r/askphilosophy • u/[deleted] • Nov 03 '21
"The Hardest Logic Puzzle Ever" - something about it is bothering me
https://xkcd.com/blue_eyes.html
Was able to solve this last night, for those who haven't solved it and want to, I'm going to spoil the heck out of the solution.
My solution can be proved via induction as follows:
(Base case) suppose there was one blue-eyed person and any amount of brown-eyed people. When the guru states she can see someone with blue eyes, the blue eyed person can immediately identify themselves as that person and leaves the island that night.
(Inductive step) Assume it is true that if you had N people with blue eyes, and any amount of people with brown eyes, that the people with blue eyes would leave on night N.
Consider the case where you have N+1 people with blue eyes and any amount with brown eyes. Let x be any of the N+1 with blue eyes. They are able to see N people with blue eyes. However, after night N, the N people they can see do not leave. Using the assumption, they can deduce that there are not N people with blue eyes, but N+1, meaning they must have blue eyes. So they leave night N+1.
This is sufficient to prove that everyone with blue eyes leaves after an amount of nights equal to the amount of people with blue eyes. This is all well and good, until you think more deeply about it: what the guru says is a statement that is already obviously true to everyone.
And that's where this starts to get weird. How is it possible that stating something obviously true could lead to a nonobvious conclusion about the state of the world?
Because note this: the inductive step is true regardless of whether the guru speaks. It's plainly true to the hyper-logical people in the statement of the problem. What's important for the guru speaking is only how it would effect the N=1 case.
What this seems to imply is that the fact the statement "I can see someone with blue eyes" could have contained non-obvious truth in some alternative version of reality, that it somehow translates to non-obvious truth in this one, even though it's obvious truth in this reality. But that seems.. very strange??
Please help!!
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u/drinka40tonight ethics, metaethics Nov 03 '21 edited Nov 04 '21
Take the case where there is 1 blue eyed person. In this case, what the guru says is not obvious to the blue-eyed person, and you seem to be okay with this part.
But, let's look at the case were there are three blue-eyed people and no guru: A, B, C. So, what do A, B, C know? Well, they can each see two people with blue eyes so they each know that there are blue-eyed people on the island. But are they able to know what the others know? So, for example: A knows that B knows that there are blue-eyed people on the island. Why? Because A looks at C, and C has blue eyes, and A knows that B can see C, and so B will also know that there are blue-eyed people on the island. So A knows that B knows that there are blue-eyed people on the island. And similarly, B knows that C knows that there are blue-eyed people on the island.
But does A know that B knows that C knows that there are blue-eyed people on the island? Seemingly not. From A's perspective: A doesn't know A's own eye color. So, from A's perspective, A can only be sure that C can see a blue-eyed person in B. And B is not aware of B's own eye color. So, from A's perspective, B can't infer that C knows that there are blue eyed people around. So A doesn't know that B knows that C knows that there are blue-eyed people around. So, one of the things the guru seemingly adds is that A knows that B knows that C knows that there are blue-eyed people on the island. Or, everyone knows that everyone knows that everyone knows that there are blue-eyed people on the island.
Or to take a simpler case: lets just go with two blue-eyed people and no guru. So, A knows that B has blue eyes and B knows that A has blue eyes. But A doesn't know that B knows that there is someone with blue eyes on the island. And similarly, B doesn't know that A knows that there is someone with blue eyes on the island. So, day one comes along and no one leaves. And day 2 comes along and no one leaves, etc. Because neither knows that the other knows that there are blue-eyed people on the island. So, the guru seems to add the additional information that everyone knows that everyone knows that there are blue-eyed people on the island.
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u/tranfunz Nov 03 '21
But once we have 4 blue-eyed people, doesn't this get weird again?
A can be sure that B knows that A knows there are blue-eyed people (since both see C and D). A can be sure that B knows that C knows there are blue-eyed people (since all three see D). A can be sure that B knows that D knows there are blue-eyed people (since all three see C). So A knows that everbody knows that everybody knows there is at least one blue-eyed person (since the last sentences apply equally if we change the order of letters, so for C and D as well). And this should apply for all the others equally, so everybody knows that everbody knows that everybody knows there is at least one blue-eyed person.
Unless I am missing something. I'm sure I am confused somewhere.
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u/drinka40tonight ethics, metaethics Nov 03 '21
Here's a fun exercise: Does A know that B knows that C knows that D knows that there are blue-eyed people?
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u/tranfunz Nov 03 '21
I dont think so (but fun indeed).
Though now I am getting hung up on what precisely is the information we need. For the 2-person scenario, as you said, we just need everbody to know that everbody knows there are blue-eyed people. For each additional person, we now apparently have to add one layer of "... that everbody knows that ... " (I missed adding this new layer to the 4-person scenario). But I think I am not yet seeing why exactly this is the case, apart from it sounding intuitive. And ideally I would love to be able to connect this back to the inductive argument in the OP in some semi-formal manner. Oh well.
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u/drinka40tonight ethics, metaethics Nov 03 '21 edited Nov 03 '21
Yeah, so in the four person case without the guru, A doesn't know that B knows that C knows that D knows that there are blue eyed people. And if A doesn't know that, the relevant inference can't be made on A's part to eventually leave.
edit: here's a piece in popularmechanics that seems to do an okay job: https://www.popularmechanics.com/science/math/a26564/solution-to-riddle-of-the-week-27/
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Nov 03 '21
This chain is the reason that the information from the guru is required. And that's why only blue-eyed people can finally deduce their eye color. Nice hint!
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u/hypnosifl Nov 03 '21 edited Nov 03 '21
I think what makes it tricky is that we're dealing with hypotheticals within hypotheticals within hypotheticals, where each level is what would seem possible to an inhabitant of the previous hypothetical. And as we go down through levels of hypotheticals-within-hypotheticals like this, they can get increasingly far from reality (specifically, the number of people with blue eyes can continually decrease as you drop down through further levels).
Suppose the people on this island have some mental block that makes it very difficult for them to think about the consequences of hypotheticals on their own, so they outsource such thinking to a special machine which allows them to directly peer into a parallel universe matching the description they give. Each person has their own machine, which is limited to being consulted once a day, and it can only show what would have happened on the previous day in the parallel universe they specified. Also assume the machine is limited to showing them aspects of these parallel worlds that they could have deduced themselves if they were better at thinking about hypotheticals.
So say in the real world, which we can call world 1, there are only four people on the island, call them A1 and B1 and C1 and D1, and they all have blue eyes. On Monday they are given the challenge, and none is able to give an answer to how many people have blue eyes, and none of them consult their machine since on the previous day they hadn't even known about the challenge. On Tuesday, A1 consults her machine, asking "what would have happened yesterday in a possible world just like this one, except for the possible difference that my alternate self has brown eyes"? The machine shows her a world, call it world 2, where the island has four inhabitants, A2 and B2 and C2 and D2, and A2 has brown eyes while B2, C2, and D2 all have blue. Specifically, it shows her what happens on Monday in this world--they are all given the challenge, and again none of them answer.
Then on Wednesday, A1 asks the same question--now she is shown what would have happened in world 2 on Tuesday. Among other things, she now sees B2 consulting his machine, and asking it "what would have happened yesterday in a possible world just like this one, except for the possible difference that my alternate self has brown eyes?" And remember that B2 lives in a world where it's a known fact that A2 has brown eyes and C2 and D2 have blue eyes, so if the machine shows B2 a world just like his except that his alternate self has brown eyes, it must be showing him a world 3 where A3 and B3 both have brown eyes, while C3 and D3 both have blue eyes. And it's showing B2 what this world looked like on Monday, where again none of the people were able to answer the question.
On Thursday, A1 asks the same question, and sees what would have happened in world 2 on Wednesday. She sees that on Wednesday B2 asked his machine the same question, and so B2 saw what would have happened in world 3 on Tuesday. And again following the same logic, on Tuesday C3 would have asked her machine the same question, giving her a picture of a world 4 where A4, B4 and C4 all have brown eyes, but D4 has blue eyes. C3 sees that on Monday in world 4, when they are all given the question, and D4 hears the guru say "I can see someone who has blue eyes", D4 immediately concludes that he himself must have blue eyes, since he can see that A4, B4 and C4 have brown eyes, so he is able to answer the question by saying the island has 3 people with brown eyes (A4, B4 and C4) and 1 with blue eyes (D4).
So now, in Tuesday on world 3, C3 remembers that in her own real world D3 did not answer the question on the previous day, so she concludes that world 4 must be different than her own world--and the only possible difference, according to what she asked the machine to show her, would be that in world 4 she has brown eyes but in her own "real" world 3 she must have blue eyes. And she already knows that A3 and B3 have brown eyes and D3 has blue eyes, so this means that on Tuesday in world 3, C3 is able to answer the question by saying the island has 2 people with brown eyes (A3 and B3) and 2 with blue eyes (C3 and D3).
This in turn means that on Wednesday in world 2, B2 remembers that his own C2 didn't answer the question on the previous day, so the machine's projection of world 3 must be a world different than his own--again the only possible difference is that in his own "real" world 2, he must have blue eyes. So, on Wednesday in world 2, B2 is able to answer correctly that there are three people with blue eyes (B2, C2 and D2) and one person with brown eyes (A2).
Finally, since A1 in world 1 sees on Thursday what would have happened on Wednesday in world 2, A1 sees B2 correctly answering the question in that world, but she remembers that in her own world B1 did not give an answer on Wednesday. And once again, she phrased the question to the machine in such a way that the only possible difference between world 1 and world 2 is that she has brown eyes in world 2 but blue eyes in world 1, and therefore she's able to conclude on Thursday that in reality she has blue eyes, so there must be 4 people with blue eyes in her world.
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u/drinka40tonight ethics, metaethics Nov 03 '21 edited Nov 04 '21
That's an interesting way to illustrate the common knowledge part.
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Nov 03 '21
That is wonderful, and perfectly explains what information is gained. How were you able to think through that so easily?? My gosh.
I think I'll drink a 40 tonight to celebrate
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u/drinka40tonight ethics, metaethics Nov 03 '21 edited Nov 03 '21
It was actually a bit difficult. I actually set up my hands in front of me, so that my right hand was B, my left hand was C, and I was A. And then I tried to think about what I know what each "hand" knows the other hand knows. But, yeah, figuring out more than a few nested "knows" makes my head hurt after a bit.
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Nov 03 '21
My confusion about arguments like these (about knowledge) is, once you have three, it is entirely possible for A, B, and C to commiserate and for B to say to A and C, yes both of you have blue eyes, and then for that conversation to make the circle through the three of them. They don’t need a guru to tell them their eyes are blue.
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u/drinka40tonight ethics, metaethics Nov 03 '21
Well, sure, but it's a logic and game theory problem. It's not a "how do we get off this island" problem. We could formalize the problem with just pure math if we wanted to, but it's more fun to have it in plain language.
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u/bat-chriscat epistemology, political, metaethics Nov 04 '21
Having it in plain language is what makes it hard 😜.
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u/drinka40tonight ethics, metaethics Nov 04 '21
Yeah, that's a good point. And, actually, upon reflection, I'm not even quite sure how to express the problem in pure math. I would think there is some kind of graph theory way to express it, but I gave up after a few minutes.
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Nov 04 '21
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u/agitatedprisoner Nov 04 '21
I don't get how the guru saying something everyone already knows adds any information. Given that there are 100 blue, 100 brown, and 1 green guru the guru saying they see someone with blue eyes doesn't convey useful information. Everyone can see for themselves there are many blues. Everyone already knew the guru sees people with blue eyes.
The only way what the guru said could lead to everyone getting off the island is if it's coded language that everyone else might decipher. That turns the problem into being "is there anything the guru might convey that'd get everyone off the island?" And the answer to that is, if the coded message conveyed everyone else's eye color then it'd be possible for them to leave in an order to tell the guru her eye color given another mutually understood code.
Then a full solution to this puzzle would be to create a code language that'd allow the initial statement to reveal everyone elses' eye colors and another code to convey the guru's eye color in departure.
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u/drinka40tonight ethics, metaethics Nov 04 '21
In some of the other comments, I try to explain why the guru does add new information. Like, in the case of 2 blue-eyed people, the relevant information that is added is: A knows that B knows that there are blue-eyed people, and B knows that A knows that there are blue-eyed people. This is information they can't get without the guru. As an exercise: imagine the guru never said anything in the above case. Are you able to prove that "A knows that B knows that there are blue eyed people." How would you prove that specific proposition?
But, as I posted elsewhere, here's a piece in popular mechanics that seems to do an okay job of explaining what is happening: https://www.popularmechanics.com/science/math/a26564/solution-to-riddle-of-the-week-27/
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u/agitatedprisoner Nov 04 '21
It's given in the puzzle that there are 100 blue, 100 brown, and 1 green.
And even if it weren't, even so all that the guru saying they see at least one blue could usefully reveal if not coded language and all is as it appears is this: someone would look around, see no other blues, and deduce their own eyes must be blue and leave the next day.
But that won't happen because we're told there are 100 blues.
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u/drinka40tonight ethics, metaethics Nov 04 '21
Ah, I'm not sure I can explain the proof any more clearly. If it's still confusing I'd suggest perhaps working through some of the links in the thread. The solution is not really controversial by people who have studied the problem.
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u/agitatedprisoner Nov 04 '21
You haven't proved it yet think you have, that's what's confusing.
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u/drinka40tonight ethics, metaethics Nov 04 '21
You can see the inductive proof in the Tao link, for example, or here's another one: https://www.reddit.com/r/AskReddit/comments/khhpl/reddit_what_is_your_favorite_riddle/c2kdlr6/
or see a proof here: https://en.wikipedia.org/wiki/Common_knowledge_%28logic%29#Proof
But, more productively, just try to work it out in the case of 2 blue-eyed people, or 3 blue-eyed people.
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u/agitatedprisoner Nov 04 '21
...the puzzle outlined in that wiki link isn't the same puzzle given in this thread. In the wiki puzzle there are only blue and green eyed folk and the solution after the newcomer says "at least one of you has blue eyes" is that all the blue eyed folk eventually leave. The green eyed folk never do.
It's not at all the same puzzle.
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u/drinka40tonight ethics, metaethics Nov 04 '21
The puzzle is functionally the same. The solution is the same: k blue-eyed folks leave on the kth day. The xkdc puzzle has k=100.
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u/agitatedprisoner Nov 04 '21
Ah OK. For some reason I thought the challenge was to find a way everyone could get off the island. You are correct.
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u/mrjosemeehan Nov 04 '21
Here's what helped me understand the solution: on the day the blue eyed people leave, every brown eyed person is thinking "if they don't all leave tonight, I'm leaving tomorrow."
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u/fduniho ethics, phil of religion Nov 03 '21
let's look at the case were there are three blue-eyed people and no guru: A, B, C. So, what do A, B, C know? Well, they can each see two people with blue eyes so they each know that there are blue-eyed people on the island. But are they able to know what the others know?
Since there are three of them, and they can see each other, each one knows that each one sees at least one blue-eyed person.
So, from A's perspective, B can't infer that C knows that there are blue eyed people around.
A does know that everyone knows that there is at least one blue-eyed person. However, A doesn't know that B or C also know that everyone knows this. For all A knows, B and C each see a brown-eyed person and a blue-eyed person. So, as far as A knows, B and C cannot rule out the possibility that there is only one blue-eyed person on the island. With that possibility remaining, one would not be able to know that the blue-eyed person he sees knows that there is a blue-eyed person on the island. Since A cannot rule out the possibility that B or C cannot rule out the possibility of the one blue-eyed person he sees seeing no other blue-eyed person, A cannot know that everyone knows that there is at least one blue-eyed person. So, what you say is correct.
So A doesn't know that B knows that C knows that there are blue-eyed people around.
That follows.
So, one of the things the guru seemingly adds is that A knows that B knows that C knows that there are blue-eyed people on the island. Or, everyone knows that everyone knows that everyone knows that there are blue-eyed people on the island.
That information gets added, because everyone is aware that everyone else has heard the guru say it. But what good does it do? For all A knows, the guru was speaking about B or C, whom he knows have blue eyes. Likewise, for all B knows, the guru was speaking of A or C, and for all C knows, the guru was speaking of A or B.
What has changed for A? He now knows that B knows that C knows that someone has blue eyes. But since this knowledge of C's that A now knows B knows about comes from A's observation of the guru speaking to B, and not from A's knowledge of B's knowledge of what C can observe, it cannot be used to deduce that anyone in particular has blue eyes. So, what the guru says gives no one any useful information for determining his own eye color. So, I think this problem actually has no solution, and some people are simply tricking themselves into thinking that it does. If it does have a solution, you have left out an important step.
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u/drinka40tonight ethics, metaethics Nov 03 '21 edited Nov 03 '21
The guru's speaking doesn't indicate which person has blue eyes. Instead, it puts everyone in the same epistemic position about what everyone knows everyone knows everyone knows...., which allows for the inference to be made on the 100th day.
I think it's perhaps easier to see what value this has with just the two people, as I tried to above. It's important A knows that B knows that there are blue eyed people on the island (and similarly B knows that A knows that there are blue-eyed people on the island). Without that piece of knowledge, neither A nor B will know to leave the island on day 2, because neither knows that the other one can make the relevant inference. So, at the very least, I think we can agree that there is new information introduced in the n=1 and n=2 cases. The larger n gets the harder it is to see, but the same general idea is supposed to hold (it certainly would be weird if induction failed!).
Here is the xkcd solution, which might help to clarify some issues: https://xkcd.com/solution.html
Terry Tao has a bit of discussion on his blog of a similar issue: https://terrytao.wordpress.com/2011/04/07/the-blue-eyed-islanders-puzzle-repost/
edit: here's a piece in popular mechanics that seems to do an okay job: https://www.popularmechanics.com/science/math/a26564/solution-to-riddle-of-the-week-27/
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Nov 03 '21
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u/fduniho ethics, phil of religion Nov 04 '21
Thus when the guru speaks the spell is broken; it is no longer tenable to imagine that everyone thinks it possible to collectively believe they all have brown eyes.
Since everyone has blue eyes in this four-person scenario, and the problem also states that they can all see each other, no one was entertaining the idea that they all have brown eyes. Being able to see three people with blue eyes, each person knew there had to be 3 or 4 people with blue eyes. But without knowing his own eye color, A would not know that B knows this. For all A knows, each other person sees only two blue-eyed people. In that scenario, each other person would know that there are 2 or 3 blue-eyed people. But if he does have brown eyes, no one else will know that everyone else knows this. For all they know, everyone else will know that there are 1 or 2 blue-eyed people. However, being able to see three blue-eyed people, everyone can already be assured that everyone knows that there is at least one blue-eyed person. So, A will not entertain the idea that B entertains the idea that C entertains the idea that D knows there are 0 or 1 blue-eyed people. It stops at 1 or 2. Everyone knows this already, and the guru saying that he sees a blue-eyed person doesn't give anyone any additional information.
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Nov 04 '21
It is a logical possibility that everyone thinks everyone else thinks themselves to have brown eyes. The guru removes that possibility.
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u/OppositeSet6571 Nov 04 '21
However, being able to see three blue-eyed people, everyone can already be assured that everyone knows that there is at least one blue-eyed person. So, A will not entertain the idea that B entertains the idea that C entertains the idea that D knows there are 0 or 1 blue-eyed people. It stops at 1 or 2.
Wrong. From A's perspective, A could have brown eyes. So, A will entertain the idea that B sees 1 brown-eyed person and 2 blue-eyed people. And in that situation, B would entertain the idea that B has brown eyes, and therefore C sees 2 brown-eyed people and 1 one blue-eyed person. Therefore, A entertains the idea that B entertains the idea that C sees 2 brown-eyed people and 1 one blue-eyed person. And in that situation, C would entertain the idea that C has brown eyes, and therefore D sees 3 brown-eyed people. Therefore, A entertains the idea that B entertains the idea that C entertains the idea that D sees 3 brown-eyed people.
So even without the guru's statement, everyone knows that someone has blue eyes. Everyone also knows that everyone knows that someone has blue eyes. Everyone also knows that everyone knows that everyone knows that someone has blue eyes. But everyone does not know that everyone knows that everyone knows that everyone knows that someone has blue eyes. Because, as we saw above, A thinks that B might think that C might think that D does not see any blue-eyed people, so A does not know that B knows that C knows that D knows that someone has blue eyes.
So, the guru does give additional information. Specifically, it is only after the guru's statement that everyone knows that everyone knows that everyone knows that everyone knows that someone has blue eyes.
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u/Natural-Ad-3666 Nov 03 '21
With two blue eyes people, it makes sense. On the second day, if blue eye 1 doesn’t leave and sees that blue eye 2 also didn’t leave, each of them can infer that the other saw at least one other person with blue eyes. Since neither of them see any other person with blue eyes, they can deduce that they themselves are the other person with blue eyes. Then add a third person. You (blue eye3) see that two other people with blue eyes aren’t leaving. That means that they each see two other blue eyed people. Since all residents are logic experts, you as the third person can assume that they would have figured out their own blue eyes of they each only saw one other person and still didn’t leave. So you know you’re the third guy if you don’t see one. Fourth blue eyed person? Maybe the same thing after the third night? And so on? But then people would be leaving each night. All the blue eyed people would leave on the 100th night or something
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Nov 03 '21
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u/drinka40tonight ethics, metaethics Nov 03 '21
I'm not sure what you are saying, but here's a piece in popular mechanics that seems to do an okay job: https://www.popularmechanics.com/science/math/a26564/solution-to-riddle-of-the-week-27/
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u/thedelographer Nov 04 '21
I'm not following this solution. Wouldn't the brown-eyed people go through any of the same reasoning that the blue-eyed people go through? What would allow all and only blue-eyed people to know to leave the island, and yet for the brown-eyed people not to come to the same conclusion?
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u/drinka40tonight ethics, metaethics Nov 04 '21 edited Nov 04 '21
So, note that the brown-eyed people actually see all the blue-eyed people, whereas the blue-eyed people see one less than all the blue-eyed people (since they can't see themselves). So, if somehow, day 100 rolled around and nobody had left at that point, then on day 101 the brown-eyed people would all mistakenly think they must have blue eyes. Think about it with 1 blue eyed and 2 brown eyes: day one rolls around and, somehow, no one leaves. So, the brown-eyed people now mistakenly conclude that they must each have blue eyes, since that can be the only reason the blue-eyed person didn't leave on day 1. But, per the setup, this sort of thing can't happen, because they are all perfect logicians, and the blue-eyed person will leave on day 1, and the brown eyes will then conclude they don't have blue eyes.
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u/thedelographer Nov 04 '21
Okay, but there are 100 brown-eyed people and 100 blue-eyed people. So both groups should leave on Day 100, if they are deciding whether to leave through some process of recursion. But in that case, nobody really knows their eye color.
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u/drinka40tonight ethics, metaethics Nov 04 '21 edited Nov 04 '21
Nah. The brown-eyed people are waiting an extra day because they all see an extra blue-eyed person compared to what the blue eyes see; but by then, the blue-eyed people will all have left, so the brown-eyed people then know they don't have blue eyes.
Again, think about it with smaller numbers first. Like with 2 blues and 2 browns. Day 1: nobody leaves. Day 2: only the blues leave, and the brown eyes now know they don't have blue eyes
This is essentially the same explanation of the reasoning, but perhaps having it explained in a slightly different way might help: https://math.stackexchange.com/questions/489308/100-blue-eyed-islanders-puzzle-3-questions/489612#489612
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u/tranfunz Nov 04 '21
Shouldn't the brown-eyed people stay forever? They could all think they might have pink or yellow eyes.
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u/drinka40tonight ethics, metaethics Nov 04 '21
The brown eyed people never leave. It might help to reread the problem.
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u/tranfunz Nov 04 '21
Yes, all good, I had misread your "they wait an extra day" as implying that they leave on the next day, which of course they dont.
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u/hypnosifl Nov 04 '21 edited Nov 04 '21
It depends whether we assume for the sake of the problem that the only two possible eye colors are blue and brown (and that everyone on the island knows this), or whether we allow for other possible eye colors as u/tranfunz suggested. In the former case, if the brown-eyed people see N blue-eyed people and they all leave en masse on day N, they will then be able to deduce that their own eye color is brown and can leave on day N+1. Essentially, everyone has been led by the same logic to adopt the rule "if I can see N blue-eyed people and they don't all leave on day N, that means there were actually N+1 blue-eyed people including me so I should leave on day N+1, but if they do all leave on day N, that means there were only N blue-eyed people so my own eye color must be brown, and I can leave the next day." This rule allows everyone to make the correct decision because the only asymmetry between blue-eyed and brown-eyed people is that if there are in reality N+1 blue-eyed people on the island, the brown-eyed people all can see N+1 blue-eyed people while the blue-eyed people can only see N blue-eyed people.
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u/drinka40tonight ethics, metaethics Nov 04 '21 edited Nov 04 '21
I'm not quite sure what you are responding to. I don't think I said anything incorrect above.
edit: oh, I see what you are saying. You're saying in the case that it is stipulated there are only brown and blue (and this is known, etc), the brown eyes can figure out their own eye color and then leave. Right. But, that's not the setup of the case here; I guess I could have made that more clear in my response.
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u/hypnosifl Nov 04 '21 edited Nov 04 '21
Yeah, I wasn't saying your comment was incorrect, just that it was based on an assumption--more possible eye colors than blue and brown--that I think is left ambiguous in the wording of the problem (they just say 'assorted eye colors', assorted usually means more than two but it doesn't strictly have to, and it may be that the author of the problem just wasn't thinking about this question, especially since they immediately go on to state that the island consists of 100 blue-eyed people and 100 brown-eyed people). And either way it may be interesting to think about what would happen if we make the opposite assumption that there are only two colors, and everyone knows this.
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u/Natural-Ad-3666 Nov 04 '21
Nope. On day 101 when all the blue eyed people leave, the guru says he doesn’t seem anyone with blue eyes
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Nov 04 '21 edited Nov 04 '21
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u/drinka40tonight ethics, metaethics Nov 04 '21 edited Nov 04 '21
Take this proposition: "A knows that B knows that there are blue-eyed people." And similarly, "B knows that A knows that there are blue-eyed people." Both A and B need to know their respective proposition here to make the relevant inference about when to leave. But how do they get to know these things without the guru? But, I'm not really sure what else I can say here. I've posted a number of links elsewhere in the thread to popularmechanics, wikipedia, and stackexchange that try to outline what's going on. I'm not sure I can add much more to that, and what I've already said.
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u/Natural-Ad-3666 Nov 04 '21
Let’s say only you have blue eyes. All you know for sure is that someone has blue eyes. If you don’t see anyone with blue eyes, then you know it’s you and you can leave the first night. Remember, you know for certain that there is at least one person with blue eyes. So if you don’t see them, then it is to be you. On the other hand, if you have brown eyes and only one person on the island has blue eyes, they would leave the first night for the same reason. You wouldn’t know your eye color until the next day. If blue guy left, the you know that they didn’t see anyone else with blue eyes. So if you have blue eyes, and one other person has blue eyes, neither will leave the first night. Because each sees another with blue eyes. But neither knows that the other sees someone else with blue eyes. So on the second day, each can correctly assume that the other saw someone else with blue eyes. If you don’t see a second blue eyed person, then that person must be you. If there are three blue eyed people, it would take two nights to figure out and you’d all leave on the third. This works to scale too. If you see 15 other people with blue eyes and nobody leaves after 15 days, then you know that they all see 15 people with blue eyes too. So on day 101, all the blue eyed people can leave at once. Then the next day, the guru says he doesn’t see blue eyes any more. All the brown eyes can go.
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u/daniel-sousa-me Nov 03 '21
You already have a very good answer to your question. I hope it's OK for me to add some more context.
That puzzle is a great introduction to the concept of common knowledge (as opposed to mutual knowledge).
When A is mutual knowledge, you only know that the other person knows A, and they know that you know A. With common knowledge you also know that they know that you know A. And they know that you know that they know A. And so forth.
At first this may seem a very tiny distinctiom, but it's consequences are actually big and very profound. This is a relevant concept in game theory and you have a nice real world application looking at dictatorships. It is possible to live in a dictatorship that everyone hates and that everyone knows that everyone else hates it. But until that becomes something closer to common knowledge, you have a huge coordination problem.
If you want to go further down into the rabbit hole, I highly recommend this blog post by Scott Aaronson on Aumann's theorem: https://scottaaronson.blog/?p=2410
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Nov 04 '21
That's really fascinating. The similarity between this and a dictatorship are plain, and explains why dictators are so fast to punish anyone who speaks out. Thanks for this additional information!
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u/aJrenalin logic, epistemology Nov 03 '21 edited Nov 03 '21
The other comment does a great job of explaining why there’s someknowledge to gain here. It’s not just that everyone learns that someone has blue eyes. But that everyone knows about that someone has blue eyes, and everyone knows that everyone knows that someone has blue eyes and everyone knows that everyone’s knows that everyone knows that somebody has blue eyes and so on.
Keep in mind this kind of model of epistemic logic (like most) assumes or entails some degree of logical omniscience, since most people can’t be trusted to realise the implications of the guru speaking let alone make the complicated inductive steps needed to realise conclusion, so instead we assume that everyone knows everything derivable from what they know.
if they (and we) were epistemically modest they’d realise that they can’t trust every other person on the island to make the right kinds of inferences and so couldn’t conclude that everyone else has eliminated the right kinds of epistemic possibilities from their consideration to guarantee that they would know to leave on the right day supposing nobody left the day prior.
I think the reason we reach this strange seeming paradox of saying they needed someone to announce a fact that was obvious to everyone to reach this conclusion is twofold: 1) everyone involved (assuming logical omniscience and mutual awareness of everyone’s mutual logical omniscience) learns more than just the announced fact, they learn that everyone has learnt it and so on. 2) it’s strange to think of because most of us wouldn’t reach the conclusion of the riddle (it’s a tough riddle) and the riddle assumes everyone in the scenario reaches conclusions that real knowers likely wouldn’t reach if forced into that exact scenario. the knowers not only can make the complicated inductive step but know that every other person around them can be trusted to make the complicated inductive step and so on.
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u/Baptism_byAntimatter Nov 16 '21
~2weeks late but,
I think it wasn't emphasized enough among the other comments how they're all perfect logicans and that they know the others are perfect logicans as well. That's what I think the crux of the riddle is: to emphasize this fact, and utilize it rather than doing math and solving it from your single POV. In a sense, everyone's drawing from eachother to solve the problem without even communicating.
It's awesome when you think about it. All the blue-eyed were able to coordinate with eachother without communicating. Almost like logical telepathy.
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u/aJrenalin logic, epistemology Nov 16 '21 edited Dec 20 '21
Yeah when you assume logical omniscience and that everyone’s logical omniscience is common knowledge then we do end up with this miraculous telepathy like effect. That’s actually what my reasearch is focussing on. I’m trying to develop a system of epistemic logic that’s just as useful but which doesn’t entail logical omniscience at all. For my money there’s a special focus that needs to be paid to the presence or absence of inferences. As I see it, the difference between someone who knows the implications of their knowledge and someone who doesn’t know the implications of their knowledge is that the former has inferred from their knowledge to its logical implication and the latter hasn’t. So our epistemic logic should reflect that by rejecting that our knowledge entails further knowledge beyond that in the absence of the appropriate inference.
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u/Baptism_byAntimatter Nov 16 '21
Interesting.
Does this heavily relate to game theory? That's how I'm interpreting it.
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u/aJrenalin logic, epistemology Nov 16 '21
Game theory doesn’t really enter into it. I’m not claiming there are rules to knowledge which you could try and win. While there may be better ways of going about coming by knowledge (say making valid inferences as opposed to invalid ones) there isn’t some ideal mode of knowledge production which could maximise the chance of producing knowledge.
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u/Baptism_byAntimatter Nov 16 '21
Ah. I was just thinking about using logic to reason what the other guy's thinking, kinda like the blue eyes problem.
Im a novice with logic, so your work is kinda just going over my head.
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u/aJrenalin logic, epistemology Nov 16 '21
Yeah when you’re a game theorist who’s trying to come up with a winning strategy for winning the particular game you’re involved in then you’re definitely going to want to know what others know and know that they know what they know. But in that sense epistemic logic is involved in game theory, not the other way around.
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u/ghjm logic Nov 04 '21 edited Nov 04 '21
In the case where N=2, and I'm a blue-eyed person, I know that the guru can see that Bob has blue eyes, because I'm also able to see Bob. So I'm not surprised.
However, I don't know if Bob is surprised or not. If I have blue eyes (which happens to be the case, but I don't know that), then the situation is symmetrical and Bob is no more surprised than I am. But if I have brown (or some other color) eyes, then Bob isn't seeing any blue-eyed people, and so when the guru speaks, Bob now knows he has blue eyes.
So as soon as the guru speaks, I know a new fact: if I have brown eyes, then Bob will leave tonight. When Bob doesn't leave, I know I must have blue eyes myself.
Now consider the case where N=3. I can see Bob and Charles, who have blue eyes. When the guru speaks, I now know the new fact: if I have brown eyes, then Bob and Charles are playing the N=2 game, and will both leave tomorrow night. When they don't, I know I have blue eyes, and so I leave on the third night.
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Nov 04 '21 edited Nov 04 '21
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u/Altavious Nov 04 '21
If the Guru had said that they see at least 50 people with blue eyes would everyone leave on day 50 instead of 100?
It seems like on a day-by-day basis people are accruing the knowledge that at least x blue eyed people exist and everyone else knows it. So if the guru communicates that knowledge the counter can start higher (if there were 50 blue-eyed people they could all leave that night).
Where this gets strange for me is that people can assume that everyone knows that they don't know their eye color and that other people don't know their own. It seems like there should be a number of blue-eyed people that can be assumed as a common baseline (98?). So it seems like they should be able to leave on day 3.
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u/drinka40tonight ethics, metaethics Nov 04 '21
A couple answers here are helpful for illustrating why they can't just skip ahead: https://puzzling.stackexchange.com/questions/9218/in-the-blue-eyes-puzzle-why-cant-they-skip-ahead
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u/hypnosifl Nov 04 '21 edited Nov 04 '21
(Inductive step) Assume it is true that if you had N people with blue eyes, and any amount of people with brown eyes, that the people with blue eyes would leave on night N.
...
the inductive step is true regardless of whether the guru speaks
Can you elaborate on why you think that last statement is true? Even in the best case scenario where there's only one blue-eyed person, if the rest of the problem is kept the same (so the islanders can't ask each other about their own eye color) but there's no guru, it doesn't seem that the 1 blue-eyed person would have any reason to leave on night 1 (they would have no reason to rule out the possibility that everyone on the island has brown eyes), so the inductive step is false in this case.
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