consciousness is an attribute of sentient beings — Wayfarer

Isn't this a truism?

And there’s no reason to believe that any collection of material components has ever been conscious — Wayfarer

There is if there are reasons to believe that I am conscious and that I am a collection of material components.

I've told you, it's probably not as simple as there being some specific $n$.

At this point, I'm just answering the question in the OP. All the blues and all the browns can, and will, correctly deduce their eye colour 100 days after they lock eyes — even if green says nothing, as they can use the reasoning above and counterfactually assume that green has said "I see blue" and "I see brown" even if she hasn't.

I said before that I wasn't going to try to explain this again, as this is as simple as I can explain it. I'm actually going to commit to that promise now. There's nothing more I can add to what I've already said.

I have deduced it, just as the people in the OP deduced it after green says "I see blue".

Our reasoning is:

P1. If green says "I see blue" and if there were 1 blue then that blue would leave on day 1.
P2. If green says "I see blue" and if there were 2 blues then those blues would leave on day 2.
P3. If green says "I see blue" and if there were 3 blues then those blues would leave on day 3.
...
PX. If green says "I see blue" and if there were X blues then those blues would leave on day X.
PX+1. Therefore, if the X blues I see don't leave on day X then I am blue.

What you don't seem to understand is that we don't need to wait for green to say "I see blue" to start this reasoning. We can start this reasoning as soon as we lock eyes with each other, or as soon as someone says "Begin!" or as soon as green says "I see brown".

Notice that you allow for everyone to assume P1 as part of their reasoning even though everyone knows that there is more than 1 blue. It is a counterfactual. And for the exact same reason we are allowed to assume P1 as part of our reasoning even if green hasn't yet said "I see blue".

True, perhaps it’s not as simple as defining some particular $n$.

Not that I think it matters to my argument. It is still the case that in the OP it is rational for all blues and all browns to counterfactually assume that green has said “I see blue” even if she hasn’t, and counterfactually assume that there is only 1 blue, even though there are more, and then counterfactually assume that there are 2 blues, even though there are more, and so on, eventually deducing that if the blues I see haven’t left by a particular day then I must be blue, with comparable reasoning for brown. This will allow all blues and all browns to leave on the 100th day, knowing their eye colour.

If you need some kind of “synchronicity” then locking eyes with everyone else will suffice (a premise in the experiment). We don’t need anyone to say anything, whether that be a blue saying “begin!” or green saying “I see blue”.

Yes

so what do you deduce? What's the rest of it? You've only given half a story here. You see 3 blue, 3 brown, 1 green, green says nothing - what do you deduce and how? — flannel jesus

The exact same thing as if green were to say "I see blue". Your insistence that I must wait for her to actually say it to start the reasoning, like runners on a track waiting for the gun to fire, is wrong. I can start right away, and just assume that she said so if it helps.

okay so you've completely bypassed all of unenlighteneds reasoning now. — flannel jesus

The reasoning is:

1. If green says "I see blue" and there is only 1 blue then that blue would leave on the first day
2. If green says "I see blue" and there are 2 blues then those blues would leave on the second day
etc.

The point I am making is that I can use this reasoning even if green hasn't actually said "I see blue". I can use this reasoning from the moment I lock eyes with everyone and come to know that everyone knows that everyone knows that green sees blue.

From the moment we lock eyes we can all just pretend that green has said "I see blue", and act as if she did, even if she doesn't.

Is that n>=4? Are you talking from the perspective of a blue eyed person? Because that's only n=4 for blue, not for brown. — flannel jesus

I might be blue, I might be brown, I might be green, or I might be other. What matters is that I see $n - 1$ (i.e. 3) of a particular colour.

Okay, so what I said here was correct:

Tommy doesn't know that everyone knows that green sees blue because if Tommy doesn't have blue eyes then the 1 blue doesn't know that green sees blues.

The reasoning only works when a) everyone knows the same thing and b) everyone knows that everyone knows the same thing. That's the essence of the problem; that everyone knows that everyone knows that green sees blue, and so her saying "I see blue" cannot possibly provide anyone with new information.

So I actually think this requires $n \gt= 4$.

If I see 3 blue, 3 brown, and 1 green, then everyone knows that everyone knows that green sees blue and brown, and that allows the blues and browns to deduce their own eye colour (e.g. by hypothesising some counterfactual scenario in which green says "I see blue" and there is only 1 blue, etc.). It is rational for them to do so from the moment they lock eyes with one another, even if green says nothing. Them all "seeing everyone" is all the "synchronicity" one needs.

Actually, ignore the above, I misread

So if Tommy sees 1 blue, 2 brown, 1 green, Tommy can safely go to the boat on the second day knowing his eyes are blue. — flannel jesus

No, because Tommy doesn't know that everyone knows that green sees blue. If Tommy doesn't have blue eyes then the 1 blue doesn't know that green sees blues.

It only works when there is shared knowledge, as I said. So in your scenario, only the browns can deduce their eye colour because only "green sees brown" is shared knowledge.

So just to be clear, this is you saying it works in the case of 2 blue 2 brown 1 green, correct? — flannel jesus

It allows the blues and browns to deduce their eyes colour if there are 2 blue, 2 brown, and 1 green.

Above doesn't include a specific scenario where it works. A specific scenario looks like "2 blue, 2 brown, 1 green". Does it work in that scenario, if the green eyed guy says nothing? If not that, what specific scenario does it work in? — flannel jesus

The shared knowledge is that green sees blue and brown. That allows the blues and browns to deduce their eye colour.

The blues will reason that if green were to say "I see blue" and the 1 blue doesn't leave on the first day then they are blue, else that if green were to say "I see brown" and the 2 browns don't leave on the second day then they are brown.

The browns will reason that if green were to say "I see brown" and the 1 brown doesn't leave on the first day then they are brown, else if green were to say "I see blue" and the 2 blues don't leave on the second day then they are blue.

The blues leave on the second day knowing they are blue and the browns leave on the second day knowing they are brown, all without needing green to say anything.

When? Everything seems so vague right now. When does it work? What does Tommy have to see for it to work? — flannel jesus

I explained it above. As per the very purpose of the puzzle, there is some shared knowledge that everyone knows (and that everyone knows everyone knows) such that if the Guru were to say "I see X", nothing new would be learned.

In the OP, that green sees blue and that green sees brown is shared knowledge that everyone knows, and that shared knowledge allows all blues and all browns to deduce their eye colour, even without green saying anything.

In your example, that green sees brown is shared knowledge that everyone knows, and that shared knowledge allows all browns to deduce their eye colour, even without green saying anything.

so you're switching back to saying it DOES work for n=2? — flannel jesus

Sometimes, but not always, as I keep saying.

But the main point still stands; in the OP, the browns and blues can reason as I said and correctly deduce their eye colour on the 100th day even if the Guru says nothing.

It would only work for n=3 if it did work for n=2. — flannel jesus

No, this doesn't follow.

The relevant difference between your example here and the OP is that green saying "I see blue" could provide new information (to Timmy, if I don't have blue eyes) in your example, but it can't provide new information to anyone in the OP. That's an important distinction, as is the whole point of the puzzle.

The reasoning only properly works when there is some shared knowledge. In the OP, not only do I know that green sees at least one blue and one brown but I also know that every blue and brown knows that green sees at least one blue and one brown. That shared knowledge allows us to assume that green says "I see blue" and "I see brown" even without her saying so, and so reason counterfactually as if she did. But I can't make this assumption in your example because I can't assume that Timmy knows that green sees blue (and I can't assume that green knows that Timmy sees green).

The only shared knowledge in your example is that everyone sees brown. From that, we can all assume that one of us says "I see brown" (even if none of us do), and reason accordingly.

Given that I see 2 brown I will reason that if the 2 brown don't leave on the second day then I must be brown, and Timmy and green will reason the same way.

Given that each brown sees 1 brown they will each reason that if the 1 brown doesn't leave on the first day then they must be brown.

Following this reasoning, the 2 browns will leave on the second day knowing that they are brown and Timmy, green, and I will remain, knowing that we are not brown.

I said it works if there are 2 brown and 2 blue. I didn't say it works if there is 1 blue, 2 brown, and 2 green.

But again, I have repeatedly accepted that it doesn't always work where $n = 2$, so I don't know how showing that it doesn't work for some $n = 2$ proves that it doesn't work for $n = 3$.

Have you deduced your own eye color? — flannel jesus

No, because this is one of those $n = 2$ scenarios that I explicitly accept doesn't always work.

In your scenario, green saying "I see blue" potentially provides new information (to Timmy, if I don't have blue eyes), and is why it is incomparable to the example in the OP.

Maybe they were literally there forever. — hypericin

Leaving aside the notion of an eternal past — which I believe to be incoherent — as I said in my first comment, if they were perfect logicians then they wouldn't have been there for endless years; the blues and browns would have left on the 100th day, even if the Guru didn't say anything.

michael can't prove it for the case of 2. — flannel jesus

That's why I explicitly said where $n \gt= 3$. There are at least some occasions where it works where $n = 2$, but I haven't claimed that it will always work where $n = 2$, because it doesn't.

As an example, if I see 1 blue and 1 brown then I can't reason anything about my eye colour without one of them saying something.

And that's because in that scenario, someone saying something does in fact provide new information for someone.

There is a hidden assumption that everyone arrives at the island at the same time, and can all see each other at that time. — hypericin

From the OP:

Everyone can see everyone else at all times [and] everyone on the island knows all the rules in this paragraph.

So it's explicit that everyone can see everyone else and knows that everyone can see everyone else, and implicit that new people don't just randomly appear or disappear (whether before or after the Guru says anything).

I am using correct deductive reasoning. Given that I know that green sees blue and that green sees brown (and that every other blue and brown knows this too) I am allowed to just assume that she says so, and reason as if she did. That gets all the browns and blues to the correct answer, and they leave on the 100th day.

I can't explain this any simpler than I already have, so I'm not going to keep trying.

This step in the reasoning doesn't ever get off the ground. — flannel jesus

As you keep saying, and yet if I were to reason in this way then I would correctly deduce the colour of my eyes.

So as I said before, either it is sound reasoning or it's just a coincidence.

spell out the reasoning then. — flannel jesus

I did here.

As I said before, if it helps we can just assume that some third party says “I see blue” and reason as if they did. We don’t need to wait for some third party to actually say this. This assumption gets everyone to the correct answer.

In the case where there's 2 blue, a blue eyed person sees one blue eyed person, 2 brown eyed people, so he thinks what? How does he reason? — flannel jesus

If the 1 blue doesn’t leave on the first day then I am blue, else if the 2 brown don’t leave on the second day then I am brown, else I am neither blue nor brown.

And regardless of whether or not you think he should reason this way, it is a fact that if he does reason this way then he will correctly deduce that he has blue eyes and will leave on the second day.

why would they imagine someone saying that? — flannel jesus

Why are they imagining, contrary to the facts, that there is only 1 blue?

Because in doing so we can deduce our eye colour. These counterfactual scenarios are a tool that allows us/them to come to the correct answer.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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?

If you want it as a step-by-step argument:

P1. Green sees blue
P2. Therefore, if I don't see blue then I must be blue
P3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
P4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
etc.

I know that P1 is true because I see 100 blue and I know that green can see them too, I know that the antecedent of P2 is false because I see 100 blue, I know that the antecedent of P3 is false because I see 100 blue, etc.

This reasoning doesn't require green to actually say "I see blue" and will allow me to correctly deduce my eye colour.

Therefore either the reasoning is sound or the correct "deduction" is a coincidence.

If green eyed person says nothing, what reason would the two blue have to leave on the second day? Without resting on the coattails of unenlightened — flannel jesus

It's the same reasoning.

Just as we can stipulate some hypothetical in which I don't see anyone with blue eyes, even though "in reality" I do, we can stipulate some hypothetical in which green says "I see blue" (and so I can know that she sees blue), even though "in reality" she doesn't.

And we can do this because we know "in reality" that green sees blue even if she doesn't say so.