The blue-eyed islanders

Question

On an island live 100 people with blue eyes and some number with brown eyes. Everyone is a perfect logician. There are no mirrors and no one may communicate about eye color. The rule: anyone who deduces their own eye color must leave the island that midnight. Everyone can see everyone else's eyes, so each blue-eyed person sees 99 other blue-eyed people, and each brown-eyed person sees 100. No one knows their own color. One day a trusted visitor announces publicly: "At least one of you has blue eyes." Everyone hears it and knows everyone heard it. What happens, and on which night?

Accepted Answer

All 100 blue-eyed people leave on the 100th midnight. The others never leave. The proof is induction on the number of blue-eyed people n. Base case, n = 1. The single blue-eyed person sees no other blue eyes. The announcement "at least one blue" must refer to them, so they deduce their color immediately and leave on night 1. n = 2. Each blue-eyed person sees exactly one other blue-eyed person. Reasoning of person A: "If I had brown eyes, then B is the only blue and B would leave on night 1 (by the base case)." Night 1 passes and B does not leave. So A concludes A must also be blue, and by sym…

The blue-eyed islanders

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