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[Darvi]

Sorry but I don't speak limericks.

Now see, theorems don't apply if the assumptions underneath them aren't true.

There are at least 99 people,
not 1 through 98

Yes, but those 99 people, if you're brown, you will ask yourself, what would those 99 think?
Those 99 would ask themselves the same thing, but with one person less.
If one of them were brown, then there are 98 people who will make the same deductions.
And so on.

[Earthquake Damage]

Having read a solution (poorly explained as it was), I don't see how the announcement is relevant in this particular version of the riddle, since everyone can already see both blue eyes and brown eyes.  Not all versions give that information (i.e. the "100 blue, 100 brown" bit), so the announcement gets the ball rolling.

[G-Flex]

Not all versions give that information (i.e. the "100 blue, 100 brown" bit), so the announcement gets the ball rolling.

The "100 blue, 100 brown" part doesn't matter, because the people on the island don't know it.

[Darvi]

The "100 blue, 100 brown" part doesn't matter, because the people on the island don't know it.

This. It's the fact that there is somebody with blue eyes on the island (as confirmed by the guru), and the fact that everybody has made a silent agreement to leave the day after the one with the amount of blue eyed people that they can see.

[G-Flex]

It's the fact that there is somebody with blue eyes on the island (as confirmed by the guru)

To be fair, they don't need this confirmed, as they all know that there are multiple blue-eyed people aside from themselves.

[Darvi]

Exactly.

You see X blue eyed people. You think "If they leave on day X, then I'm not blue. If they don't, then I am blue and we can leave on day X+1"

I've already established that this is true for X=2. So for X bigger or equals 3, this is true. (also for X=0 or X=1, but that's been already proven)

[Leafsnail]

Actually, the announcement is completely vital.  There's definitely no way you can construct the proof without it.

[G-Flex]

Yeah, but the question remains why. Specifically, the question is what information she actually imparts upon the people that they didn't already know.

[Darvi]

Right. You could deduce down to two people, but the last person couldn't be sure whether he's blue or not. X=0 and stuff.

Maybe I should've formulated the beginning of my last post better.

Yeah, but the question remains why. Specifically, the question is what information she actually imparts upon the people that they didn't already know.

Mathematical sequence. for x>0, people will leave on day X+1, as established. But for x=0 (i.e.  just one person there, who doesn't see any blues), you'd need the guru.

[Jreengus]

That's why I prefer to describe the solution using mathematical induction, you can clearly see how the announcement is vital for the base case of x=1 and without it the whole proof falls apart.

[Leafsnail]

You can, but trying to logically say what she adds is an interesting challenge.  Mathematical proofs can't pull something out of nothing...

[Darvi]

Any forum where you can dissect the allegedly hardest puzzle on the web is a good forum

[Earthquake Damage]

derp

I get it.  The announcement proves the n=1 case.

Though via more general reasoning (i.e. not strict logic) the brown-eyed islanders could also leave the island.  Everyone sees at least 99 people with brown eyes, so while the n=1 case may not be true in general (no premise supports it), we know it's true in our particular case (someone has brown eyes -- at least 99 someones, in fact) so we can accept it.  That doesn't satisfy a logic puzzle, obviously. 

[Leafsnail]

No.  That doesn't work.

Let's say one of the brown eyed people was replaced by someone with red eyes.  If he used that same logic, he'd say his eye colour was brown and he would be... wrong.

I'm not entirely sure why, but the announcement is necessary...

[Earthquake Damage]

No.  That doesn't work.

not strict logic

Also, everyone else would have left the day before Mr. Red decided to leave.