Solution
The Blue-Eyed Islanders
Show the problem again
On an island, 100 people have blue eyes and 100 have brown eyes. Everyone can see everyone else's eyes but not their own, no one may communicate about eye color, and anyone who deduces their own eye color must leave the island that midnight. All are perfect logicians, and all of this is common knowledge. One day a visitor announces to everyone: "At least one of you has blue eyes." Counting that day as day 1, on which midnight do the blue-eyed islanders leave?
Worked solution
All 100 blue-eyed islanders leave together on the 100th midnight. Induct on the number of blue-eyed people b. If b = 1, that person sees no blue eyes, deduces it must be them, and leaves night 1. If b = 2, each blue-eyed person waits: when the other doesn't leave night 1, both learn there are two and leave night 2. Generally, b blue-eyed people leave on night b. The announcement matters despite seeming redundant: it creates common knowledge: everyone knows that everyone knows (to depth 100) that a blue-eyed person exists, which no one had before.
Source: Common-knowledge puzzle in wide circulation for over a century; a version appears in Littlewood's 'Mathematical Miscellany' (1953). Statement written for AxiomIQ.