If you haven't seen it yet, we recommend trying to solve it before reading further (as with any logic puzzle).
Here's the puzzle as worded by Randall, here's its solution, and here is the main thread discussing its solution. This post by Xias on that thread inspired me to create this particular thread, intended to serve as a kind of FAQ for the puzzle.
Following are some common objections to the solution, each one in bold. Underneath the bolded objection is an elaboration in italics, meant to provide support for the objection's argument and add more detail as to what someone making this objection might be trying to say. Both the bold and italics part are therefore being stated by the same hypothetical person, not different ones. In reality, I myself wrote the whole thing — there are no actual quotes of anyone else intended, just paraphrases of common ideas.
After each objection and elaboration comes my response. Each response is in a spoiler, more to avoid a wall of text than to hide any actual "spoliers"; I presume you won't read this until having learned the problem and solution (by one means or another) beforehand.
Within these responses I often use italics for emphasis and bolding for a "stronger" emphasis. Sometimes, italics are used to help distinguish some parts of a sentence from others, which might be helpful in understanding things like "A thinks that B thinks that C thinks…". (This style of mine has nothing to do with the bolding of the original objections and italicizing of their elaborations.)
There's enough text in this thing as a whole that I've split it into two parts. Part 1 deals with seven objections, Part 2 with the remaining ten. You can read the responses in any order you like, but since I wrote them just the way you see here (EDIT: Except for edits, of course), they'll probably make the most sense in that order.
In general, this is meant to address the problem whereby new posts and/or new threads raise the same points again and again, by people who were unwilling to read 25 pages of posts. So I read 'em for you, and I hope I managed to distill everything decently here. No more excuses, yes? (Until, of course, this thread manages to sink out of view. Even then it can still be linked to, of course.)
Feel free to post on this thread if you wish to add another common objection which you don't see here, or if you think I can be clearer with some of my responses, etc. If you wish to argue against the actual content of one of my responses (and by extension, to defend an objection), you may do so, but don't expect others here to be very happy about it, or to be convinced.
1. The Guru's statement, "I can see someone who has blue eyes", does not provide new information to the islanders.
Everyone already knew that. No one will ever leave.
Everyone already knew that everyone knew that someone had blue eyes.
But we cannot add an arbitrary number of "everyone knew" to this statement, and that fact is crucial. Specifically, prior to the Guru's statement, the following statement is not true:
"Everyone knows that [repeat the last three words before this bracket such that they appear exactly one-hundred times] at least one person with blue eyes exists."
And yet the following statement is true:
"Everyone knows that [repeat the last three words before this bracket such that they appear exactly ninety-nine times] at least one person with blue eyes exists."
This really does make a big difference. It's rather like how "I will finish this novel the day after [repeat the last three words outside this bracket three hundred times] tomorrow" really does mean something different than "I will finish this novel the day after [repeat the last three words outside this bracket two hundred ninety-nine times] tomorrow". I'm referring to two different days. It just so happens that we have a nice shorthand to say this in the English language (I can simply say "in three hundred days" or "299 days"), and we don't have such a shorthand to refer to the "everyone knows that everyone knows" situation, at least not one that non-logicians use regularly.
Logicians do have a useful phrase for the type of infinitely-recursive situation the puzzle addresses: "Common knowledge".
If "X is true" is common knowledge for a group of people, that means that any statement of the following form is true: "Everyone in this group knows that [repeat the last six words before this bracket any number of times you want to] X is true." To understand this better, read the Wikipedia article on the subject.
Before the Guru's words, the statement "At least one person on this island has blue eyes" was not common knowledge. Then, after she said it, it was. The reason for this is not merely that she told everyone, but that she told everyone publicly, so that Sasha could see that James could see that Briana could hear the Guru's statement, and likewise for every possible ordering and permutation of the two hundred islanders.
2. I knew what common knowledge was (or, now I do). In this situation, "At least one person has blue eyes" was already common knowledge.
We can easily deduce that everyone already knows that everyone knows that everyone knows, ad infinitum.
Here's a longer reply, which should also help you see what's wrong with the next (and very similar) objection, #3.
Imagine an island with just two people, blue-eyed Yolanda and brown-eyed Zoe. As usual, Yolanda doesn't know what color her eyes are. But suppose she performed the following syllogism:
1. I don't know what color my eyes are.
2. If I were Zoe, I would observe that I have blue eyes.
3. Therefore, my eyes are blue.
That's nonsense, obviously. Even though we know what color the real Zoe sees on Yolanda, we also know that Yolanda can't somehow deduce it.
However, if we nest Yolanda and Zoe within the nested thoughts of other people, then keeping track of the distinctions between what we know and what hypotetical islanders believe is much harder.
Consider the following statement, made about an island of exactly four blue-eyed people:
that Xavier thinks
that Yogi thinks
that Zach believes
that at least one islander has blue eyes.
It sounds innocuous enough. Four is plenty of blue-eyed people, right? How could Zach possibly not know that someone has blue eyes? But there's a problem.
The problem arises when we try to answer the question: Which islander or islanders does this hypothetical Zach believe has blue eyes? First of all, it can't be Zach himself; like Yolanda, he can't "see" other people seeing that his own eyes are blue. So perhaps it is Wendy? Ah, but remember that it is Wendy herself who is supposing all this. For it to be Wendy, she would have to be doing the same magic that Yolanda did.
Well, how about Xavier? Doesn't Wendy know that Xavier is blue-eyed? Wouldn't Wendy be able to see that Yogi can see that Zach can see Xavier? Yes, that's all correct. The problem is that in the original sentence, Xavier was another one of the people doing the "thinking". So in order for it to be true that "Wendy knows that Xavier thinks that Yogi thinks that Zach sees that Xavier has blue eyes", then Xavier would have to already know his own eye color, or Wendy would have to think that he does.
At this point, I hope I don't have to explain why it also couldn't be Yogi.
So there you go. Even with four blue-eyed islanders, the existence of blue eyes cannot be common knowledge by the logical definition of common knowledge (until something happens, such as the Guru's announcement). Now, if you imagine a statement like the one above, but one hundred names long, it should be clear that the same problems will arise; each islander's ignorance of his own eye color means that the "last" islander, nested inside numerous hypotheticals, cannot see who has blue eyes. And of course, the ordering of the names is arbitrary, so this principle applies equally to all the islanders.
Some may insist that "Which islander does hypothetical-Zach think is blue?" is irrelevant or even a trick question. Various attempts have been made to show why the "innermost" islander in any chain like this can be inferred by the outermost islander to be aware of blue eyes, even if we can't name which ones the outermost islander thinks he sees. It has been argued that because the situation can be modeled as a net, and not just a chain, then we can "recurse" or "criss-cross" the lines of information. And because of this criss-cross, we can find that if A knows that C knows "Q", and A knows that B knows "Q", then A knows that B knows that C knows "Q" (or at least, this is somehow true if there are four agents instead of three).
Yes, the situation can be modeled as a net or graph, but any path on this graph will still be a chain, and must still be described in the same terms as the example here. For example, if it really is the case that W can "know" that X knows Y knows Z sees someone with blue eyes, then there must be an answer to the "Who does Z see?" question. No islander can "absorb" the abstract information "at least one blue-eyed islander exists" by some sort of osmosis or critical mass of islanders, any more than Yolanda could when it was just her and Zoe. The can only learn this fact by seeing it for themselves. And all islanders will know this about all other islanders.
Another point some have raised is: sure, W might suppose that X supposes that Y supposes that Z sees no blue-eyed people, but why suppose all that? Isn't it arbitrary? W may as well suppose that X supposes that his own eyes are blue. The answer to this is… yes, that's perfectly true. In fact, no islander is assuming that his own eyes either are or are not blue, or that other islanders are making such an assumption. That's not what's going on here. Each islander is simply considering both possibilities, and knows that everyone else is considering both possibilities, etc. These various possibilities multiply out. When a chain is extended to include every single islander, than every possible combination of eye colors is accounted for. For example, W knows that if:
• his own eyes are blue, and
• X (by coincidence, correctly) imagines that his own eyes are blue, and
• X supposes that Y (incorrectly) imagines that his own eyes are brown, this means that:
• W would suppose, in this one example, that X would suppose that Y would suppose that Z would see two blue-eyed people (W and X) and one brown-eyed person (Y).
That's just one of the many, many possibilities of nested beliefs. (I calculate 192, counting all the possible orderings of four islanders thinking about anoe another thinking about various combinations of blue and brown eyes.) All that matters for this particular problem is that "seeing no blue-eyed people" is at least one of these nested possibilities, and thus our logical islanders are considering it, along with all the other ones.
3. Yes, with just one blue-eyed person, it wasn't common knowledge, nor with two, or possibly three. But with over ninety blue eyes, it's obviously common knowledge.
Hypothetical blue-eyed islander A sees 99 blues. A knows that B sees at least 98 blue eyes. Clearly, A should know that B knows that C sees at least 98 blue eyes, because A can see C seeing the 98 other blue eyes, and A can see that B can see C seeing the 98 other blue eyes. Combine these two facts, and it is clear that the minimum possible number of blue-eyed people whose existence is common knowledge is 98, not 0 or 1.
A should know that B knows that C sees at least 98 blue eyes, because A can see that B can see C seeing the 98 other blue eyes.
— to be true…
Then either A, B, or C must somehow know the color of their own eyes, or be believed by the others to know his/her own color. Neither of these can be the case, so 98 is too high a lower bound.
Every time you add the word "knows" to a statement of the form "A knows that B knows that… D sees at least N blue-eyed people", you have to subtract 1 from N. You must account for the fact that each person is ignorant of their own eye color, and is accounting for the next person's ignorance of their own eye color, and is accounting for the next person accounting for the next person's ignorance, etc.
Without any statement from the Guru, this process of "A knows that B knows…" will have to end with "that Z (here the 100th person on this list, not the 26th) knows that at least 0 blue-eyed islanders exist." For it to stop at a number greater than 0, someone has to know something which can't be known, or believe something which can't be believed, absent anyone's information.
After the Guru's statement, though, that number immediately goes to 1, and then increases by 1 every midnight thereafter. This leads to the eventual situation of all 100 blue-eyed people leaving on Day 100.
If you're confused (which is perfectly normal!), then again, we advise you to contemplate the situation with just one blue-eyed islander, then two, etc.
EDIT: Here's another thought about what's going on with the (correct) statement that "A knows that B knows that C knows there are at least 97 blue-eyed people." It's true that, darn it, A can just plain see 99 pairs of blue eyes, and that A can see that B sees 98, and that A can see that C sees 98. However, when A wonders about B's understanding of C's knowledge, A isn't simply curious about how many blue-eyed people there are. Nor is A simply curious about how many B thinks there are, or how many C thinks there are.
A really is curious about B's knowledge of C's knowedlege, and likewise for all other such orderings. A has to deduce this meta-knowledge in order to work out the whole situation. This deductive process can't just "stop" with three, so apart from 0, there is no "common knowledge minimum" of blue-eyed people before the Guru's statement establishes a minimum of 1.
EDIT 2: Here may be a simpler way to put it, a point raised by many dealing with this problem: If you think the blue-eyed people can nonverbally agree on some sort of common minimum which is greater than zero, than surely the browns should agree to the same minimum, or else it's not really common to the islanders. But how? Suppose each blue thought "I see 99 blues. 99 minus 2 is 97, so 97 is the absolute common minimum to which we all agree." In that case, a brown-eyed person would go "I see 100 blues. 100 minus 2 is 98, so 98 is the absolute common minimum". If a brown instead came to conclude that the minimum is 97, that means he went one step further than a blue would. Why would he? See, this is the (initially) surprising result of the existence of a one-person difference between two people's counts of eye color, all thanks to the commonly-known fact that no one knows the color of her own eyes.
4. Who on that island could possibly believe that no one with blue eyes lives there?
In order for the puzzle to work, each islander is hypothesizing an islander who sees no blue eyes— but they all already know that no such islander exists.
No one is hypothesizing such an islander, despite appearances. Rather, they are hypothesizing that everyone else is hypothesizing that [repeat an arbitrary number of times] there is such an islander. This person is nested within numerous nested hypotheticals. Don't feel bad if that doesn't seem to make sense right away.
Again, try it with three blue-eyed islanders, which seems to be a point many people get stuck on. If there are three, then it is true that everyone knows that everyone else can see at least one person with blue eyes. So Objection #4 seems to apply when there are just three blues total — no one would possible imagine someone who imagines someone who sees no blue eyes.
But this is not sufficient to establish blue eyes as common knowledge. If A, B, and C all have blue eyes, the only way for A to know that B knows that C sees at least one pair of blue eyes is if A knows A's own eye color to be blue, or if A thinks that B knows B's own eye color to be blue. Neither of these can be the case.
This brings us to the next objection…
5. Yes, it works if there are just one or two blue-eyed people, but not three or more.
Induction cannot prove everything.
Imagine that you are one of the islanders. The Guru has made her statement. You happen to see exactly two pairs of blue eyes. How many blue-eyed people are there altogether? Either two (just them) or three (including you); you don't know yet.
Now, you accept the logic that if the total number of blue-eyed islanders is two, they will both leave on Day 2. You know that there is no way for this not to occur; no one is going to forget to leave, or forget to deduce that the other one can see his eyes. Nor will anyone leave sooner for some reason. If there are just two, then the exodus will occur on Day 2 and Day 2 only.
You wait until Day 1. Nothing happens, which doesn't surprise you (but may have surprised the two blue-eyed folks you see).
You know now that if there are just two blue-eyed folks, they will each have figured this out, and prepare to leave the next day.
Day 2 arrives. You wake up, look around, and see… that they are still here. Why would that be? Why would they not have left yet?
The only possible answer is that each of them also saw more than one person with blue eyes. Which means they each saw some blue-eyed person you couldn't see, and the only person that could have been is you. In short, they were each waiting for you (and one another) to leave first. So, now you know the color of your eyes: blue.
This applies equally to all the blue-eyed islanders, of course. So we have now established that if there are exactly three blue-eyed people, they will all leave Day 3. An absolute, undeniable rule. If three blue eyes, then all of them leave on Day 3.
6. I knew that it works for three (or, now I realize that). But it breaks down at four.
I have a very good reason to say it breaks down at four. (Different people provide different reasons here.)
Imagine that you are one of the islanders. You happen to see exactly three pairs of blue eyes. How many blue-eyed people are there altogether? Either three (just them) or four (including you); you don't know yet.
Now, you accept the logic that if the total number of blue-eyed islanders is three, they will all leave on Day 3. There is no way for this not to occur; none of the three is going to forget to leave, or forget to deduce that the other two can see his eyes. Nor will any of them leave sooner for some reason. If there are just three, then the exodus will occur on Day 3 and Day 3 only.
You wait until Day 1. Nothing happens, which doesn't surprise you. There was no way for anyone to leave that soon.
Day 2 arrives. Nothing happens, which doesn't surprise you — however, you know now that if there are just three blue-eyed folks, they will now figure this out, and prepare to leave the next day.
Day 3 arrives. You wake up, look around, and see… that they are still here. Why would that be? Why would they not have left yet?
The only possible answer is that each of them saw more than two people with blue eyes. Which means they saw some blue-eyed person you couldn't see, and the only person that could have been is you. In short, they were each waiting for you (and for each other) to leave first. So now (and not a day before) you all know the color of your eyes: blue.
This applies equally to all the blue-eyed islanders, of course. So we have now established that if there are exactly four blue-eyed people, they will all leave Day 4. An absolute, undeniable rule. If four blue eyes, then all of them leave on Day 4.
7. It works for one, two, three, or four. It stops working at five.
I know the induction thing is looking good for you so far, but trust me, it doesn't keep going forever.
This means it works for up to at least 45 billion blue-eyed islanders, after which we don't know, because 45 billion is the largest number known to mathematics.
EDIT: Here's another useful way of putting it, paraphrased from several posts in the original solution thread. If you claim that N blue-eyed people will deduce correctly, but N+1 will not, then you are arguing that different outcomes will occur depending on which islander's perspective we take. For example, say you think it works with four, but not five. That means that if you, an islander, see exactly four pairs of blue eyes, then you know that:
• there are either four or five blues total (a basic induction)
• if you don't have blue eyes, then the four you see will all leave on Day 4 (because it works with four)
• If you do have blue eyes, then even after seeing no one leave on Day 4, no one will ever deduce their own eye color (because it somehow doesn't work with five)
The last premise must be wrong.