# Talk:Blue Eyes

Is it really incomplete on the grounds that Joel hasn't be identified? Explanations of comics 57-59 leave no more explanation of "Scott" than that he appears to be Randall's friend. The fact that we don't have a last name for him doesn't make either Scott or those comic explanations incomplete. Similarly, not have a full identifier for "Joel" in this one doesn't, in my opinion, warrant an incomplete tag. I'm removing the tag. If anyone object, revert it. Djbrasier (talk) 19:22, 22 May 2015 (UTC)

The proof for this puzzle is incomplete, if not wrong. The theorem is too weak, it should be: "Theorem: N blue eyed people with Nth order knowledge of all N people being logicians, N people having blue eyes, and any blue eyed person will leave as soon as possible after deducing they have blue eyes, will be able to leave on the Nth day." This may seem pedantic, but it really gets to the heart of the problem, which is trying to illustrate the use of orders of knowledge. In the theorem as stated, just N blue eyed people will leave on the Nth day, the proof for the inductive steps does not hold. You need to further assume that the person is able to deduce the hypothesis (which should be proven). In other words, you say X-1 people would leave on the (X-1)th day by hypothesis, so the Xth person knows he can leave on the Xth day. But you did not prove that the Xth person can actually deduce this, namely that he has all the information necessary to do so. In the correctly stated hypothesis, you then need to show that N + 1 people with (N+1)th order knowledge of all those things can deduce that the N people would leave if it was just them, and further that N+1 people have (N+1)th order knowledge of all these things. This is very important, and holds true (Since N+1th order knowledge is equivalent to knowing the N people have the Nth order knowledge necessary to fulfill the hypothesis, and by symmetry if the N logicians can figure it out the (N+1)th can too. Also, they have (N+1)th order knowledge of people leaving as soon as they can and everyone being a logician since in the proper statement of the puzzle it should be noted this is common knowledge, and the guru makes the knowledge of someone having blue eyes common knowledge.). Then you have a full proof, since you have now included that they can actually deduce the inductive step. Again, this may seem pedantic, but is really necessary both to be correct and as it illustrates the key of the puzzle, namely the guru gives 100th order knowledge of someone having blue eyes (this is the main problem people have, realizing the concrete piece of information the guru gives). Jlangy (talk) 00:29, 9 July 2015

What I don't follow here is that there's no clarification that the Guru is talking about someone different each time. Just because she says "I see someone with blue eyes" N times doesn't mean that there are N people with blue eyes; she could be talking about the same person every time, or each of two people half the time, etc. Can anyone clarify this?
Thanks - 108.162.218.47 13:20, 28 October 2015 (UTC)

(EDIT: Observe the process of comprehension in action...or don't? I've been thinking about my own brain, with itself, long enough for one day, I'm tired.)
So, maybe I am indeed just "dumb", as the wiki insists. Clearly, I do not have a perfect understanding of formal logic. But frankly, my read of this puzzle is that "formal logic" just enables you to jump to ridiculous conclusions.
Let's theorize a simpler version of this puzzle. There are now only two people besides the Guru on the island, both with blue eyes. We'll call them Bill and Ted (totally bogus, I know). No matter how logical Bill and Ted might be, when Bill hears the Guru say "I see a person with blue eyes" to himself and Ted, and Bill has seen Ted's blue eyes himself, why would Bill assume anything about his own eye color? It would seem to Bill that Guru was just talking about Ted's eyes, and Ted would believe the reverse. Even knowing* that Ted would leave that night if Ted deduced he had blue eyes too, I still don't see why Bill would jump to the conclusion that the Guru was talking about him - he remains in the dark, as does Ted, and neither of them can be any more certain of anything than they previously were. Adding 98 more blue-eyed people, let alone doubling the island's population with irrelevant brown-eyers, hardly reduces the confusion.

- This was the point at which I began to think I had understood it, but then I became unsure again. Like I said in the "edit", my brain is tired.

--So, that settles it, I do not understand how the puzzle can be true, and I'm not convinced that it actually is. Knowing Randall is, in general, smarter than me...I still do not have the ability to completely accept that he's always right, or that I'm always wrong to ignorantly question his rightness. I have long maintained that certain well-respected "systems of knowledge", of which formal logic is a textbook example, have been respected too well for too long for not-good-enough reasons. To me, they seem to be founded on an assumption which is itself founded on nothing. I'm not trying to insult Randall or anyone else, I'm just utterly failing to comprehend. I will appreciate if anyone else attempts to educate me on the subject, but I may prove an intractable student, since I am unable to extend much faith or trust (or even, on a day where my mood is worse than today, the moderate degree of politeness as I've already managed) to a teacher. 173.245.54.52 19:18, 30 October 2015 (UTC)

In your simplified version of the puzzle, Bill sees Ted has blue eyes.

Here's Bill reasoning:

- Either my eyes are blue or not.

- If my eyes are not blue, then Ted knows that his eyes are blue, because the Guru said at least one of us has blue eyes, and he'll leave the island tonight.

- Let's wait. If Ted doesn't leave tonight, that means he doesn't know his eyes are blue, and therefore my hypothesis is false.

When Bill sees Ted doesn't leave that night, he can deduce that he has blue eyes.

Ted can do the same reasoning.

After that first night, both will know they both have blue eyes.

--108.162.228.5 14:09, 14 December 2015 (UTC)

- Superrationality

The solution relies on the fact that "at least 1 blue" is new information which triggers a cascade.

Wouldn't the entire population of the island be able to conclude that everyone else on the island knows there is at least 1 blue eyed individual already?

For example, every person on the island will see at least 99 blues and 99 browns. From this, they can assume that everyone else on the island can see at least 98 blues and 98 browns. Of course, the actual numbers will differ, but 98 is the lower limit for all perspectives.

A blue will see 99 blues and 100 browns, so he will assume that all other blues can see at least 98 and all browns can see at least 99 blues. Similar logic for a brown or any observer.

Flewk (talk) 09:26, 26 December 2015 (UTC)

The solution here is different to Randall's solution, and I think is actually incorrect for two reasons that add confusion and prevented me from understanding the solution until I'd thought about Randall's solution and realised these are actually different.

- It seems to falsely presume that the Guru is speaking to them each day, when this is explicitly not the case in the puzzle.
- I also believe it is incorrect to state that the brown-eyed people can be disregarded. The solution is actually dependent on a *combination* of hypothesis testing and on theory of mind; not just one or the other. It matters that everyone is also thinking about what the brown-eyed people around them must be thinking, otherwise you can't explain why mistakes will not happen with brown-eyed people getting on the ferry when they're not supposed to, and screwing up everybody else's logic.

- If you're on the island and you have blue eyes, there are two hypotheses: either there are 99 people with blue eyes or 100. If there are 99, then everyone one of those 99 people is thinking "either there are 98 people with blue eyes, or there are 99" (and therefore you do not have blue eyes). Blue-eyed people also know that if there are 99 of them, then the brown-eyed people are thinking, "Either there are 99 blue eyed people, or 100." If there are 100, then the brown eyed people are thinking, "Either there are 100, or 101". To summarise, blue eyed people are deciding between 99 or 100, and presuming that other blue eyed people are either suspecting there could be 98/99, or 99/100, while presuming that brown-eyed people are either suspecting there are 100/101, or 99/100. - If there are 99, then blue-eyes are thinking 98/99, and brown-eyes are thinking 99/100. Blue eyes will plan to leave if the 98th day passes and nobody has left, brown-eyes will plan to leave if the 99th day passes and nobody has left. - If there are 100, then blue-eyes are thinking 99/100, and brown-eyes are thinking 100/101. Blue eyes will plan to leave if the 99th day passes and nobody has left, brown-eyes will plan to leave if the 100th day passes and nobody has left. - So you know that if you have brown eyes, you'll watch all the blue-eyes leave on the 99th day. And you know that if you have blue eyes, you'll watch all the brown-eyed people hold back in case their day is the 101st. If you're allowed to leave, there will be no situation where brown-eyed people mistakenly leave on the 100th day, thus confusing things. If you're not allowed to leave, there'll be no reason for you to mistakenly make an attempt to leave on the 99th day. - Thinking about this fact - what the brown-eyed people are thinking - also reveals why the Guru's comment matters, and adds information, even though it should seem to most people as if no information is being added (because they can all already see that blue-eyed people exist). I think this is a key part of why the problem is so tricky. 108.162.249.155 07:42, 10 March 2016 (UTC)

The new information the guru gives is nothing more than a common marker (the day of the announcement) to use as a starting point for counting days. Before the announcement, being unable to communicate with each other, they were unable to coordinate a means of figuring out their own eye color. 172.68.59.105 21:52, 22 September 2016 (UTC)

That's wrong. In that case the browned eyed people would do the same, but they can't.

What I'd like to know: If there were 100 blue-eyes, 200 brown-eyes, 300 grey-eyes and 400 red-eyes, and the Guru says "I don't see anyone with a unique eye color", would that permit everyone to leave (except the Guru herself) using the same logic? Meaning the blue-eyes leave again on day 100, the brown-eyes on 200, the grey-eyes on 300, and the red-eyes on 301.

I think it would actually be days 99, 199, 299, and 300, because the 'what if there were only two blue-eyes' case would be solved on day 1 - i.e. both would see only one blue-eye and deduce that they are also a blue-eye, and both would leave - so everything gets moved up by one day.141.101.76.16 13:52, 4 January 2018 (UTC)

This was bugging me today - specifically the guru doesn't seem to actually give any information, because with at least 3 blue-eyed people, everyone on the island knows that the guru sees people with blue eyes, and also everyone knows that everyone knows the guru sees people with blue eyes. So for a while I thought the brown-eyed people must have as much information as the blue-eyed people, and either they both could leave or neither could leave.

After a bunch of reading and testing possibilities I think I've actually figured it out now and why only blue-eyed people leave, but I haven't seen an actual good explanation for it yet, so here's my explanation: The information the guru gives is that, NO MATTER HOW MANY BLUE-EYED PEOPLE THERE ARE, they can figure out they have blue eyes. It is important that it is possible for blue-eyed people to be able to solve it for EVERY number, even if everyone knows there is more people than that number (basically because, everyone doesn't know that everyone knows there is more than that number and there's a gap in their logic without knowing that). If there is any number for which blue-eyed people cannot figure it out, then any solution (namely, what I thought before testing possibilities) would require that there is a number N of blue-eyed people that cannot leave, but a number N+1 of blue-eyed people that can leave. This is self-contradicting though. If N blue-eyed people can't figure it out, than N+1 people (regardless of eye color) can't get meaningful information from the action of those N blue-eyed people. And since they can't get meaningful information from N people's actions, N+1 people can't tell if there are N people and that individual is not blue eyed, or N+1 people and they are blue-eyed. It would be logically incorrect for them to assume an eye color at that point, which means they don't know if they can leave, and then N+2 people are similarly unable to get meaningful information from N+1 people's actions, and so on. Because a single blue-eyed person cannot figure it out, more blue-eyed people (regardless of number) cannot make any assumptions without additional information. Then the guru effectively states a single blue-eyed person could leave immediately (which means 2 could leave the next night confidently, and thus 3 the next, and so on in an UNBROKEN chain). Kejardon - 162.158.214.82 11:15, 9 January 2020 (UTC) (I doublechecked and edited/corrected my post kind of at the same time, so your reply doesn't make sense anymore, sorry Lupo. Feel free to delete these two sentences if you change/delete your reply)

- You bring up 2 points. First about the common marker. This is true, but it contains more information than "start counting from today," because every blueeyed person has 2 scenarios: With 99 blue eyed people and with 100. The numbers are not important, and it would also be needed with 1,2,3,etc. blue eyed people. The point about the brown eyed people: The brown eyed people have no way to conclude (remember, they are "perfect logicans" and their task is to figure out, not to make an estimated guess) that their own eyes are in fact brown, and not red, or green just like the Gurus. If the guru just said: "Start counting", then noone woul leave at any given night. --Lupo (talk) 11:05, 9 January 2020 (UTC)

I figured this out in less than a minute... there were so many warnings about the solution being convoluted that I thought I couldn't possibly have it right. It's not really that confusing and I've seen waaaaaaay harder logic problems.

This image is listed under "My Hobby" for some reason, despite not even being a comic, let alone a "My Hobby" comic. 162.158.111.151 00:58, 9 September 2019 (UTC)How sign edit

## Unstated assumption (about motives)[edit]

While trying to solve this, I started questioning my assumptions about the motives of the islanders. It turns out my assumptions were correct, but I think they deserve to be explicitly stated:

- Everyone wants to leave the island ASAP.
- An islander only "figures out" their eye color through logical deduction (no guessing or playing the odds)
- The Guru's statement is meant to help others leave the island. (but as stated above, it may not be the optimal means of helping others leave -- "nobody has a unique eye color")

- Regarding the 1st point: Yes, that needs to be stated.--Lupo (talk) 11:37, 9 January 2020 (UTC)
- Regarding the 2nd point: It should be mentioned, but could be implied by the words "figure out" and by the fact that everyone is a "Perfect logican". --Lupo (talk) 11:37, 9 January 2020 (UTC)
- Regarding the 3rd point: If it was not meant to help people leave, it would still do the same job. Also the statement from the Guru "nobody has a unique eye color" would be wrong and misleading! In that case everyone would wrongly assume: "I must have green eyes, as otherwise the Guru would have an unique eye color." - The alternative statement "I see noone with a unique eye color" would work. --Lupo (talk) 11:37, 9 January 2020 (UTC)

Storming lighteyes (not sorry) SilverMagpie (talk) 17:50, 10 January 2020 (UTC)