The Blue-Eyed Islanders

Question

On an island live 100 blue-eyed and 100 brown-eyed perfect logicians. There are no mirrors and eye color is never discussed; everyone sees everyone else's eyes. Anyone who deduces their own eye color must leave at the next midnight, visibly. One day a visitor truthfully announces to everyone at once: "At least one of you has blue eyes." What happens, and why did an announcement of something everyo

Accepted Answer

All 100 blue-eyed islanders leave together on the 100th midnight. The brown-eyed never leave. Induction on the number of blue-eyed, k: - k = 1: the lone blue-eyed sees zero blue eyes, so the announcement is about them, leaves night 1. - k = 2: each blue sees one blue and reasons: "if I'm not blue, they're alone and leave tonight." Nobody leaves night 1 → both deduce their own color → both leave night 2. - Generally each blue-eyed sees k-1 blues and waits: if night k-1 passes without departures, the hypothesis "I am not blue" is refuted, and all k leave on night k. With k = 100: night 100. What the announcement added, common knowledge. Everyone already saw 99+ blue-eyed people; but a blue-eyed islander didn't know that a chain of 100 nested "X knows that Y knows that…" statements held, before the announcement, the depth-100 statement "everyone knows^100 someone is blue-eyed" was false (each blue sees 99, and reasons about someone who might see 98, who reasons about 97…, bottoming out at a hypothetical person who sees none). The public, simultaneous announcement makes the fact common knowledge to infinite depth at a stroke, which starts the synchronized countdown, the nightly non-dep

The Blue-Eyed Islanders

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