gathers the population, and says out loud that at least one islander has blue eyes. The local rule is old and absolute: anyone who works out their own eye colour leaves that night. Nine mornings pass and nobody has gone. On the tenth morning, people leave. How many?

Ten. The reason this deserves an article rather than a sentence is that the stranger appears to have said nothing whatsoever. If ten islanders have blue eyes, then every single person listening can already see at least nine of them. His announcement is old news to his entire audience, and it still starts a clock that runs for exactly ten days.

The objection that has to be answered first

State the objection at full strength, because half of it is correct. Let $B$ be the proposition that at least one islander has blue eyes. With ten blue-eyed islanders, a blue-eyed one sees nine and a brown-eyed one sees ten, so the worst informed person in the room is looking at

blue-eyed neighbours. Everybody knew $B$ before the stranger opened his mouth. No fact changed hands. If facts were the only currency, the announcement would be theatre and nobody would ever leave.

That conclusion is false, and it is false in a way you can check rather than argue about. Model the island explicitly, run it without the announcement, and no departure ever occurs at any population. Run the identical model with the announcement and the tenth morning empties. True premise, false conclusion. The gap between them is the whole subject.

The induction, done slowly

Begin with one blue-eyed islander. He sees nobody with blue eyes, and the stranger has just asserted that somebody has them. There is exactly one person he cannot inspect, so it is him, and he leaves the first night. This is the one case where the announcement is real news: before it he had no reason to think blue eyes existed at all.

Now two, call them A and B. A sees one blue-eyed islander and reasons: if my own eyes are brown, B is the only one, B sees nobody, and B goes tonight. Morning comes and B is still there. So my eyes are blue. B runs the same argument in reverse and both leave on the second night.

The general step is that same paragraph with a variable in it. Suppose that $n-1$ blue-eyed islanders would first depart on night $n-1$. With $n$ of them, each blue-eyed islander sees $n-1$ others and reasons: if I am not one of them, the count is $n-1$ and those people go tonight. Nothing happens. So the count includes me:

Each islander is counting mornings against what he can see, and the arithmetic is that thin. Nine silent mornings mean everybody was waiting for the ninth, which means everybody sees nine, which means

What the stranger added

Write $K_i \varphi$ for the statement that islander $i$ knows $\varphi$, and let $E\varphi$ mean that all of them do. Iterating $E$ gives the ladder that matters here:

The precise situation before the stranger speaks is this. With $n$ blue-eyed islanders, $B$ is known to depth $n-1$ and to no greater depth:

Check it at $n=2$. A knows $B$, since A is looking at a blue-eyed islander. Does A know that B knows $B$? A has to allow the world in which A is brown-eyed, and in that world B sees nobody and does not know $B$. So $E^{2}B$ fails while $E^{1}B$ holds. Push the same argument one level deeper for each extra islander and you get the general statement.

A public announcement makes $B$ common knowledge in one stroke, supplying every level from $n$ upward. At $n = 10$ that is nine levels of nested knowledge nobody had, delivered by a sentence whose plain content everybody had already. Informally, each silent morning is the island cashing in one of the new levels, which is why the wait is as long as it is. The case $n = 1$ is the odd one out: there, and only there, somebody learns a fact.

Counting the worlds instead

All of this can be settled without trusting the induction at all. Fix a population, say twelve residents. A possible world is just the subset of them with blue eyes, so there are

worlds. An islander can tell two worlds apart exactly when they differ somewhere other than his own eyes, and nothing further needs to be said about anybody's reasoning. The announcement deletes the single world in which nobody is blue-eyed. Every silent morning after that deletes more, and the pattern is exact: after $t$ silent mornings the worlds ruled out are precisely those with $t$ or fewer blue-eyed islanders,

which runs 13, 79, 299, 794, 1586, 2510, 3302, 3797 and 4017 for $t = 1$ through $9$. Seventy-nine worlds survive the ninth silent morning, and they are exactly the worlds with ten or more blue-eyed islanders. A blue-eyed islander was choosing between nine and ten, and nine has just been deleted, so he knows.

A brown-eyed islander is choosing between ten and eleven. Both survive, so he learns nothing and stays home. That asymmetry is why ten people leave rather than the whole island.

Ten mornings, ten people

A ladder drawing invites the reading that one person leaves per morning for ten mornings. That is not what happens. Nine mornings pass with nobody going anywhere, and then ten islanders leave together, because every blue-eyed islander occupies the identical position: each sees nine, each has waited nine, each finishes the same deduction at the same instant.

There is a tail to it. The islanders left behind watch the departure, conclude that the count was ten rather than eleven, and so learn their own colour as well, at least where there are two colours to choose between. The interesting part ends on the tenth morning.

The assumptions, and how to break them

The result is exact and also brittle, which is a combination worth respecting. It needs every islander to reason perfectly and to know that the others do, and it needs mornings to be shared events so that silence is observable. Remove any of that and the count stops meaning anything.

The most instructive failure is a private announcement. Suppose the stranger takes each islander aside and whispers the same sentence. Now $E B$ holds, but $E^{2}B$ does not, because no islander knows the others were told. The ladder is never built and no departure ever happens, from identical content delivered in a different room. One islander who cannot carry out the induction does the same damage: the chain stalls at his rung and the island stays put.

One assumption people expect to matter does not. The answer counts blue-eyed islanders, not residents. Nothing in the argument ever asks how big the island is, which is why the puzzle can be posed without telling you.

Sources and further reading

• The ladder of iterated knowledge — Common knowledge (logic)
• The modal machinery behind the operators used above — Epistemic modal logic
• Possible worlds as the model doing the counting — Kripke semantics
• Common knowledge as a working tool, after Aumann's 1976 paper — Aumann's agreement theorem
• The induction itself, in the form used above — Mathematical induction. The puzzle is treated at length, under the name of the muddy children, in Reasoning About Knowledge by Fagin, Halpern, Moses and Vardi.

The numbers here were checked against an explicit possible-worlds model of a twelve-person island, run over all 4096 worlds. First departure night equals the number of blue-eyed islanders for every population from one to twelve, that many leave together each time, and with the announcement removed the model never produces a departure at all.