I see five sets of blue eyes.
Those people with blue eyes either see 4 sets of blue eyes or five sets of blue eyes (depending on my eye color).
I know no one is leaving on day one because I see five sets of blue eyes.
I also know that everyone already knows that no one is leaving on day one, because at minimum they are seeing four sets of blue eyes.
I’ll answer what appears to be the most basic mistake in what you typed:
As has been explained, you absolutely do not know that everyone knows the same things you know. I know B’s eye color, but not everyone knows B’s eye color. I know that everyone knows that C sees at least one pair of blue eyes, but not everyone knows that everyone knows that C sees at least one pair of blue eyes. (For example, B does not know that A knows that C sees at least one pair of blue eyes. And, by the same token, A does not know that B knows that C sees at least one pair of eyes. See my blue/brown eyes diagram linked to above. It illustrates exaclty how this can be.)
If I see 5 sets of blue eyes, how do I not know that everyone else sees at least 4 sets of blue eyes? Explain how it is possible for me to see 5 sets of blue eyes and for someone else to see 3 or fewer sets of blue eyes.
If you see five, each of those five might only see four. When those five think about those four, the realize each of those four might only see three. When those five think of those four thinking of those three, they see they’d think they only see two. etc.
It’s not that each night is telling you something you didn’t know with regards to how many people there are. It’s placing a wall beyond which you don’t need to consider the possibilities of other blue eyed people considering the possibilities.
Imagine you’re on the island. Pick the largest number of people leaving on night N that makes sense to you. Now imagine that they didn’t leave that night. You’d now know for sure that you have blue eyes. You couldn’t have left earlier than that 'cause you thought they’d all leave the night before.
Don’t take this the wrong way, please, but I’m not going to talk about 6-person scenarios til we’ve got you on board with the correct analysis of the 3-person scenario.
All that happens in the 6-person scenario is that things get more nested in levels of “x thinks that y thinks that z thinks that etc…”
ETA: Hrm, having said that, the post I was responding to does have an error I can address. You said “How do I not know that everyone else sees at least four sets of blue eyes.” But you do know that. What you don’t know is that everyone knows that everyone else sees at least four sets of blue eyes.
But that isn’t how it works. I can put an absolute minimum on the number of sets of blue eyes that people see, and so can everyone else. I explained it above. Everyone in the group can follow that logic, and collectively exclude the possibility of N-2, where N is the number of blue eyes you see.
I’ve done it with the generic case. If I see N number of blue eyes, then I know what everyone else is thinking. If they have blue eyes, it is N-1, N, or N+1. If they have brown eyes, it is N, N+1, or N+2.
If you can see n blues, you know that the others must see at least (n-1).
But if they could indeed only see (n-1), then they would only know for sure that the others could see (n-2).
In the case of five, from your point of view it is therefore possible that they think it’s possible that the others can only see three, even though you know yourself that it’s not possible.
You can put a minimum on how many anyone can see. But you can’t guarantee that everyone knows the same minimum. You see 5, so you think there might be 5 or 6 blue eyed people. Also, you know people see a minimum of 4.
If people see that minimum (4), they think there might be 4 or 5 blue eyed people. And that there’s a minimum of 3 people seen.
Etc.
Okay, I’m not a perfect logician, but with a few hours to think about it I can express myself much more clearly.
Thesis: the Guru did not tell us anything we did not already know.
Proof:
I can see a number of blue eyed people, and we shall assign that value to N. I know that the number seen by other people cannot be smaller than N-1.
As we are all perfect logicians, anything I could deduce from a given value of N can also be deduced by all the others, if they see the same number.
Therefore, anything I could deduce from seeing N-1 blue eyed people is something that everyone knows.
If N=1, then I know there is at least one blue eyed person on the island.
If N=2, then I know that everyone knows that there is at least one blue eyed person on the island.
If N=3, then I know that everyone knows that everyone knows.
Therefore, for all values of N >=3, everyone knows that the number of blue eyed people is non-zero, and everyone knows that everyone knows that.
Everybody already knew that there was at least one blue eyed person, and that everyone knew that, so the Guru’s statement contained no information that we all hadn’t already gotten by logical deduction.
You and not he. He will leave on the first night for he learned something new.
And neither of those two people know that. So they leave on the second night.
Unless they are one of those three blue eyed people. They know the two blue eyeds they see can see at least one. They have to deal with the possibility that there are only those two blue eyed people. When those two people don’t leave (as you expect they would if there really were only two), they know there are three.
Four people leave when the above doesn’t happen.
Ok, let me think about that a little bit more.
[QUOTE=treis (addressing the case of 3 islanders with blue eyes)]
If I had brown eyes, they would have left day one.
[/QUOTE]
This looks like a clear mistake.
If you had brown eyes, the total of islanders with blue eyes would have been two. There’s no way either can be sure of their eye color until the first night passes without a departure - they must wait until the second night to leave.
So it’s only when they don’t leave on the second night that you learn your eyes are in fact blue.
Why have you stopped at “everyone knows that everyone knows that…”, only two layers deep?
Maybe because it’s bloody hard to think about when it gets more indirect than that. It’s a bit like trying to imagine a four-dimensional object. But that’s what makes this puzzle so hard to understand.
What you’re saying is true but not relevant, because you’re continuing to talk about how many blue eyes different people think may be on the island, when what’s relevant is how many blue eyes different people think different people think may be on the island. (This is for the three-person case. It gets more nested with more people.)
Have you looked at the diagram?
Once more, this diagram illustrates how it is true, on the three-person island, that:
Before the guru speaks, A thinks it’s possible that B thinks it’s possible that C thinks it’s possible there are no blue-eyed islanders.
(After the guru speaks, A can’t think this anymore because he knows that B knows that C knows there is at least one blue-eyed islander.)
But I’m pretty sure for N > 3 you can start to loop the recursion.
A, B, C, D and E have blue eyes. There are some number of brown-eyed follows as well, but (According to the prevailing theory) they’re irrelevant.
A knows that B Guru can see a blue eyed person, because B can see C. A knows that B knows that A knows that Guru can see a blue-eyed person, because A and B can see c. And A knows that C knows that B knows that A knows because A, B, and C can see D.
Here is where I think the logical error is introduced in the solution: A knows D knows that C knows that A knows, because A and D can both see B and C. And A knows that D knows that A knows that D know… because A and D can both see B and C. And so on.
You don’t need to refer to a NEW blue eyed person with every hypothetical loop. Once your recursive logic starts recursing blue-eyed individuals, no longer make any progress.
Four-person diagram is here. It works for five, six, and every other number besides. Please don’t ask me to draw all the diagrams though 
The diagram shows how, on a four-blue-eyed island, before the guru speaks, D thinks it’s possible that A thinks it’s possible that B thinks it’s possible that C thinks it’s possible that there are no blue-eyed islanders.
I stopped two layers deep because the Guru’s statement stops there too.
And what functional difference does adding layers at that point make? If everyone knows something, and everyone knows that everyone knows it, does it matter that everyone knows that everyone knows that everyone knows? The point is, when I look at someone and wonder if they know, the answer is yes, and when I wonder if they know that I know it too, the answer is also yes.
And in the problem as defined, everybody knows that there is someone with blue eyes, and knows that everyone knows, to 90+ layers, so the Guru saying that didn’t give anybody any information they didn’t have before.
I’ll just pop in here to say that the following two points (which have been mentioned several times in this thread) are the start of understanding the solution:
There are different observations about the number of blue eyes, which lead to different knowledge about the number of blue eyes, and which leads to different hypotheses about what others know about the number of the blue eyes, which lead to different hypotheses about what others know about what others know about the number of the blue eyes, all of which leads to uncertainty about the things above. And, it is this uncertainty (along with all the rules of the puzzle) that results in everyone staying on the island.
The Guru’s announcement provides information that eliminates one piece of that uncertainty, and which starts the chain of conclusions which eventually results in all the blue-eyed residents concluding that they have blue eyes.