Dammit, I saw this little puzzle over at Metafilter and now I’m fascinated by it. Here’s the text of the puzzle for your own enjoyment, which I thoughtfully ganked from the original site.

There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).

Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).

One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.

One evening, he addresses the entire tribe to thank them for their hospitality.

However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.

What effect, if anything, does this faux pas have on the tribe?

There are two arguments to be satisfied here:

Argument 1. The foreigner has no effect, because his comments do not tell the tribe anything that they do not already know (everyone in the tribe can already see that there are several blue-eyed people in their tribe).

Argument 2. 100 days after the address, all the blue eyed people commit suicide.

Oh crap, does this bake your noodle. The solution (well the solution that I think makes the most sense) can be found in the comments on that page. I should warn you that some of the comments get rather math-y, but the solution that feels right to me isn’t all that hard. I will post it here tomorrow.

***Update

Ok, so here’s the answer (among many that are available, it seems) that makes sense to me.  It’s up to commenter “Saad” to explain it best (or so it seems to me).

If there is only one blue-eyed person (BP) on the island, it’s obvious.

(**)
So say there are two BP. Then the reasoning will be as follows:

The day of the stranger’s address, each BP will expect the other BP to commit suicide the next day (that is, one day after the stranger’s address). The next day will come. Nobody will commit suicide, so each BP will know he has blue-eyes. Therefore, according to the custom, they’ll both commit suicide the next day (that is, two days after the stranger’s address).

(***)
Say there are three BP. Then the reasoning will be as follows:

The day of the stranger’s address, each BP will look at the other two BP and think that if it there are only two BP on the island (the two he’s looking at), then they’ll follow the (**) argument above and two days after the stranger’s address, they’ll be dead. If they’re not dead after the two days, then I’ll know I’m a BP. However, EACH of the three BP is thinking this about the other two BP, so none of them will be dead after two days. Therefore, the next day (that is, three days after the stranger’s address), all three will know they’re BP and will kill themselves.

(****)
Say there are four BP. Then the reasoning will be as follows:

The day of the stranger’s address, each BP will look at the other three BP and think that if it there are only three BP on the island (the three he’s looking at), then they’ll follow the (***) argument above and three days after the stranger’s address, they’ll be dead. If they’re not dead after the three days, then I’ll know I’m a BP. However, EACH of the four BP is thinking this about the other three BP, so none of them will be dead after three days. Therefore, the next day (that is, four days after the stranger’s address), all four will know they’re BP and will kill themselves.

And so on for five BP, six BP, all the way up to a hundred BP. Therefore, on the 101st day after the stranger’s address, the 101 BP will all kill themselves.

So there you have it.  I like that puzzle a lot….