[ErrantX]

Suppose X is one, then the brown eyed person appears to be stuck

Yes, the problem specifically doesn't work if the number of people with an eye colour is below a certain amount.

The case for X>2 seems similar to the case for X=2, there's no magical point at which they can suddenly all say

In a scenario where there is, say, just brown and green (regardless of whether they know this or not), then X>3 is enough.

Where X=3 or less there is not enough information to go on (you successfully point out the flaws) and they could conceivably be any other eye colour. But above that you know it is impossible for there to be an individual who can see only one or two browns, therefore it can be inferred properly.

One of the main problems comes from the explanation of the issue which involves the suggestion that you recurse the problem back to X=1, this is inaccurate because there is always a known absolute minimum people that could have brown eyes (n, the brown eyed people you can see) and a known minimum number that any one person could see (n - 1). So long as for a brown eyed person (n - 1) > 2 the the inference is possible.

Remember; it does not matter what alternative eye colour they might be (and whether they can see it or not). Only that they are either brown or not brown. The logical inference is based on what the others must be seeing.

Try jotting it down and then reasoning it out as an individual in the tribe. Once it clicks it will seem simple :)

[andrew1]

OK, so we agree that if there are X <= 3 people then they're stuck and cannot kill themselves.

You claim that if X = 4 and everyone has brown eyes then on day 4 they will all kill themselves. Suppose I am one of these islanders. I can see three people with brown eyes. On days one, two and three no one kills themselves. On day four I get up and kill myself because I know that I have brown eyes.

Fine, now let's consider the case where there are three people with brown eyes, and one person with green eyes:

As I think we've agreed above, the three people with brown eyes cannot infer that they have brown eyes (i.e. the X < 3 case). So they are not able to kill themselves.

But consider the green eyed person: he can see three people with brown eyes and no one kills themselves on days one, two or three.

He is in exactly the same situation at this point as the brown eyed person we considered in the four-brown-eyed-people case. So at this point, by your logic, he must know that he has brown eyes. Which is a contradiction.

I think you are wrong that induction is not involved in establishing eye colour. (unless you can convince me in the X = 4 case that is :) ).

This isn't a problem I'm unfamiliar with it, a colleague asked me it when I was being interviewed for my current job, and it gets rediscussed periodically. I really do think you're wrong I'm afraid.

[ErrantX]

Umm, we defined X as the number of people with brown eyes. So on an island with 4 people, three of whom have brown eyes, then X=3 and, yep, there is a problem.

Where X=4 (i.e. there are 4 people with brown eyes) it works.

In your case, where the number of people is 4, but X=3, then the fourth possibly incorrectly infers that he has brown eyes and so, on his own, kills himself. On the other hand they are "highly logical" so I argue they would realise that there were too few people to know.

[andrew1]

But that's precisely the point, if they're all "highly logical" then in the island-wtih-3-brown-and-1-green case the person with green eyes can't correctly know their eye colour so can't kill themself. But they are in exactly the same situation as a person with brown eyes in the island-with-four-brown-people case. So if the green eyed person in the first case can't deduce their eye colour, then neither can the brown eyed person in the second case.

If you don't agree, please explain to me what extra piece of information the brown eyed person in the second case has which means he can kill himself on day four. As far as I can see they are both in the situation that they can see three people with brown eyes, and no one has yet killed themselves. What is the extra piece of information that allows the brown eyed person in the second case to deduce that he has brown eyes?

[ximeng]

I disagree with you.

Try jotting it down and then reasoning it out as an individual in the tribe.

(1) Suppose there are 4 browns, b1 to b4, and b1 is trying to figure out if his eyes might not be brown.

(2) b1 wants to know if it's logically possible for his eyes not to be brown and for the others not to know their eye colour.

(3) b1 therefore supposes b1 has green eyes. If this were the case, b2 would see b1's green eyes and both b3 and b4 having brown eyes.

(4) If then b2 imagined b2 had green eyes, b3 would see b1 and b2's green eyes, and b4's brown eyes.

(5) If then b3 imagined b3 had green eyes, b4 would see three sets of green eyes.

---

(6) So b1 can imagine a situation, where b2 imagines a situation, where b3 imagines a situation, where b4 sees three sets of green eyes.

(7) If everybody on the island knew at least one person had brown eyes then in b1's imaginary scenario, b4 would know they had brown eyes and would commit suicide on day 1.

(8) If this didn't happen then we'd know that at least 2 people had brown eyes by day 2, and the induction takes off.

Since they don't know that at least one person has brown eyes, then b1's imaginary scenario is possible, and it's therefore possible that b1 doesn't have brown eyes.

[ErrantX]

Damn. You're right :S

I didn't manage to make that final leap... and hence see to importance of the foreigner. Subtly brilliant!

Consider me trouted.