Apparently, you can't comprehend what you wrote, so I've bolded the relevant information below. Everything else is a red herring.

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?

So, if you have blue eyes, you'll only be able to see 99 people with blue eyes, 100 people with brown eyes, and 1 person with green eyes. This means that you must have blue eyes in order for there to be 100 people with blue eyes.

But, if you have brown eyes, you'll only be able to see 99 people with brown eyes, 100 people with blue eyes, and 1 person with green eyes. This means that you must have brown eyes in order for there to be 100 people with brown eyes.

Since everyone on the island is logical, everyone is able to come up with the same conclusion. Therefore, everyone knows what color eyes they have.

The answer to this riddle isn't hard.

you're clearly confused.

Just because it's true that there's 100 blue eyed people doesn't mean any individual blue eyed person knows there's 100 blue eyed people. They don't have that fact available to them. There could be 101 brown eyed people for all they know.

You can't use information available to us, from outside the island, as if it's necessarily available to them.

Here's what they know:

A blue-eyed person knows there are either 99 or 100 blue-eyed people.
A brown-eyed person knows there are either 100 or 101 blue-eyed people.

No-one knows the number of eye-clolours, but they do know it's either 2 (the ones they can see), or 3 (if their own isn't among the ones they can see). Therefore:

A blue-eyed person knows there's:
(a) 99 blue-eyed people and 101 brown-eyed people (and their eye-colour is brown)
(b) 100 blue-eyed people and 100 brown-eyed people (and their eye-colour is blue)
(c) 99 blue-eyed people, 100 brown-eyed people, and themselves (with a unique, unknown eye-colour)

If (a) and (b) were the only options, Michael's rule would work from a logical point of view, but (c) messes up things, here. We know that [# of blue]+[# of brown]=200; they don't: [# of blue]+[# of brown] could be 199.

So what changes when the Guru tells them something they already know?

That's the problem. There's a wedge between a logical sequence and an empirical reality I find hard to reconcile:

See, the catch is this: If an islander sees no blue-eyed person, then all other islanders see exactly one person with blue eyes. So all of the logic here is counterfactual: you don't really have to go see if someone leaves; you know nobody will. But at the same time, this logic spins forward until Day 99 or Day 100, depending on wether the person is blue-eyed or brown-eyed (unbeknowest to themselves), and then it's supposed to work in the world we live in. How?

It's the event on Day 100 of all blue-eyed people leaving that tips the brown-eyed people off that they're not blue-eyed (but not that they're brown-eyed). All that hinges on the fact that systematically brown-eyed people see more blue-eyed people than blue-eyed people do, and thus their set-off point is later. And their set-off point is only later, because the Guru talked about blue-eyed people.

The important fact seems to me this:

Every Islander knows that blue-eyed people see one fewer blue eyed person than non-blue eyed people.

If there had been 100 blue-eyed people, 90 brown-eyed people, and 10 green-eyed people, they'd still have had to wait 99 days before making the decision, because both brown-eyed and green-eyed people form the relevant group of people whose eyes aren't blue.

If there had been 20 blue-eyed people and 180, the game would be over much sooner if blue-eyed people were "seen" by the guru, or much later if brow-eyed people were "seen".

It's obvious to me, matemathically, that the announcement matters. I just don't know how to interpret this in pragmatical terms. It's baffling.

See, the catch is this: If an islander sees no blue-eyed person, then all other islanders see exactly one person with blue eyes. So all of the logic here is counterfactual: you don't really have to go see if someone leaves; you know nobody will. — Dawnstorm

That's not the catch, it's the hook on which the whole thing hangs. If the guru says he sees blue eyes but I see no blue eyes then I must be the blue eyed one and I leave that night.

But if I see 1 and only 1 person with blue eyes and they do not leave that night, then they too must see blue eyes and since I only see him, I must be the other that he can see with blue eyes, So the next night we will both know we are blue eyed and leave.

But if I see 2 and only 2 people with blue eyes and they do not leave the 2nd night, again there must be another blue eyed person that is me, and they will be reasoning the same way and so we will all leave together on the 3rd night.

etc.

And so everyone, whatever colour their eyes (because no one knows their own eye colour), is waiting to see if after n nights (where n is the number of blue eyed people they see) the blue eyed people leave, and if they don't, they can conclude they also have blue eyes, and if they do then they conclude they have eyes of some other colour.

And so everyone, whatever colour their eyes (because no one knows their own eye colour), is waiting to see if after n nights (where n is the number of blue eyed people they see) the blue eyed people leave, and if they don't, they can conclude they also have blue eyes, and if they do then they conclude they have eyes of some other colour. — unenlightened

Yes, that's all perfectly clear to me. What's not clear to me, for example, is why they can't skip forward to day n. I know they can't, but it makes no sense other than in purely logical terms. That's what I find so nuts about this riddle. There's a rift between logic and experience here I don't know how to bridge.

I'd call it a matter of coordination. What's unnatural is that there's no communication.

n is not the same for everyone. So there's no skipping to n possible because we have to all reach every step of the argument together. I'm waiting for my nth day, and you're waiting for yours and we don't know yet if they are the same day or not. If one of us has blue eyes and the other doesn't, our n is different and we find out on the nth day of the person with blue eyes when that one of us leaves, along with all the other blue eyed folk. And the other is no longer waiting because there are no blue eyes left and no more argument to be made and their n was never reached.

n is not the same for everyone. — unenlightened

Ah, yes, of course. I missed that (thought of it in another context, but somehow didn't make the connection on the practical front). Thanks.

I've got here late and just read the first and last pages, but I'd agree that this version of the puzzle is not logically sound.
If we follow it through, then if I'm an islander with red eyes, I will still erroneously conclude on day 100 that I have blue eyes and get thrown off the ferry.

If we follow it through, then if I'm an islander with red eyes, I will still erroneously conclude on day 100 that I have blue eyes and get thrown off the ferry. — Mijin

based on what?

Doh! My mistake.
I meant the bit about when all the brown eyed people realize they have brown eyes...but on checking, that wasn't from the OP, that was from someone's solution.

You're right that in the OP as stated, the blue eyed people all leave on day 100 or whatever, and no-one else.

ah gotcha

Here's my list of limitations of this puzzle:

1. The guru spoke "I can see someone who has blue eyes" when speaking in front of the islanders. The crowd I imagine has a mixture of blue and brown eyes. So, what is the point of the guru's comment? The guru spoke only once in many years and this is the sentence?

2. Can the islanders not know by counting how many islanders present and how many blue eyes and brown eyes? ( I get it that each one of them will end up counting 99 and 100) But is it just us who know this fact?

3. In what context is the "on what night" the islanders leave? Do we respond, the first, the second, the third, and so on?

So, what is the point of the guru's comment? — L'éléphant

That's half the puzzle.

Can the islanders not know by counting how many islanders present and how many blue eyes and brown eyes? ( I get it that each one of them will end up counting 99 and 100) But is it just us who know this fact? — L'éléphant

They don't know their own eye colour. They can count everyone's colours except their own, but counting 99 and 100 doesn't tell them their own. They can't just assume it's an even split.

3. In what context is the "on what night" the islanders leave? Do we respond, the first, the second, the third, and so on? — L'éléphant

You could say, "the islanders don't rely on the guru saying anything, everyone leaves the island on the third night" or "the islanders leave the island on the second night after the guru speaks." The thing that happens every night, once a night, is that the ferry comes. Maybe they can't figure it out the first night, but somehow waiting a day gives them extra information

Here’s my solution:

From the point of view of any given blue-eyed person on the island they must be either the 101st brown-eyed person, the 100th blue-eyed person, or neither blue nor brown-eyed. From the point of view of any given brown-eyed person, they must be either the 100th blue-eyed or 101st brown-eyed person, or neither blue nor brown-eyed.

If any given islander realizes that it is actually a 100/100 split brown/blue (the guru not being included in that count) they will deduce that they must be either the 100th blue-eyed person or 100th brown-eyed person because they see 99 people with the eye color that corresponds to their own; they must be the hundredth for everything to add up. Therefore, everyone but the guru would leave the island on the first night.

I will now show why this is the ultimate outcome:

We can assume that everyone deduces everything in the first paragraph of this solution and thus they can check their own possible deductions/considerations against everything the other islanders could deduce. Any given brown-eyed person must consider that:

- They could be the 101st blue-eyed person
- They could have neither blue nor brown eyes
- They are the hundredth brown-eyed person

Any given blue-eyed person must consider that:

- They could be the hundredth blue-eyed person
- They are neither blue nor brown-eyed
- They are the 101st brown-eyed person

From here we check each possible deduction/consideration against the other: a given brown-eyed person cannot correctly reason themselves to be the 101st blue-eyed person if a blue-eyed person reasons that they are the hundredth blue-eyed person because, given the guru is not blue-eyed, that would add up to 202 people on the island. The same goes for the reverse. So those possibilities can be thrown out.

Next: if there were two or more islanders that had neither blue nor brown eyes, then there would have to be 98 or less people with either brown or blue eyes instead of 99 (other than the guru), and any islander could see that that is not the case.

We are then left with the possibility of a brown eyed-person reasoning that they are the 101st blue-eyed person or a blue-eyed person reasoning they are the 101st brown-eyed person while the other has neither blue nor brown eyes. Any given islander can see that this is clearly not the case because they are not seeing anyone with eyes that are not brown or blue (other than the guru).

Thus, we are left only with the possibility of it being a 100/100 split between brown and blue, and, deducing this, the islanders all leave on the first night and the guru stays behind. I guess forever.

bruh what?

Is it really that crappy of a solution?

there's steps in there that you didn't really explain

there's steps in there that you didn't really explain — flannel jesus

Like what? Maybe I can explain it. If you are confused about the discussion of possible considerations/deductions being measured against each other, it comes from this:

Any given brown-eyed person must consider that:

- They could be the 101st blue-eyed person
- They could have neither blue nor brown eyes
- They are the hundredth brown-eyed person

Any given blue-eyed person must consider that:

- They could be the hundredth blue-eyed person
- They are neither blue nor brown-eyed
- They are the 101st brown-eyed person — ToothyMaw

Next: if there were two or more islanders that had neither blue nor brown eyes, then there would have to be 98 or less people with either brown or blue eyes instead of 99 (other than the guru), and any islander could see that that is not the case. — ToothyMaw

I don't get this paragraph. There's a green eyed person, and everyone who doesn't have green eyes sees her.

Next: if there were two or more islanders that had neither blue nor brown eyes, then there would have to be 98 or less people with either brown or blue eyes instead of 99 (other than the guru), and any islander could see that that is not the case.
— ToothyMaw

I don't get this paragraph. There's a green eyed person, and everyone who doesn't have green eyes sees her. — flannel jesus

I'm only considering the reasoning of brown or blue-eyed people about potentially blue-eyed or brown-eyed people.

As such, what I'm saying there is that there would be a group of islanders that would be part of the whole but would also not have brown or blue eyes, which would mean there being less than 99 of either blue or brown (disregarding the guru). Any given brown-eyed person or blue-eyed person would see this is not true and rule out the corresponding possibility that there are at least two (relevant) islanders with non-brown or blue eyes. The guru doesn't really have to factor into this part, although I understand your concern. I could have phrased it better.

so can you phrase it better now? Because I still don't get what reasoning you're offering.

so can you phrase it better now? Because I still don't get what reasoning you're offering. — flannel jesus

Alright. If a brown-eyed islander reasons that it is true that they have neither brown nor blue eyes, and a blue-eyed islander also reasons in parallel that they have neither brown nor blue eyes, then from the point of view of a brown-eyed islander, there would be 98 brown-eyed islanders and one with non-blue or brown eyes and from the point of view of a blue-eyed islander there would be 98 blue-eyed islanders and one with non-blue or brown eyes. We know that this cannot be the case, however, because in the problem it is stipulated that both blue-eyed and brown-eyed islanders know that there are at least 99 islanders of each eye color.

so all that says is that, other than the guru, there can't be 2 non brown non blue eyed people. So? There can still be 1.

so all that says is that, other than the guru, there can't be 2 non brown non blue eyed people. So? There can still be 1. — flannel jesus

It appears that that is the one possibility I left out, of course. And I doubt I could account for it with the approach I took. Whatever.

unfortunately that's always been the only possibility that matters anyway. From the beginning, we already know everyone can see everyone else's eye colours, just not their own - the only thing that matters is that one unknown.

It's a great puzzle, really counter-intuitive. I've done up to the case where there are 4 islanders (see below), and it works, so I can see it would work for n islanders. It still just feels weird though.

yeah, nice nested logic there. I think that's right.

If I were one of the islanders, I would just use the probability because my need to get out of the island is more important. So if I risked guessing that I had blue eyes, then the probability is 100/200, if I counted 99 blue eyes. ( I am not including the guru here so I also guessed that there's only one guru and her eye color is green).
If I counted 100 brown eyes, and I guessed that I am brown-eyed, then the probability of me being a brown-eyed is 101/200, which is roughly the same as the probability of blue-eyed.
But if I guessed that I was neither blue-eyed or brown-eyed, then the probability is 1/200. Which amounts to a very small chance that I was neither blue-eyed or brown-eyed.
In conclusion, I would pick that my eyes were blue if I counted 99 blue eyes.
Because if my eyes were red, then that would put me in a unique position as the guru.
The guru is the only one with a unique eye color.

how are you getting those probabilities?

how are you getting those probabilities? — flannel jesus

Though your explanation of the rules of this puzzle.
The islanders can know the number (sans himself) the number of blue, brown, and green eyes. Isn't it?