skeptical scientist wrote:Imagine you were a cave man living in 40,000 BCE, and you saw lightning for the first time. What would you think was going on? Now you're actually not a cave man, and you presumably know what lightning is, but when you imagine yourself as a caveman, you deny yourself access to that information, and simulate reasoning without it. That's exactly what's going on here, and why the information the top person has is irrelevant when determining how the hypothetical person will reason. All that's relevant is the information the hypothetical person has.

Exactly. A's actual knowledge is not fully available in the imagined brain of the person he's hypothesizing about. A imagines what B's analysis might look like, and in doing so restricts the knowledge to what both A and B know. A's imagined B then hypothesizes about person C, which reduces it to knowledge that A, B, and C all share. And so on with persons D, E, F, G, H, etc. The eye color of each and every one of these people is unknown to at least one of the people involved, so the eye color of any person in the list at any given level of the nested hypotheticals is treated as unknown, and the minimum number of blue-eyed people is treated as the count of blue-eyed people who are not in that list.

Anyway, the nested hypotheticals thing is a rather difficult to understand explanation and is not necessary to solve the problem. Basic induction can do it quite easily.

Base step: Clearly, if there is exactly 1 blue-eyed person on the island, he will leave on day 1.

Inductive step: Assume that if there are n blue-eyed people on the island they will leave on day k

_{n}, and this can be logically deduced (the manner of deduction is irrelevant, only that it can be done). Suppose that there are, in fact, n+1 blue-eyed people on the island. Each of them will deduce that if his own eyes are not blue then all blue-eyed people will leave on day k

_{n}. When day k

_{n}passes without anyone leaving, he will realize that his own eyes must be blue and leave the next day, day k

_{n+1}. This works no matter what value n has, and gives that k

_{n+1}= k

_{n}+ 1.

For any number n, if there are n blue-eyed people on the island they will all determine their eye color and leave on day n.