"The famously difficult green-eyed logic puzzle" - I disagree with the proposed solution

[MaxTheVool]

@Sage_Rat

Which is why we only skip the amount that we can all be known to agree on.

Repeating this from my previous post:
So… let’s take smaller sample. There are 7 total people. 4 of them have green eyes. So 3 people think that it’s either 4/5 and 4 people think that it’s 3/4. What precisely happens in your solution?

And is there a triggering piece of information at all? Because I don’t see how any solution could ever work without a triggering piece of information.

[MaxTheVool]

@White SIFL

Can someone please walk me through the outcome? Note that I am not trying to fight the hypothetical and I do understand the math tree as stated above but I’m having trouble applying it to this hypothetical prison and prisoners.

I understand Chessic Sense’s interviewer solution but aside from stepping out of the hypothetical, it doesn’t help the first interviewee once you’ve moved on (and again, the original puzzle makes no provision for a third party interviewer).

So given the stated rules:
“[T]here are no reflective surfaces or cameras etc on the island. Communication between prisoners is strictly forbidden, hence, and importantly, no one has ever learned their own eye color.”
What happens next in Max’s above example? Hearing the news that (at least) one of them has green eyes, they both wordlessly stand and walk out? The logic doesn’t hold hold for me.

OK, there are only two green-eyed people, Alex and Bob. Each one of them has thought for ages “there’s only one green-eyed guy, that guy over there… plus maybe I’m green-eyed. But that guy over there doesn’t know if he’s green-eyed, and of course I don’t know if I’m green-eyed”.

Then one day, the external guy comes along and says “there’s at least one green-eyed person”. How does Alex react? Well, he looks around, he sees only a single green-eyed person, Bob. He thinks “poor Bob, he’s never realized that he was green-eyed. But now this new information has arrived, and Bob now knows with certainty that there’s at least one green-eyed person. So… am I green-eyed? If I’m not green-eyed, then Bob sees that there are no green-eyed people other than himself. So, if I’m not green-eyed, and Bob is told that there is at least one green-eyed person, then Bob will know that he is green-eyed. So, tomorrow morning, Bob may have left the island.”

Then tomorrow morning comes along and Bob has not left the island.

Now Alex thinks “well, Bob did NOT realize that he was the only green-eyed person. Why not? if he saw only brown eyes, and was told there was at least one green-eyed person, he would have left. But he didn’t. So there must be another green-eyed person. Since I can see that all 98 other prisoners have brown eyes, the other green-eyed person must be ME!”. So the next morning comes, and Alex and Bob have both departed (because Bob was thinking exactly the same things, with the names reversed).
I’ll take this one step further, up to three people. To do so, I find it most easy to talk as if everyone assumes they do not personally have green eyes. It doesn’t change the logic, but it changes the mindset a bit.

So Charles also has green eyes. But he has always believed he doesn’t. So as far as he’s concerned, there are two green-eyed people, Alex and Bob. He hears the announcement. He then immediately runs through all of the above logic and says “well, there are two green eyed people on the island, therefore in two mornings they will both leave”. Two mornings later they do NOT both leave. So he’s thunderstruck, and he reconsiders all of his suppositions, and the only possibility remaining is that he, Charles, ALSO has green eyes. So the next morning, Alex and Bob and Charles all leave (and again, Alex was certain the whole time that Bob and Charles were the only two green eyes, and was shocked when they had not left on the second morning, and so forth).
Make sense?

[MaxTheVool]

@John_DiFool

My issue with that is, what logic in the world compels you to focus on just ONE person at a time like that? Why should I privilege Bob, or anyone else, any single individual, by focusing on him (and his thoughts) first, then on the next person, and the next, etc. etc. etc.? Why would speculating on their minds matter, at all, vs. the simple physical brute facts?

Wouldn’t you instead just focus on them all (the blues) as a group, simultaneously, and simply say, “OK, I see 99, but they might see 99, but perhaps only 98, full stop?”

(a) I’m not quite sure what you’re saying… so what DO you think would happen? Do you think that the accepted solution is correct, but that the arguments are incorrect? Or do you think there’s a better/different/no solution?

(b) I think you’re misinterpreting the “privileging”. The puzzle isn’t solved by Alex thinking about Brian thinking about Charles thinking about David, etc. The puzzle is solved by induction. The reason people have brought up the Alex thinks that Brian thinks that Charles thinks that David thinks stuff is to demonstrate how it is that the public announcement does insert new information into the system, even though it appears not to. The new information it inserts can be described as a very long chain of “thinks that”. But it’s not a singular one, it’s all of them. So with 3 people, it simultaneously relates to what Alex thinks that Brian thinks that Charles thinks, and what Brian thinks that Alex thinks Charles thinks, and what Charles thinks that Brian thinks that Alex thinks, etc.

[John_DiFool]

@MaxTheVool

(a) I’m not quite sure what you’re saying… so what DO you think would happen? Do you think that the accepted solution is correct, but that the arguments are incorrect? Or do you think there’s a better/different/no solution?

There appears to be a subtle, and perhaps fallacious, insertion of a second level of truth above the first level. It simply seems like a reductio ad absurdum conclusion must be drawn, once your “logical” argument appears to demonstrate that one person could perceive zero green (/blue) eyed people. i.e. there is a hidden premise there which must be false. Not sure why a r.a.a. conclusion is not allowed by these so-called “perfect” logical thinkers.

(b) I think you’re misinterpreting the “privileging”. The puzzle isn’t solved by Alex thinking about Brian thinking about Charles thinking about David, etc. The puzzle is solved by induction. The reason people have brought up the Alex thinks that Brian thinks that Charles thinks that David thinks stuff is to demonstrate how it is that the public announcement does insert new information into the system, even though it appears not to. The new information it inserts can be described as a very long chain of “thinks that”. But it’s not a singular one, it’s all of them. So with 3 people, it simultaneously relates to what Alex thinks that Brian thinks that Charles thinks, and what Brian thinks that Alex thinks Charles thinks, and what Charles thinks that Brian thinks that Alex thinks, etc.

See-that’s just it. I don’t think there is any chain, because, again, by doing so you privelege one viewpoint at a time, and not the communal viewpoint. I think everyone past the 1st person in said chain are utterly imaginary, and superfluous.

I think I did find a flaw in the setup tho (and not one of the “if this was the real world” kind, either). If everyone does indeed look at everyone else, then they would also see everyone looking at everyone else, thus if Graham sees Stephen & David look at Judy and her luminous blue eyes, Graham will know right then and there that they have indeed seen a blue-eyed person.

I simply think an unneccesary epicycle has been introduced into the works, and nobody still has yet to explain the core rationale (#99 didn’t).

[MaxTheVool]

@John_DiFool

There appears to be a subtle, and perhaps fallacious, insertion of a second level of truth above the first level. It simply seems like a reductio ad absurdum conclusion must be drawn, once your “logical” argument appears to demonstrate that one person could perceive zero green (/blue) eyed people. i.e. there is a hidden premise there which must be false. Not sure why a r.a.a. conclusion is not allowed by these so-called “perfect” logical thinkers.

I’m still not quite sure where you disagree.

I claim that with only one green-eyed person, he will leave the island the next morning (after the announcement).

I claim that with two green-eyed people, they will leave the island two mornings later.

I claim that with three green-eyed people, they will leave the island three mornings later.

Etc., to infinity.

I also claim that with no announcement, no one will ever leave, for any possible number of green-eyed people.

Which, if any, of the above do you disagree with?

Or do you agree with all that, but disagree with some of the meta-discussions about how making an announcement does or does not introduce new information, etc?

I think I did find a flaw in the setup tho (and not one of the “if this was the real world” kind, either). If everyone does indeed look at everyone else, then they would also see everyone looking at everyone else, thus if Graham sees Stephen & David look at Judy and her luminous blue eyes, Graham will know right then and there that they have indeed seen a blue-eyed person.

I consider that to be fighting the hypothetical.

I simply think an unneccesary epicycle has been introduced into the works, and nobody still has yet to explain the core rationale (#99 didn’t).

I still have no idea precisely what you’re asking.

[Dr.Strangelove]

@John_DiFool

I don’t think there is any chain, because, again, by doing so you privelege one viewpoint at a time, and not the communal viewpoint. I think everyone past the 1st person in said chain are utterly imaginary, and superfluous.

Ultimately, all this talk about viewpoints and A thinks that B thinks that C thinks that stuff is just our intuitive explanation for what’s going on, and is irrelevant to the actual solution.

Chronos presented the inductive proof already. If you’re going to argue against the problem, you must find a flaw in this proof–either that the base case isn’t correct, or that there’s something wrong with the inductive step. Otherwise, you’re (effectively) just claiming that logic doesn’t work in general. Or fighting the hypothetical.

The proof is trivial, so it seems unlikely to contain hidden flaws. These are the basic lemmas:

• If there is one person with green eyes, he will leave on the first day
• Anyone that sees N green eyes knows that the total must be either N or N+1
• If the N case is excluded, then the count must be N+1
• If no one has left on night N, then that case is excluded
• If the count is N+1, then you have green eyes and may leave at the next opportunity, which is the next night

[Chessic_Sense]

@Dr.Strangelove

Ultimately, all this talk about viewpoints and A thinks that B thinks that C thinks that stuff is just our intuitive explanation for what’s going on, and is irrelevant to the actual solution.

Chronos presented the inductive proof already. If you’re going to argue against the problem, you must find a flaw in this proof–either that the base case isn’t correct, or that there’s something wrong with the inductive step. Otherwise, you’re (effectively) just claiming that logic doesn’t work in general. Or fighting the hypothetical.

The proof is trivial, so it seems unlikely to contain hidden flaws. These are the basic lemmas:

• If there is one person with green eyes, he will leave on the first day
• Anyone that sees N green eyes knows that the total must be either N or N+1
• If the N case is excluded, then the count must be N+1
• If no one has left on night N, then that case is excluded
• If the count is N+1, then you have green eyes and may leave at the next opportunity, which is the next night

That’s one way of looking at it. Quite honestly, I find Chronos’s proof to be the intuitive explanation that’s irrelevant to the solution. Which true and accurate explanation is intuitive and which is “relevant” are matters of opinion.

Specifically, I find “If no one has left on night N, then that case is excluded” to be intuitive, not rigorous. It’s true, no doubt…just lacking in detail.

[Chessic_Sense]

@John_DiFool

You say, “Well, think of what Bob thinks, then what of Bob thinks of what Suzie thinks”, ad nauseum.

My issue with that is, what logic in the world compels you to focus on just ONE person at a time like that? Why should I privilege Bob, or anyone else, any single individual, by focusing on him (and his thoughts) first, then on the next person, and the next, etc. etc. etc.?

Wouldn’t you instead just focus on them all (the blues) as a group, simultaneously, and simply say, “OK, I see 99, but they might see 99, but perhaps only 98, full stop?”
…

Why would perfect logicians stop, when there are so many more things to realize? How about “OK, I see 99, but they might see 99, but perhaps only 98, and thus would think each other sees 97, minimum.” How about “I see 99, but they might see 99 or 98, and thus I can’t be sure they are reaching this exact same deduction right now.”? How about “Since I’m a perfect logician, I understand that it’s not only possible to assess what everyone else is thinking, but also extremely fucking important to me getting out of here, so I’d better consider it!”?

I know I will just get yet another round of the “Oh no you’re wrongs” and lectures and additional pointless, flawed, irrelevant and too-subtle analogies that will likely go whooshing right over my head again. Fine. I simply think you all have been butts of some sly joke.

I don’t know why you’re being so rude about this, especially when you call our perfectly accurate explanations “flawed” and “irrelevant” while admitting you haven’t spent the time to think about them.

How about this puzzle: There’s a room full of perfect logicians, discussing things. In walks a drunk, less-than-perfect logician, insulting everyone. If a train leaves Chicago at 95 miles/hour and one conductor always tells the truth while the other lies, and halfway through, the mayor offers to let you switch or stay, should the perfect logicians still bother to explain anything to the drunk?

@White SIFL

Can someone please walk me through the outcome? Note that I am not trying to fight the hypothetical and I do understand the math tree as stated above but I’m having trouble applying it to this hypothetical prison and prisoners.

I understand Chessic Sense’s interviewer solution but aside from stepping out of the hypothetical, it doesn’t help the first interviewee once you’ve moved on (and again, the original puzzle makes no provision for a third party interviewer).
…
What happens next in Max’s above example? Hearing the news that (at least) one of them has green eyes, they both wordlessly stand and walk out? The logic doesn’t hold hold for me.

The announcement isn’t the only piece of information introduced. It’s also that no one leaves the first, second, …, 99th night. Each night, new information is introduced.

More in a bit…

[MaxTheVool]

@Chessic_Sense

The announcement isn’t the only piece of information introduced. It’s also that no one leaves the first, second, …, 99th night. Each night, new information is introduced.

More in a bit…

I’m not disagreeing with you, but the thing that makes the puzzle so counterintuitive is that the information added by the announcement isn’t what it seems to be. That is, once you have 3 or more green eyed people, everyone knows that there’s at least one green eyed person, and everyone knows that everyone knows that there’s at least one green eyed person. And then an announcement is made, “there is at least one green eyed person”. Which sounds like it doesn’t add information. After all, that’s nothing that anyone didn’t already know.

So people start thinking about that as a “starting point”, and wonder why some other starting point wouldn’t do, can’t we all just agree that today is the starting point, etc.

When in fact, there is information added, it’s just very subtle information that has to be discussed via confusing chains like “A thinks that B thinks that C thinks that D thinks that…”

[Dr.Strangelove]

@Chessic_Sense

Specifically, I find “If no one has left on night N, then that case is excluded” to be intuitive, not rigorous. It’s true, no doubt…just lacking in detail.

That particular lemma is derived from the inductive assumption. Recall that the inductive step is “if you assume P(N), then P(N+1) must be true”.

Quoting Chronos, the statement that we’re trying to prove is:
For all values of n, if there are n green-eyed people, they will all leave on night n

Deriving the lemma above is just applying modus tollens:
if there are N green-eyed people => they will all leave on night N

they have not all left on night N => there are not N green-eyed people

[Dr.Strangelove]

@Chessic_Sense

Which true and accurate explanation is intuitive and which is “relevant” are matters of opinion.

BTW, by “intuitive” I did not mean “this is the argument which is most convincing to me.” That is indeed a matter of opinion. I meant “this is a non-rigorous explanation which may be helpful in understanding but is not a formal proof.” That’s not really subjective; the only rigorous proof presented so far is the inductive one. The others are essentially hand-waving.

[Chessic_Sense]

@John_DiFool

once your “logical” argument appears to demonstrate that one person could perceive zero green (/blue) eyed people.

No one ever argued or demonstrated that. Good thing, too, because it’s completely incorrect.

@Chessic_Sense

More in a bit…

[QUOTE=Dr. Strangelove]
the only rigorous proof presented so far is the inductive one.
[/QUOTE]

What perfect timing! There’s no rigorous proof because it takes forever to type out, and no one’s got the time or inclination. More over, people won’t read it. They might even call it “flawed” because they only read half of it and get confused. So why bother? You won’t get a “formal” logic proof, but how about a rigorous one? I find myself with some time and a bit of inclination…here goes nuthin’!

In a prison with 5 green-eyed people (A, B, C, D, and E) and 232 brown-eyed people, the following thoughts (and thoughts about thoughts) exist:

How many green eyed people are in this prison?
A: 5 or 4, the latter if I have brown eyes.
B: 5 or 4, the latter if I have brown eyes.
C: 5 or 4, the latter if I have brown eyes.
D: 5 or 4, the latter if I have brown eyes.
E: 5 or 4, the latter if I have brown eyes.

How many green-eyed people do any of green-eyed people (that you know of) think there are?
A: Either “5 or 4” or “4 or 3,” but not both. Just one of those pairs, the latter if I have brown eyes.
B: Either “5 or 4” or “4 or 3,” but not both. Just one of those pairs, the latter if I have brown eyes.
C: Either “5 or 4” or “4 or 3,” but not both. Just one of those pairs, the latter if I have brown eyes.
D: Either “5 or 4” or “4 or 3,” but not both. Just one of those pairs, the latter if I have brown eyes.
E: Either “5 or 4” or “4 or 3,” but not both. Just one of those pairs, the latter if I have brown eyes.

How would a green-eyed person answer that last question?
*(I will now use the ‘/’ and ‘-’ symbols to stand for the word “or” for readability)
A: “5/4-4/3” or “4/3-3/2” but not both. Just one of those quads.
B: “5/4-4/3” or “4/3-3/2” but not both. Just one of those quads.
C: “5/4-4/3” or “4/3-3/2” but not both. Just one of those quads.
D: “5/4-4/3” or “4/3-3/2” but not both. Just one of those quads.
E: “5/4-4/3” or “4/3-3/2” but not both. Just one of those quads.

How would a green-eyed person answer THAT last question?
A: [5/4-4/3 or 4/3-3/2] possibly. Could’ve been [4/3-3/2 or 3/2-2/1].
B: [5/4-4/3 or 4/3-3/2] possibly. Could’ve been [4/3-3/2 or 3/2-2/1].
C: [5/4-4/3 or 4/3-3/2] possibly. Could’ve been [4/3-3/2 or 3/2-2/1].
D: [5/4-4/3 or 4/3-3/2] possibly. Could’ve been [4/3-3/2 or 3/2-2/1].
E: [5/4-4/3 or 4/3-3/2] possibly. Could’ve been [4/3-3/2 or 3/2-2/1].

How would a green-eyed person answer THAT last question?
A: [5/4-4/3 or 4/3-3/2] [4/3-3/2 or 3/2-2/1]…or
[4/3-3/2 or 3/2-2/1] [3/2-2/1 or 2/1-1/0]
B: [5/4-4/3 or 4/3-3/2] [4/3-3/2 or 3/2-2/1]…or
[4/3-3/2 or 3/2-2/1] [3/2-2/1 or 2/1-1/0]
C: [5/4-4/3 or 4/3-3/2] [4/3-3/2 or 3/2-2/1]…or
[4/3-3/2 or 3/2-2/1] [3/2-2/1 or 2/1-1/0]
D: [5/4-4/3 or 4/3-3/2] [4/3-3/2 or 3/2-2/1]…or
[4/3-3/2 or 3/2-2/1] [3/2-2/1 or 2/1-1/0]
E: [5/4-4/3 or 4/3-3/2] [4/3-3/2 or 3/2-2/1]…or
[4/3-3/2 or 3/2-2/1] [3/2-2/1 or 2/1-1/0]
It is that final question whose answer changes immediately after the announcement. No one, not A or B or C or D or E, would change their answer to the first four questions. They would, however, have to answer differently the final question. That last question is basically "A, what would B assess that C would assess about what D would say that E could be thinking about the number of green eyed people on the island?**

After the latter case, A would say that B would have to realize that C must know that D must assume that E (equivalent to any other chain, such as E-B-C-D-A) could be leaving the prison that very night.

Why? Because of that ending pair that’s no longer a pair:
[5/4-4/3 or 4/3-3/2] [4/3-3/2 or 3/2-2/1]…or…
[4/3-3/2 or 3/2-2/1] [3/2-2/1 or 2/1-1]

The next day, when everyone is still there, the number “1” disappears from the solutions:
[5/4-4/3 or 4/3-3/2] [4/3-3/2 or 3/2-2]…or…
[4/3-3/2 or 3/2-2] [3/2-2 or 2]

See the two all alone, without a pair? That would force any fifth-order thinking to say “OK, either no one is thinking 2 or the pair is leaving tonight. I already know that won’t happen, but I can’t be sure that it’s common knowledge.” Night two comes and goes, and this is the reduction:

[5/4-4/3 or 4/3-3] [4/3-3 or 3]…or…
[4/3-3 or 3] [3]

Hey look! A 3 all by itself. That means 3 people will leave tonight. But of course they won’t…

Can I skip 3 and 4 yet? That leaves us with:

[5].

Now everyone knows that everyone knows that everyone knows that everyone knows that everyone knows there are five green-eyed people in the prison. Well, all the greens, anyway.

**This is part most struggling puzzlers won’t bother to parse. They’ll just brain-dump it and pretend like it’s not there or not meaningful.

[Lucas_Jackson]

@MaxTheVool

OK, there are only two green-eyed people, Alex and Bob. Each one of them has thought for ages “there’s only one green-eyed guy, that guy over there… plus maybe I’m green-eyed. But that guy over there doesn’t know if he’s green-eyed, and of course I don’t know if I’m green-eyed”.

Then one day, the external guy comes along and says “there’s at least one green-eyed person”. How does Alex react? Well, he looks around, he sees only a single green-eyed person, Bob. He thinks “poor Bob, he’s never realized that he was green-eyed. But now this new information has arrived, and Bob now knows with certainty that there’s at least one green-eyed person. So… am I green-eyed? If I’m not green-eyed, then Bob sees that there are no green-eyed people other than himself. So, if I’m not green-eyed, and Bob is told that there is at least one green-eyed person, then Bob will know that he is green-eyed. So, tomorrow morning, Bob may have left the island.”

Then tomorrow morning comes along and Bob has not left the island.

Now Alex thinks “well, Bob did NOT realize that he was the only green-eyed person. Why not? if he saw only brown eyes, and was told there was at least one green-eyed person, he would have left. But he didn’t. So there must be another green-eyed person. Since I can see that all 98 other prisoners have brown eyes, the other green-eyed person must be ME!”. So the next morning comes, and Alex and Bob have both departed (because Bob was thinking exactly the same things, with the names reversed).
I’ll take this one step further, up to three people. To do so, I find it most easy to talk as if everyone assumes they do not personally have green eyes. It doesn’t change the logic, but it changes the mindset a bit.

So Charles also has green eyes. But he has always believed he doesn’t. So as far as he’s concerned, there are two green-eyed people, Alex and Bob. He hears the announcement. He then immediately runs through all of the above logic and says “well, there are two green eyed people on the island, therefore in two mornings they will both leave”. Two mornings later they do NOT both leave. So he’s thunderstruck, and he reconsiders all of his suppositions, and the only possibility remaining is that he, Charles, ALSO has green eyes. So the next morning, Alex and Bob and Charles all leave (and again, Alex was certain the whole time that Bob and Charles were the only two green eyes, and was shocked when they had not left on the second morning, and so forth).
Make sense?

Thanks for taking the time. Yes, with two it makes sense now. I guess I fall into the not being able to juggle more than that in my head camp. For now, I’ll take your word but I’m sure it will drive me crazy in the next few days.

[Garula]

I admit that this puzzle confuses me, in large part because it seems to require someone believe that someone believe that someone believe something illogical. I’ve tried to follow the thread as best I can, but I admit I may have overlooked something that would clear up my confusion.

The three person scenario makes complete sense to me. But I can’t follow it to four.

Supposing a scenario with 4 people: Adam, Betty, Cate, and Dan.

Adam sees 3 people with green eyes.
Therefor, Adam knows that every other person sees at least 2 people with green eyes.
Therefor, Adam knows that every other person knows that every other person sees at least 1 person with green eyes.

Adam sees 3 people with green eyes.
Adam knows that Betty can see 2-3 people with green eyes.
Adam knows that Betty knows that Cate can see 1-3 people with green eyes.
Adam knows that Betty knows that Cate knows that Dan can see 0-3 people with green eyes.

Except, since Adam knows that every other person knows that every other person can see at least 1 person with green eyes, he knows that it’s illogical for Cate to think it’s possible for Dan to think 0 people have green eyes and therefor illogical for Betty to think that Cate thinks it’s possible for Dan to think it’s possible for 0 people to have green eyes, right?

I suppose that if we’re applying induction, shouldn’t the crucial piece of information that needs to be stated at the beginning not be “At least one person has green eyes.” and rather be “At least (total number of green eyes - 2) people have green eyes.”?

[Chessic_Sense]

@Garula

Except, since Adam knows that every other person knows that every other person can see at least 1 person with green eyes, he knows that it’s illogical for Cate to think it’s possible for Dan to think 0 people have green eyes and therefor illogical for Betty to think that Cate thinks it’s possible for Dan to think it’s possible for 0 people to have green eyes, right?

No, not right. That “at least one person” that Dan sees has a name - it’s Betty. Adam knows Dan sees Betty. Adam knows that Cate is up to speed on that, too. But Adam also realizes that Betty couldn’t know that. So while Adam knows that Cate is keeping pace with him, he knows that Betty can’t say that Cate knows the same thing.

Fact: “Dan sees Betty’s green eyes.”
Who knows this fact?
Adam does.
Cate does.
Dan does, obviously.
Betty does not.

What do the people know of this list?
Adam, Cate, and Dan all know Betty’s not on the list.
Adam knows that Cate and Dan are on it.
Cate knows that Adam and Dan are on it.
Dan knows that Adam and Cate are on it.
But Betty has no clue. She only knows she’s not on the list, since she obviously doesn’t know the fact in the first place.

Finally, Adam, Cate and Dan all know that last part - that Betty doesn’t know if Dan sees Betty’s green eyes. They know other third-stage facts, but I’ll skip those cuz they’re not relevant right now. So in other words, everyone but Betty knows that Dan sees Betty’s green eyes, and everyone but Betty knows the other two know. They could even talk about it out loud, so long as Betty doesn’t hear.

therefor illogical for Betty to think that Cate thinks it’s possible for Dan to think it’s possible for 0 people to have green eyes, right?

Yes, it is illogical, but Adam doesn’t know that. Adam thinks it’s just the three people - B, C, and D. You’re having trouble following it because you keep giving Adam knowledge that he doesn’t have. He thinks it’s a 3-green island, though we know it isn’t. So he does indeed think that Betty thinks Cate thinks that Dan sees no green eyes. Who could Dan see? Not himself. Not Adam- Adam thinks he’s a brown. Not Betty - Adam thinks that Betty will count herself out. Betty, for her part, is assuming Cate counts herself out. So yes, there’s the misapprehension in Adam’s head that everyone else has a misapprehension about Dan’s count.

[Chessic_Sense]

By the way, we’re using the word “know” as shorthand. In reality, you should be saying “thinks it possible.” Adam doesn’t ever dare to declare “I know what X knows” exactly. Adam, or anyone, can only ever confirm or deny possibilities. If they ever knew exactly what another person thought, they would know their own eye color. That’s actually how they get off the island/out of prison in the end…they deduce what others know, realize it matches their perception, conclude their eye colors match, and thus treat their neighbor as a mirror.

The brown eyes, btw, realize their perspectives don’t match up and thus have to wait an extra day to be told as much.