It seems to be a simple mathematical fact that for all n>=3

>=
3
, if I see n−1

−
1
people with X-coloured eyes and if they don't leave on day n−1

−
1
then I have X-coloured eyes. — Michael

And why n >= 3, rather than n >= 2?

And why n >= 3, rather than n >= 2? — flannel jesus

Maybe also when $n = 2$.

There are 2 brown, 2 blue, and 2 green.

Each brown reasons that if the 1 brown doesn't leave on day 1 then he is brown, that if the 2 blues don't leave on day 2 then he is blue, and that if the 2 greens don't leave on day 2 then he is green.

So when the other brown doesn't leave on day 1 he correctly deduces that he is brown.

And then the same each for blue and green, all deducing the correct answer.

Each brown reasons that if the 1 brown doesn't leave on day 1 — Michael

This is the part that logically fails. Why would 1 brown leave on day 1 anyway, if guru says nothing?

This is the fundamental part that fails in the logic, Michael.

There's no reason whatsoever for 1 person, brown eyed or blue eyed, to leave on day one unless the guru says something. If the guru doesn't say anyhing, then all you know is there's X blue eyed people, Y brown eyed people, Z green eyed people and absolutely no way to know your own eye color. There's no mechanism in your logic Michael. You keep on riding the coattails of unenlightened's logic, but throwing out the fundamental premise of unenlightened's logic. You're trying to have your cake and eat it too.

Without the guru saying anything, there's no mechanism whatsoever short of magic for a single blue-eyed or brown-eyed person to know what color their own eye is. You can't skip past step 1.

Damn it. I buy it now. @unenlightenedhad it worked out before I even typed anything.

This brings up a related question I had thought of before: if it wasn't given in the question, I would have said, no one leaves, end of story. Even after seeing the answer, I had a hard time accepting it.

Given that cases like this exist, how do we even trust our own reasoning? I think the answer is, we can't (except maybe unenlightened!)

Given that cases like this exist, how do we even trust our own reasoning? — hypericin

You just have to accept that you aren't a perfect logician. Is that so bad?

You just have to accept that you aren't a perfect logician. Is that so bad? — flannel jesus

The point is, usually when things feel logically certain, we think we at least know that much. That feeling of logical certainty amounts to a kind of psychological "proof". How else do we ultimately know anything logically follows?

Here, I was tripped up by the idea that the guru can't possibly be giving new information. But, amazingly, despite that feeling, she is, no matter how many blue eyed people there are.

Of course this forum, and philosophy in general, is a quagmire of mistakes. But it is probably much worse than we suspect. If our intuitions are that uncertain, even when they feel totally certain, it seems we are always on logical quicksand.

course this forum, and philosophy in general, is a quagmire of mistakes. But it is probably much worse than we suspect. If our intuitions are that uncertain, even when they feel totally certain, it seems we are always on logical quicksand. — hypericin

I guess I don't feel that way about this, because this is an especially contrived scenario, deliberately built to be counter intuitive. I don't think my failure here necessarily hints at a more wide, general failure at logic or thinking.

This is a stark example, but there have definitely been others, where it felt like something clearly was one way, when it turned out to be another. Surely you have experienced this as well, that the "clearly" feeling just isn't as reliable as it feels.

Surely you have experienced this as well, that the "clearly" feeling just isn't as reliable as it feels. — hypericin

I actually think that's a good thing. I mean, we already have situations where two groups of people feel clearly that the other side is wrong - having examples where most people's "clearly" feelings are off base at least forces everyone to be a little more rigorous in their reasoning than just "it feels wrong".

having examples where most people's "clearly" feelings are off base at least forces everyone to be a little more rigorous in their reasoning than just "it feels wrong". — flannel jesus

I mean, maybe, if everyone went through this problem, or similar, and perfectly internalized that lesson. But, they won't, and frankly we will probably forget this too, sooner or later. But the deeper quandary to me is, how can we ever really be certain? No matter how rigorous we are, or think we are, there can always be some error.

as philosophers, surely we've known that the whole time anyway. That there can always be an error in our thinking.

If you want to get better at stuff like this, you need to learn from the master, Raymond Smullyan.His puzzles are wonderful and such talent as I have is down to reading a couple of his puzzle books, a long time ago. I almost didn't answer because I suspected this puzzle was one of his, and the answer came so intuitive and so quick, I thought it was one I knew but had forgotten. Mind, at one point @Michael got me so confused I said something completely wrong about brown eyed people leaving later.

Anyways, Smullyan - The Lady and the Tiger - or any of his logic puzzles are recommended to all.

thanks!

now I get why you got the answer so fast and so cleanly as well. Very cool

There are 2 brown, 2 blue, and 2 green.

Each brown reasons that if the 1 brown doesn't leave on day 1 then he is brown — Michael

Just to follow up on this.

In the case if 2 2 2 like you laid out, from the point of view of a brown, here's what he knows:

"There's 1 brown, 2 blue, 2 green and 1 unknown -me"

Now you're saying "if 1 brown doesn't leave on day 1..." But there's no reason for 1 brown to leave on day 1. If he was the only brown eyed person, he would see 5 people with non brown eyes, and nobody is saying anything in this scenario, so... what information is this hypothetical guy supposed to have that his eyes are brown? His eyes don't have to be brown. There's no rule that says "there's at least one brown eyed person on the island". For all he knows, there could be no brown eyed people. His eyes could be green, blue, yellow, grey - anything.

This is why the canonical answer does in fact rely on the guru saying something. You need that to get the logic rolling.

Why would 1 brown leave on day 1 anyway, if guru says nothing? — flannel jesus

He wouldn't, but that's irrelevant. It can be demonstrated that if everyone just follows the rule: 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, then they will correctly deduce their eye colour (unless they have a unique eye colour).

Knowing this fact is all it takes for everyone on the island to deduce their eye colour (except those with a unique eye colour). And perfect logicians would know this fact.

So say I see 4 blue, 5 brown, and 6 green

I reason:

1. If the 4 blues don't leave on day 4 then I am blue
2. If the 5 browns don't leave on day 5 then I am brown
3. If the 6 greens don't leave on day 6 then I am green

If I did have blue eyes then the others with blue eyes would reason:

1. If the 4 blues don't leave on day 4 then I am blue
2. If the 5 browns don't leave on day 5 then I am brown
3. If the 6 greens don't leave on day 6 then I am green

And the browns would reason:

4. If the 5 blues don't leave on day 5 then I am blue
2. If the 4 browns don't leave on day 4 then I am brown
3. If the 6 greens don't leave on day 6 then I am green

And the greens would reason:

4. If the 5 blues don't leave on day 5 then I am blue
2. If the 5 browns don't leave on day 5 then I am brown
3. If the 5 green don't leave on day 5 then I am green

All 5 of us with blue eyes would deduce after day 4 that we have blue eyes and leave on day 5.
And at the same time the 5 with brown eyes would deduce after day 4 that they have brown eyes and leave on day 5.
And then finally the 6 greens would deduce that they have green eyes after day 5 and leave on day 6.

Everyone deduced the correct answer without anyone having to say anything.

they will correctly deduce their eye colour (unless they have a unique eye colour). — Michael

But since they do not know their eye colour they might all have unique eye colours and none of them can deduce their eye colour at all. Guess and hope is not deduction.

But since they do not know their eye colour they might all have unique eye colours and none of them can deduce their eye colour at all. — unenlightened

That's why I said: for all $n \gt= 3$ if I see $n − 1$ people with X-coloured eyes...

So if there are at least 3 people with X-coloured eyes and at least 3 people with Y-coloured eyes (and at least 3 people with Z-coloured eyes, etc.) then everyone can deduce their eye colour without anyone saying anything.

Except that you can't because you might have Z coloured eyes and although you can see that the others don't have Z coloured eyes, they don't know that, and so they cannot make the deduction that you rely on them making, to make your deduction

Except that you can't because you might have Z coloured eyes and although you can see that the others don't have Z coloured eyes, they don't know that, and so they cannot make the deduction that you rely on them making, to make your deduction — unenlightened

Which is why I also said "unless they have a unique eye colour", and is the Guru in the original example. She cannot determine the colour of her own eyes but the 100 blue and 100 brown can all determine their own eye colour by the 100th day, even without the Guru saying anything.

No they cannot because they cannot determine that they do not also have a unique eye colour. You are talking nonsense.

They don't need to know that they don't have a unique eye colour. If they don't have a unique eye colour then the reasoning will work, as demonstrated in the post here.

Everyone does in fact correctly deduce their eye colour.

Although I think the particular reasoning in that post only works if nobody has a unique eye colour. If somebody does have a unique eye colour then they can apply the reasoning in my original post.

Everyone does in fact correctly deduce their eye colour. — Michael

No they don't because they could have a unique colour and being, unlike you, perfect logicians they know that, and therefore do not make the fallible guess that they do not have a unique eye colour, and so none of your predicted leavings happen and you will conclude that you must have eyes of every colour.

No they don't because they could have a unique colour and being, unlike you, perfect logicians they know that, and therefore do not make the fallible guess that they do not have a unique eye colour, and so none of your predicted leavings happen and you will conclude that you must have eyes of every colou — unenlightened

They don’t assume that they don’t have a unique eye colour. Rather, they infer it based on what the others don’t do. Notice that each step is a conditional. The implicit final step is “if everyone else has left then I have a unique eye colour”.

Just as in your example; the blues and browns don’t assume that their eyes aren’t green or red. They figure it out.

The point I am making is they that don’t need to wait for green to say anything. They already know that she sees blue and brown. If it helps they could just imagine her saying “I see blue” and “I see brown” and apply the same reasoning.

It is bizzare to suggest that our perfect logicians all know this but must wait with bated breath for her to verbally express what they already know before they can start.

They can start the moment they arrive at the island.

He wouldn't, but that's irrelevant. It can be demonstrated that if everyone just follows the rule: for all n>=3

>=
3
, if I see n−1

−
1
people with X-coloured eyes and if they don't leave on day n−1

−
1
then I have X-coloured eyes, then they will correctly deduce their eye colour (unless they have a unique eye colour). — Michael

But if 1 wouldn't leave, then you can't correctly deduce your eye colour at n=2. You already can't.

That's why I said: for all n>=3 — Michael

You said it applies to n=2 as well, that's what we're talking about. Here's the context:

https://thephilosophyforum.com/discussion/comment/1002914

At n=2, I see only 1 person with blue eyes. I see he doesn't leave on day one. This is the scenario where the guru didn't say anything, so IF he's the only person with blue eyes, he wouldn't leave on day 1 anyway, since he would have no way of knowing he or anybody else had blue eyes.

So he wouldn't leave on day 1 if he was the only one, and he wouldn't leave on day 1 if he wasn't the only one.

So him not leaving on day one gives me no new information, because it would have happened regardless, and I can't use that non information to leave on day 2.

That's the state of play, that's how I and unenlightened analyze this situation.

The point I am making is they that don’t need to wait for green to say anything. They already know that she she’s blue. If it helps they could just imagine her saying “I see blue” and apply the same reasoning. — Michael

You made that point before. but you are wrong. I have already also explained before why you are wrong.

It is bizarre but it is true because the puzzle was set up like that. The act of saying it changes the situation despite giving no new information in its content.

Furthermore, the reasoning cannot work for the brown eyed, because it begins:

—If there was only one brown eyed person,and someone said "I see brown eyes" that person would know they had brown eyes.

But if no one said it, as no one did in this puzzle, then that unique brown eyed person could not have any idea of their eye colour, and therefore the whole chain of reasoning could not get started, and so no brown eyed people leave. Instead, they reason along with the blue eyed except that as they see 100 blue eyes, they will wait an extra day and learn that they do not have blue eyes because all the blue-eyed have gone. But they still won't know if their own eyes are brown, green or pink.

Just as we can imagine a counterfactual scenario in which there is only one brown we can imagine a counterfactual scenario in which green says “I see brown”.

We’re doing it right now. This island doesn’t exist and there is no real Guru saying anything. And yet we can still say “if there was one brown and if the Guru says ‘I see brown’ then…”

And just as we’re allowed to say this, so too are our hypothetical islanders.

So if I was to be magically transported onto this island and see 99 blue, 100 brown, and 1 green, then even though I don’t know if I am blue, brown, green, or other, knowing what I know I don’t have to wait for green to say anything. I know that if the 99 blue don’t leave on the 99th day then I am blue, else if the 100 brown don’t leave on the 100th day then I am brown, else I’m either green or other.