I’ve skimmed a lot of this thread. If someone already said this, I missed it,but I’ve deduced that it doesn’t matter what color the guru’s eyes are in this puzzle. And if the guru happens to have blue eyes, she can leave too when the time comes. If I’m wrong about that, please tell me why.
I’ll admit I got a bit lost trying to follow the “everybody knows that everybody knows that everybody knows…” logic up above. But there seems to be consensus that it works. Therefore:
After all these logical deductions, we’ve got this theorem: If the guru tells everyone in public on Day 0 that there is at least one Foo-eyed person on the island, I (one of the islanders) count up the number of Foo people I see, and I get X. I can then wait till day X and if I still see those X Foo people still hanging around, I know there are really X+1 Foo people, I must be that extra one, and that night I can go to the boat and leave. (yeah, it’s funky for the case of just 1 Foo-eyed person, but you can count the empty set and it works)
Fine. I’m taking that as a given now; it’s been proven. No need to rehash the logic.
BUT: We’re all perfect logicians. Everyone’s going to consider other possibilities. We’re not only going to consider just what the Guru said, but what the Guru could have said. What if the Guru had said, “I see at least one Bar-eyed person” instead of Foo?
Well, that’s easy. All N Bar-eyed people will be gone on day N (counting the day the Guru said something as day zero, remember), We have that theorem in hand, it doesn’t matter which eye color was named.
Then I make the next leap. I look around at all the other eye colors I see that the Guru could have named instead of Foo. Then I imagine: Why did the Guru have to name that one? I’m seeing all those other eye colors; the Guru could have named one of those instead, why not? What if she did? And all those other perfect logicians will be thinking the same thing I am. Maybe other eye colors might be able to figure things out too, now that the Guru has named a starting day zero by saying anything at all.
I can only consider cases where I see at least two people who share a given other eye color. Because if I see only one person, then it’s possible they’re the only one, and they won’t even know that color is available to choose. But for every other color with at least two visible examples, we can follow the same logic just as if the Guru had named it instead. Everyone already knows which and how many colors that exist except their own…
Oops. I goofed. If I see two Bar-eyed people, each of them is possibly looking at only one other Bar person. And if they only see one, they’ll have to be considering the case where that single one they see is the only example available, and the singleton won’t know their own eye color exists, so scratch the case of counting two Bars. Sucks to be them.
So, I have to see at least three other Bar-eyed people to consider Bar. Those three will count what they see and know two other Bars exist. And they won’t know their own, but they’ll see those two Bars looking at each other and counting that color, but those Bars might only count one, not two, and if they count one, we’re back where we started…
Hmm. Oh well, if there are only three, guess they can’t figure anything out either. Too bad.
Fine. I find a different color and there are four of them that I see. I know there are at least four of them, could be five. I know they are also counting, and being unable to see their own color, they’ll count only three if I’m not one of them. They’ll count three others of that color… Ahh… [censored] and I’ve already eliminated the cases of if they only see three, and two, and one…
:smack:
I thought I’d seen a way for more than just the named color to escape as well. I think it’s wrong, and just proved why. And I think I’ve also just proved why it’s necessary for the Guru to name a color when she speaks. This post started out in a very different place than it ended up…
*** Ponder
