There is an island known for its people with blue eyes, yet there is at least one green-eyed person on the island. No one knows the color of their own eyes, as there are no reflective surfaces on the island and discussion of eye color is forbidden, but they can see everyone else’s eye color. If any islander were to come to know that they do not have blue eyes, they would leave the island in shame before the next sunrise.

One day, an outsider visits the island and remarked how there was at least one islander with green eyes. Within the day, every islander had heard and understood this new information.

Assuming departures from the island are noticed by everyone by the next day, and assuming each islander is highly logical and is able to keep track of all other islanders’ eye colors and actions – what happens to the islanders and does it depend on the number of green-eyed islanders?

This is a classic logic puzzle, also known under a different story (but same core logic) as Josephine’s Problem.


Let’s say there are n green-eyed islanders. After the outsider’s remark, nothing will happen until the n-th night, at which point all n green-eyed islanders will leave the island before sunrise.

Why? Some clever logical induction:

  • n = 1: when that islander learns there is at least one green-eyed islander, they know that all the other islanders have blue eyes, so they must be the one with green eyes, so they leave on the 1st night.
  • n = 2: both green-eyed islanders see one green-eyed islander. After the 1st night, both of them will see that the other green-eyed islander has not left. But since the other green-eyed islander would have left the 1st night if they were the only green-eyed islander (see above), both green-eyed islanders realize they must also have green eyes, so they both leave the 2nd night.
  • n = k: all green-eyed islanders see (k – 1) green-eyed islanders. After the (k – 1)th night, all of them will see that the other (k – 1) green-eyed islanders have not left. But all of them know that the (k – 1) green-eyed islanders would have left on the (k – 1)th night if there were in fact on (k – 1) green-eyed islanders. Thus they all realize there are in fact k green-eyed islanders including the one they could not see (themselves), so they all leave the k-th night.

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