It's the same reasoning — Michael

Same reasoning as unenlightened, who was assuming the green eyed person said something?

Make your reasoning explicit. You're still riding on coattails, you can't have your cake and eat it to. If you want to disagree with his premises, show your reasoning yourself.

P1. I know that green sees blue
P2. Therefore, if I don't see blue then I must be blue and will leave on the first day — Michael

This doesn't work if green doesn't say anything. If green doesn't say anything, p2 isn't the case. If green doesn't say anything, and you don't see blue, your eyes could literally be any colour.

This doesn't work — flannel jesus

It does work given that it allows me to correctly deduce my eye colour. What more proof do you need other than the results?

Or is it just a coincidence?

I'm done with arguing here, but looking around there are plenty of wrong answers about, but here is a fairly decent run through.

https://xkcd.com/solution.html

you keep begging the question. You're just declaring yourself correct repeatedly, with incorrect premises. Premise 2 is incorrect.

yeah that's the canonical solution, and matches yours. I'm still endlessly impressed you figured that out on your own, not everyone can do that. I couldn't.

Premise 2 is incorrect. — flannel jesus

So you say, and yet if blues were to follow this reasoning and browns were to follow comparable reasoning then they would all correctly deduce their eye colour and leave on the 100th day — without green saying anything. I think the results speak for themselves.

I've explained it as clearly as I can, so there's nothing else to add.

So you say, and yet if blues were to follow this reasoning and browns were to follow comparable reasoning then they would all correctly deduce their eye colour — Michael

What you're not understanding is that they could just add easily incorrectly deduce their eye colour. It's a coin flip at best, because the "deduction" isn't based on sound premises.

What you're not understanding is that they could just add easily incorrectly deduce their eye colour. — flannel jesus

No they won't. Let's take the example with 3 blue, 3 brown, and 1 green.

Each blue's reasoning is:

A1. Green sees blue
A2. Therefore, if I don't see blue then I must be blue
A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue

B1. Green sees brown
B2. Therefore, if I don't see brown then I must be brown
B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown

Given that each blue sees 2 blue and 3 brown, they can rule out some of these premises:

A1. Green sees blue
A2. Therefore, if I don't see blue then I must be blue
A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue

B1. Green sees brown
B2. Therefore, if I don't see brown then I must be brown
B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown

Then, come the third day, they can rule out one more:

A1. Green sees blue
A2. Therefore, if I don't see blue then I must be blue
A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue

B1. Green sees brown
B2. Therefore, if I don't see brown then I must be brown
B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown

Given that one of the As and one of the Bs must obtain, and given that only A4 is left of the As, blue knows on the third day that A4 must obtain and so that they are blue.

The complimentary set of arguments will have all browns deducing that they are brown on the third day.

All without green having to say anything.

A1. Green sees blue
A2. Therefore, if I don't see blue then I must be blue — Michael

These pair of premises don't make sense together. If green hasn't said anything, then the only reason you could possibly know green sees blue is precisely because you see blue.

"Therefore if I don't see blue, I also don't know that green sees blue, so therefore A1 is no longer applicable"

These pair of premises don't make sense together. — flannel jesus

Yes they do. Given that I know that green sees blue, I can just assume that she says so even if she doesn't, and so if helpful I can stipulate that in some hypothetical world in which I don't see blue (even though in reality I do see blue) that she says "I see blue" (even though in reality she doesn't say "I see blue").

And again, the proof is in the pudding; as the above shows, all blues and all browns correctly deduce their eye colour on the third day.

I can't explain this any clearer than I already have. So if you disagree then we're just going to have to agree to disagree.

I think you're giving up too quick.

If there's only 2 blues, then each blue DOESN'T have justification to think that if the one blue they see wasn't blue, green would still see blue.

There's a canonical answer, and you're disagreeing with it. You have a bit of a burden of proof here. You can cop out if you want with "agree to disagree", but it is that, a cop out.

Ok, this one is really tricky, and I couldn't figure it out on my own. But it is still not adding up for me, something is off.

From the start, everyone knows there is not just one person with blue eyes, so why are these perfect logicians waiting the first day?

From the start, everyone knows there is not just two people with blue eyes, so why are these perfect logicians waiting the second day?

...

fantastic question!! Seriously. That's what makes this such a great puzzle.

That's why you have to break it down into pieces. If everything were the same, (including what the guru says), except only one person has blue eyes, how and when could he figure out his eyes are blue? And then do the same question, but two blue eyed people.

What's tripping me up is this:

If only one person has blue eyes, the guru's statement is clearly informative: the person with blue eyes doesn't see any blue eyes.

If only two people have blue eyes, the guru's statemen is clearly informative, since no one leaving rules out the 1st case for each of the two blue eyed people.

But at three people, the Guru may as well not have spoken. Everyone knows that there is at least one blue person, and everyone knows that everyone knows that there is at least one blue person. Once you move beyond two blue people, the scenario shifts, yet you are relying on the one and two blue people cases to reason about it.

But at three people, the Guru may as well not have spoken — hypericin

But here's the trick. You've agreed with the case of two blue eyed people. Which means, unambiguously, if there were two blue eyed people, they would leave on the second day, right?

Which implies, unambiguously, if there WEREN'T only two blue eyed people, they wouldn't leave on the second day.

Right?

Right? — flannel jesus

I mean yeah, but... but...

Why should the step "If there were one blue, they would leave on the first day" appear in the brains of perfect logicians who already knew before the guru spoke that this was not the case?

If that is not an active possibility, which it is not when blue >2, the failure of anyone to leave on the first night also provides no information.

Whereas if blue = 2, blue = 1 is still an active possibility, so its disconfirmation on the first night provides new information.

You are a logical person who does not know their eye colour; so is everyone else.
You know everyone's eye colour except your own, and everyone else knows your eye colour but not in each case their own.
So to know the colour of your own eyes you need to compare what you see to what they see. Now as each day passes with no blue eyed people leaving, the minimum number of blue eyes each blue eyed person must be seeing increases by one. [Whereas a brown or green eyed person will see one more.] So if you are seeing 2 blue-eyes and they haven't left on the second night, they must also be seeing (at least) 2 blue-eyes, which means you must have blue eyes since they cannot see their own eyes and you can see everyone's but your own.
So when the days have passed that equal the number of blue-eyes that you see, that minimum requires that you have blue eyes too, otherwise the blue-eyes would have already gone. In which case you, and of course all the others remaining must have brown, grey, violet, green, or some other colour eyes, though as it happens you know as they each don't that they all have brown eyes except the guru.

Why should the step "If there were one blue, they would leave on the first day" appear in the brains of perfect logicians who already knew before the guru spoke that this was not the case? — hypericin

It's a counterfactual conditional from which valid deductions can be made thus:

If wishes were horses, then beggars would ride.
But beggars do not ride, but have to walk.
Therefore wishes are not horses.

that's why you have to take it one step at a time. Start by ONLY imagining the scenario with two blue eyed people. The case of "what would happen if there are only one?" would naturally occur to them. Right?

t's a counterfactual conditional from which valid deductions can be made thus:

If wishes were horses, then beggars would ride.
But beggars do not ride, but have to walk.
Therefore wishes are not horses. — unenlightened

It's a valid deduction, but we already know from the outset that wishes are not horses, it tells us nothing new. Similarly, the blue would have left if b=1, but we already know b>1, so their not leaving also tells us nothing new.



We agree that if b=1 or b=2, we MUST have the guru's statement to get the ball rolling. But if b>=3, then @Michael's reasoning seems to apply. We may as well just imagine the guru making the statement, which means we may as well just imagine the guru, and this imaginary guru can make the statement about blue or brown, and so everyone would have left long ago. But if this works with b>=3, surely it works with b=1 or b=2. But it does not.

I don't see why, at any stage, one can just imagine the guru. If you can't imagine him at stage one or two, then you can't simply imagine him at stage 3 either.

At b=3, everyone can make the guaranteed true statement, "everyone must see at least 1 blue"
At b=2 or b=1, this is not a true statement.
So at b=3, but not b=1 or b=2, anyone can say of the guru, "she could truly say, 'I see a blue'", and so anyone could say "if there were a guru, she would say, 'I see a blue'".

why not at b=2? Think about it. At b=2, guru could say I see a blue, and everyone knows guru could say that, because everyone sees a blue who isn't themselves and also isn't guru

No, at b=2, every blue sees one other blue, and for all they know, that blue does not see a blue.

oh wait...

No, I was right, at b=2, a guru must see a blue, but it is not true that everyone else must also reach that conclusion. But at b=3, not only must everyone know that a guru must see a blue, everyone must arrive at the conclusion that everyone else knows that a guru must see a blue.

Focus on what everyone does not know which in each case is only the colour of their own eyes. Now as soon as anyone does know, they are gone, and you know that they knew. Imagining and guessing are not ever allowed in deductive logic so if nothing is said, no one can ever know and thus nothing can ever change.

So someone says I see blue. And now we all know that if anyone did not see blue, they would be blue, and they would know they would be blue and be gone tonight. We know that we can see 99 blues, but that doesn't change the logic, because it's counterfactual conditional.
So tomorrow, we know that everyone can see at least one blue because no one left. But if anyone could only see 1 blue, they would know that, since that one blue did not leave, they themselves must be blue too. And in that case they would both leave that night. And so as each day passes, the counterfactual argument gets augmented by "but no one left therefore everyone must see one more blue", until it gets just exactly to the number of blues (which remember no one exactly knows, because they do do not know their own colour) So after 99 days you know that all the blues are seeing 99 blues, and you are seeing 99 blues and therefore you must be the extra blue that all the other blues must be seeing - because they cannot see themselves. You can see that no one else is.

And at this point I really cannot be arsed if anyone still doesn't get it. I done my bestest and thunked hard how to explain it - Over and out.

did you edit the prior post to this one? The fact that you did willingly edited this post makes me wonder if you might have edited the one prior as well...

Don't edit posts, it confuses everything. If you made a mistake just admit the mistake in your next post. I know you edited a post, and that unfairly makes this whole conversation more confusing than it has to be.

I don't think so, but fair enough.

you made a post, and then you edited it to say "oh wait", and then you edited it again after that. It's making the whole conversation incredibly difficult to follow.

yeah that's the one I edited, I won't do that anymore.

We may as well just imagine the guru making the statement, which means we may as well just imagine the guru, and this imaginary guru can make the statement about blue or brown — hypericin

I don't even think we need to do that.

It seems to be a simple mathematical fact that for all $n \gt= 3$, if I see $n - 1$ people with X-coloured eyes and if they don't leave on day $n - 1$ then I have X-coloured eyes.

So not only is the green person saying "I see blue" a red herring, but the green person being there at all is a red herring.

Anyone who applies the above reasoning will correctly deduce their eye colour without anyone having to say anything, or even imagined to have said anything.

I see 24 green, 36 blue, and 4 red.

Therefore if the 4 red leave on day 4 but the 24 green don't leave on day 24 then I have green eyes.