No one’s expecting you to imagine someone to look around and see two. We’re just saying you might imagine that someone else might imagine that someone looks around and sees only two.
“Bob thinks it might be the case that (Fred thinks it might be the case that X)” is NOT the same as “Bob thinks it might be the case that X”.
And there’s the error that everyone makes at first, and that some in this thread are still making. You’re not being asked to imagine one of those three looking around and seeing two. That would be an A-C thought train, which isn’t what’s being asked. You’re being asked about the A-B-C thought train.
Our human minds are not geared to imagine down more than three levels, which is why you’re finding it so hard. That’s why my explanation used writing instead. It’s a permanent record of thoughts that can be passed from person to person. It helps your mind to cope with the difficult process of digging down multiple layers.
Go back and read my post where the islanders write down their answers. Tell me if it makes sense to you. If not, tell me where you fail to understand. Because at some point you’re saying “No, they can’t be thinking that, because I know it to be false,” which is as erroneous as saying “No, Johnny can’t answer that 2+2=5 because I know that to be false.” Sure, but Johnny might not.
To clarify, I think you all are confusing my understanding of the situation with my willingness to play along.
I fully get A thinking through B’s eyes, thinking through C’s eyes, thinking through D’s eyes and seeing just one blue eyed person. I get it.
But I’m not D. I’m not C. I’m not B.
I’m A.
I can look around and see four other blue eyed people and I know each of them can see at least three other people. So while it’s possible to play the game “what does D see” I end up with an answer I, as A, know to be factually incorrect. I know I see four blue eyed people. I know B, C, D, and E all see at least three blue eyed people. I am certain everyone knows there’s at least one blue eyed person.
I’d also like to point out that you’re failing the false-belief test and the appearance-reality task. We’re showing you blue eyes and you’re refusing to believe that others might think they’re brown.
I can imagine pink unicorns but that doesn’t mean I get to make a wish.
Again, I get that I, as A, can imagine B as he imagines C as he imagines D as he imagines E and come up with an answer based upon that scenario. It may be logical, but it’s not factual.
The Guru seeds the base case.
The solution relies on a base case: the theoretical situation with one inhabitant who only sees brown eyes needs the Guru to provide “I see at least one person with blue eyes” to deduce that he has blue eyes.
What is tripping people up is that this base case does not actually exist, but what The Guru does is inform this base case upon which all other cases infer their solution. Without this theoretical base case, the islanders can not deduce their own eye color.
Correct. That’s the trick. I totally get why people are rejecting the solution, which is what makes the problem interesting. The solution depends on an elaborate fiction that all islanders accept. Reject the fiction and well, have fun being stuck on your little island.
I actually agree with you and I like the counting down point of view, but this method doesn’t really address the issue many are having with the solution. That is, you should know that Indistiguishable doesn’t think there could be 97 blue-eyed people because you yourself can see 99 blue eyes so Indistinguishable should be able to see at least 98 (if Indistinguishable has blue eyes) or at least 99 (if he has brown eyes). You know this!
The problem hinges on FICTIONAL subcases. Everyone has to believe in the fiction for it to work. So instead of naming a particular individual on the island, we have to consider all the fictional cases, with fictional islanders with fictional bits of information and what they would do with that information even though they aren’t real.
Okay, I think I’m starting to get it a little bit – the problem is that A is unable to presume about what anyone thinks about her eye colour (because she doesn’t know her own eye colour), and that’s why she doesn’t know anything about what everyone thinks about anyone’s eye colours or about anyone’s opinion about anyone’s eye colours.
And the reason that’s important – and here it’s dim for me, what I’m trying to do is understand why everyone doesn’t just go “no shit, Sherlock” when the guru makes her pronouncement, and even here it’s not helpful – is because the win condition isn’t having blue eyes, it’s knowing you have blue eyes. And your only way of doing that is deducing what everyone else knows about your eyes, and the only way of doing that is deducing what everyone knows everyone knows (x99) about your eyes, so you have to start with something that everyone knows, and everyone knows everyone knows, and so forth at all levels, because you have to start breaking down the hypotheses.
Is that right? Like I said, I’m wording this informally because I already read the proofs but I’m trying to conquer my intuition (trying to figure out why it’s true as opposed to that it’s true.)
(" I know. You know I know. I know you know I know. We know Henry knows, and Henry knows we know it. We’re a knowledgeable family.")
Yes, you can be sure everyone knows. You know that B knows there’s at least one blue. You know that C knows. You know that D knows. You know that E knows.
But you don’t know that B knows that C knows, etc. Why? Because you don’t know your eye color, so you can’t count on the one blue being you. And B doesn’t know his eye color, so you can’t count on him assuming it’s him. And you both know that C can’t assume it’s him. And so on down the line. In the end, no one is willing to fess up and say “I’m the blue-eyed person everyone can see”.
Seriously. I typed it out earlier. What step did you fail to understand? I’m not being beligerent, I really want to know what part you didn’t get so I can explain it.
Heh, I was about to link to that a while ago myself; I have it open in another tab and everything. But I refrained, on grounds it might be taken as insulting. 
[Also, I hereby acknowledge that you thought to actually play the game a page before me. And still no one took it up!; the most basic way of definitively settling the question about what people would do by seeing what people would do…]
I think that I should switch doors …
This logic puzzle seems like the converse of the Monty Hall problem, where the solution is more obvious if you increase the number of doors. In the “Blue eyes” puzzle, the solution is more obvious if you decrease the number of islanders, and even more so if you decrease the number of islanders who have blue eyes.
The logic puzzle as presented on the XKCD site includes the following paragraph:
Superficially, it seems that the paragraph is clarifying the puzzle by providing some specifics and some examples, but it contains useless and misleading information – deliberately, I’m sure. It doesn’t matter how many brown-eyed people there are and it doesn’t matter what color eyes the Guru has. (A red herring, perhaps?) In fact, it doesn’t matter what color their eyes are, other than the fact that they are not blue. There is no additional information provided by the Guru that tells us anything about brown-eyed or green-eyed people, or about anything other than the presence of at least one person with blue eyes. And, it doesn’t matter in the puzzle how many blue-eyed people there are as long as it is a number that we can accept as being capable of being processed and tracked by the islanders. 100 is fine, but so is 400 or 4,000, or 4. I think that the puzzle uses 100 because it is big enough to cause confusion but small enough to appear manageable. (Also, having just one blue-eyed person in the puzzle would make the solution trivial, although it is the start of the solution for larger numbers.)
So, at its simplest, the puzzle involves some islanders who have blue eyes and some islanders who don’t have blue eyes. (And, of course, all the details specified in the first paragraph of the puzzle.) Then someone announces: I see someone with blue eyes. Why does this announcement change anything? Several posters have already answered why. I started to formulate one more explanation but I have nothing new to add. I’ll just emphasize that the key scenario is where there are two people with blue eyes, and understanding why, shortly after midnight following the Guru’s announcement, each blue-eyed person will determine the color of his own eyes. And, then understanding how the same logic applies to larger numbers of blue-eyed islanders.
It’s the difference between logic and fact.
If I’m A I can logically go through the stages. What does B see? OK, now if I’m B imagining C, what do I see? Now I’m I’m B imagining I’m C imagining I’m D what do I see?
I can do that. I can be completely consistent and logical in my deductions. But the end result of that thought chain is an answer that I, as A, know is not factual. It may be logical based upon the constraints I’ve placed within the problem but it’s not factual and so I must reject the answer as incorrect.
Suppose there’s a five-letter password. You (Enderw24 or whoever wants to play along) know all the letters of the password except for the third one. Mario knows all the letters except the first one, Luigi knows all the letters except the second one, Toad knows all the letters except the fourth one, and Princess Peach knows all the letters except the fifth one.
Specifically, what you know is that the password looks like HO*SE
You think the password might be HORSE and it might be HOUSE. You don’t know which it is.
You think the password might be HOUSE, in which case Mario would see *OUSE.
You think the password might be HOUSE, in which case Mario would think the password might be HOUSE and might be MOUSE.
So you are unable to rule out the possibility that (Mario is unable to rule out the possibility that the password is MOUSE).
You know damn well that the password isn’t MOUSE. But you think it might be the case that Mario thinks it might be the case that the password is MOUSE.
Is that clear so far? Let’s keep going.
You think the password might be HOUSE, in which case Mario would see OUSE and think the password might be MOUSE. If the password were MOUSE, then Peach would see MOUS, and think the password might be MOUSY.
So you think Mario might think that Peach might think the password is MOUSY. This is even though you know damn well that no one thinks the password is MOUSY; you yourself know the password is of the form *OUSE, and you know everyone has the password correct to within 1 letter, which MOUSY would not be.
But that does not conflict with the fact that you think Mario might see OUSE, in which case Mario would think that Peach might see MOUS, in which case Peach would think the password might be MOUSY. You think Mario thinks that Peach thinks the password might be MOUSY, even though you know damn well that Peach does not think the password might be MOUSY.
Mario does not have as much information about what Peach sees as you do. You happen to know that Peach sees an ‘H’ at the beginning of the word. Mario does not know that. You know that Mario does not know that. So you know that Mario might consider possibilities for Peach’s guesses which include 'M’s at the beginning of the word, even though you know that Peach would never do that.
Everything you’ve said here is true, but let me explain why it’s not relevant.
No one’s saying A should accept what he knows B knows C knows D knows E knows about eye colors as correct. Instead, we’re pointing out the fact that what A knows B knows C knows etc… changes after the guru speaks–and the way it changes is exactly what lets A deduce his own eye color after 5 days. (Assuming five people here instead of a hundred…)
It’s not so much, actually, about what A knows B knows C knows etc… rather, it’s about What A knows B might think about what C might think about what D might think about what E might think. That, hopefully, reinforces my point that no one’s saying A should agree with his imaginary-to-the-fifth-power E. The whole thing is just about what people might think anyway. The guru’s utterance ends up eliminating some of those possibilitiesl.
Sounds like you’ve got it. But if you want to phrase it informally, you should phrase it this way: “I believe it’s possible that I have brown eyes, while Bob believes it’s possible he has brown eyes, while Charlie believes it’s possible he has brown eyes…Zed believes it’s possible he has brown eyes. (Also, a lot of other possibilities).” Never once do we say anyone actually does have brown eyes. Merely that no one’s eye color is common knowledge. And once the announcement comes, that possibility is no longer there. It’s not possible for everyone to simultaneously think they have brown eyes.
The last line here should be “…damn well that no one thinks the password could possibly be MOUSY”.
Put another way: Suppose we were to go around asking first:
And suppose, just for the sake of argument, that everyone knows Mario is so vain, he likes to use the letter ‘M’ whenever he makes a random guess. And everyone knows Peach tends to use the letter ‘Y’ whenever she makes a random guess.
Would it be reasonable for you to guess “MOUSY” in 3), even though you know the password is actually of the form HOSE and Peach’s guess will actually be of the form HOS*? Sure! Why not?
And to be perfectly, brutally clear, when we say “might think”, we mean “considers it as a possibility” or “cannot rule out”.
And we should also point out that Mario may not ever be thinking that the password is actually MOUSE. He might think it’s MORSE or HORSE and you’ve been wrong the whole time with your HOUSE guess. You think it’s either HOUSE or HORSE. Mario thinks it’s either MOUSE or HOUSE. So any outside observer could conclude that the password must be HOUSE. Yet you’re unable to conclude that. You’d think of Mario’s thoughts “He either thinks [MOUSE HOUSE] or [MORSE HORSE]. But not both sets. One or the other.” Then in considering Peach, you’d say “Mario thinks Peach is thinking of [MOUSY HOUSY MOUSE HOUSE] or [MORSY HORSY MORSE HORSE].” And that’s how the error builds. See it clearer now?
Yes, exactly.
Huh? B and C can both see blue-eyed D and blue-eyed E. B knows C can see D and E.
Or are you proposing a different scenario from Enderw24’s 5x blue-eyed scenario?
Let’s suppose the islanders are actually rather vain logicians, and furthermore, that green-eyes are considered by far the most beautiful.
Even though every islander knows they might possibly have blue eyes, they hope that they instead have green eyes. Being logicians, they never hope for anything they know is impossible, but that’s ok; they all think it’s possible that they personally have green eyes.
Alice and Bob both have blue eyes. But Alice knows that Bob hopes that Alice sees Bob to have green eyes. Alice knows that Bob thinks it might be the case that Alice thinks Bob has green eyes. This is so EVEN THOUGH Alice knows that Bob totally doesn’t have green eyes.
Once you understand what just happened with Alice and Bob, you can understand how Alice hopes Bob hopes they both have green eyes. And in the same fashion, how Alice hopes Bob hopes Carl hopes Derrick hopes Erasmus hopes all five of them have green eyes. EVEN IF Alice knows everyone else has blue eyes, she might still plausibly hope that Bob plausibly hopes that Carl plausibly hopes that Derrick plausibly hopes that Erasmus plausibly hopes all five of them have green eyes, by the same exact sort of phenomenon that happened in the paragraph above.
Let’s just back up and re-state the original reasoning. Anyone who has a problem with the original reasoning, tell me at what step it fails.
A: If there were exactly one blue-eyed person, we can logically deduce that he would leave on Night 1.
B: We were able to logically deduce this, and all islanders are perfectly logical, so every islander is also able to deduce this, and so all islanders know A.
C: If there were exactly two blue-eyed people, then on the first night, each one might stay put, thinking that the other one might be the only one. So nobody would leave on Night 1.
D: But everyone knows that if there were only one blue-eyed person, he would leave on Night 1. Given that nobody left, everyone knows that there isn’t only one blue-eyed person. Each of the blues would then look around and see only one blue, and realize that they must be the other blue. So all blues would then leave on Night 2.
E: But this is a logical deduction, which all the islanders are capable of. So all islanders know that if there were only two blues, then they’d both leave on Night 2.
F: If there were exactly 3 blues, then on the 2nd night, they would all stay put, since they would each think that the other two blues might be the only blues.
G: But once night 2 passes with nobody leaving, each of the blues would know that there are not exactly 2 blues, because if there were, they know they would have left. So after that point, each of the blues knows there must be at least 3 blues, and they must be one of them.
H: So if there are 3 blues, then they will all leave on the 3rd night.
I: But this is a logical deduction, and so the logical islanders will deduce it. So all islanders know that if there are exactly 3 blues, then they will all leave on the 3rd night.
J: So if there were 4 blues, they’d all stay put on Night 3, because they might think that the other 3 blues are the only ones, and are going to leave
K: But after Night 3 passes without anyone leaving, all the blues would know that there are at least 4 blues, because if there were only 3, then they would have left.
L: So now all four of the blues must know that they themselves are blue-eyed.
M: So on the 4th night, if there were exactly 4 blues, all of them would leave.
N: But this is a logical deduction, and so all of the islanders, being logical, will know that if there were 4 blues, they’d all leave on the 4th night.
I can keep going with this, but I don’t think it’d be productive to carry it any further; the pattern is the same. So, if there’s a flaw in the reasoning, which step is it in?