Is there no solution to the blue-eyed islander puzzle?

Argument 1 is clearly wrong.

Consider the island with only two blue-eyed people. The foreigner arrives and announces "how unusual it is to see another blue-eyed person like myself in this region of the world." The induction argument is now simple, and proceeds for only two steps; on the second day both islanders commit suicide. (I leave this as a crucial exercise for the reader.)

Now, what did the foreigner tell the islanders that they did not already know? Say that the blue-eyed islanders are $A$ and $B$. Each already knows that there are blue-eyed islanders, so this is not what they have learned from the foreigner. Each knows that there are blue-eyed islanders, but neither one knows that the other knows this. But when $A$ hears the foreigner announce the existence of blue-eyed islanders, he gains new knowledge: he now knows that $B$ knows that there are blue-eyed islanders. This is new; $A$ did not know this before the announcement. The information learned by $B$ is the same, but mutatis mutandis.

Analogously, in the case that there are three blue-eyed islanders, none learns from the foreigner that there are blue-eyed islanders; all three already knew this. And none learns from the foreigner that other islanders knew there were blue-eyed islanders; all three knew this as well. But each of the three does learn something new, namely that all the islanders now know that (all the islanders know that there are blue-eyed islanders). They did not know this before, and this new information makes the difference.

Apply this process 100 times and you will understand what new knowledge was gained by the hundred blue-eyed islanders in the puzzle.

This isn't a solution to the puzzle, but it's too long to post as a comment. If one reads further in the post (second link), for clarification:

In response to a request for the solution shortly after the puzzle was posted, Terence Tao replied:

I don’t want to spoil the puzzle for others, but the key to resolving the apparent contradiction is to understand the concept of “common knowledge”; see http://en.wikipedia.org/wiki/Common_knowledge_%28logic%29

Added much later, Terence Tao poses this question:

[An interesting moral dilemma: the traveler can save 99 lives after his faux pas, by naming a specific blue-eyed person as the one he referred to, causing that unlucky soul to commit suicide the next day and sparing everyone else. Would it be ethical to do so?]

Now that is truly a dilemma!

Added: See also this alternate version of same problem - and its solution, by Peter Winkler of Dartmouth (dedicated to Martin Gardner). See problem/solution $(10)$.

Easiest case to show what's wrong with the solution offered here and all other solutions like on wikipedia etc is to consider case of 4 blue eyed people.

Proposition 0: If there are 4 blue eyed people than everyone sees other 3 and also all of them can conclude(even without knowing their own eye color) that all of them know(not a typo) that there are at least 2 blue eyed people on the island. So if blue eyed people are A,B,C & D than A can say that B knows that C knows that D knows that A knows that there are at least 2 blue eyed, B can make similar assumption etc.

This would be enough for a pure mathematical proof, since if all of them know that all of them know that there are at least 2 blue eyed people than visitors anouncement that there is one blue eyed among them does not introduce new knowledge.

Let's find out what's wrong with the proof:

General argument for any kid of proof: If proof/solution assumes something that is impossible or contradicts to initial conditions than every conclusion based on such an assumption is useless.

All solutions I've seen so far include some variations of :

Suggestion A: Suppose/if there was only person (same as suppose n=1)
Suggestion B: Suppose/If there was only one blue eyed person

Clearly if there are 4 blue eyed person we can't suggest that there is only one person on the island so Suggestion A is clearly wrong and all proof based on such a suggestion are wrong. This also invalidates all recursive proofs that assume that n=1 / day one . One can abstract from the concrete example and say suppose n=1/day one, but than you can't imply the knowledge that on day one you knew that there is one blue eyed person.

Now what's wrong with the suggestion B? remember proposition 0. Everybody knows that all of them know that there are at least 2 blue eyed people on the island. But suggestion B says suppose there is only one blue eyed person. This Suggestion is also wrong since we know that there are at least two of them. So, all profs/conclusions based on suggestion B are also wrong.

That's it.

If you find a proof that does not use some variation of A or B than we can reopen this discussion.

Here's my answer as to why the outsider gives new info. I'm considering the situation of (A)lice, (B)ob and (C)athy as mentioned in a post above, where all 3 have blue eyes. For argument's sake, I'll be Alice.

• I know there are at least two people with blue eyes
• I know Bob and Cathy know at least one person has blue eyes
• Here's the tricky part: What do I know about what Bob knows about Cathy's knowledge?
• As far I know, I may have brown eyes, and Bob may think he has brown eyes. So when Bob looks at Cathy, he may think she sees no one with blue eyes. So from Bob's point of view, Cathy may not know there are blue eyed people.

Simply put, I know there are blue eyed people. I know that Bob and Cathy know. But I don't know that Bob knows that Cathy knows. When the outsider announces it, I now know that everyone knows, and everyone knows that everyone knows.

It's also easier to imagine that everyone assumes they have brown eyes. So when Alice thinks about Bob's thoughts on Cathy's thoughts, everyone down the line assumes they themselves have brown eyes. When Alice thinks of Bob's thoughts, she's assumes he thinks they both have brown eyes, and only Cathy has blue. When Alice thinks of Bob's thoughts on Cathy's thoughts, she imagines Bob will conclude that Cathy will not see blue eyes.

This scales up to four or more, and it's easier (for me) to think of the fourth being 'above' A (Alice), let's say Omega. So Omega must imagine what Alice thinks that Bob is thinking about what Cathy thinks, not just what Cathy herself thinks.

So when Omega imagines Alice's thoughts, he's assuming she thinks they both have brown eyes. When he imagines Alice's thoughts on Bob's thoughts, same thing, everyone assumes they have brown eyes in their own mind. That's why Omega concludes that Alice may imagine Bob could think that Cathy may not know there are blue eyed people.

Very deep and difficult, but the 'everyone assumes they have brown eyes' POV helped me wrap my head around it.