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.

Sometimes, but not always, as I keep saying. — Michael

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

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.

I explained it above — Michael

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?

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.

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

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.

OK, thank you for clarifying that, last time this came up you said "No, because this is one of those n=2 scenarios that I explicitly accept doesn't always work."

So it DOES work.

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. Right? That's your explicit position. It's not a "n=2 that doesn't always work" scenario, this is a scenario where you're saying explicitly, the logic DOES work, Tommy CAN deduce his eye color when he sees 1 blue, 2 brown, 1 green, and he sees that the blue eyed guy doesn't leave on day one.

It's clear, explicit, no "maybe sometimes doesn't work", you agree unambiguously that you think Tommy can deduce this.

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.

Actually, ignore the above, I misread

okay no problem.

So Tommy can deduce it?

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.

okay so you've completely bypassed all of unenlighteneds reasoning now. You're entirely on your own here and nothing you say relies on his logic at all.

So I actually think this requires n>=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 — Michael

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.

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.

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?

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.

The exact same thing as if green were to say "I see blue". — Michael

Okay, so... you wait for day 4, you think "the blues would have all left on day 3 if there were only 3", and then you leave on day 4 as a blue with all the other blues?

Yes

"the blues would have all left on day 3 if there were only 3"

Doesn't that rely on the logic working for n=3?

So I actually think this requires n>=4. — Michael

If you think the logic works for n=4, but the logic relies on a premise that pretty much explicitly says "the logic also works for n=3", then... n=4 can only work if n=3 works, no? And n=3 doesn't work. We both agree n=3 doesn't work. So if n=3 doesn't work, can you rationally say "the blues would have all left on day 3 if there were only 3"? That IS what n=3 is saying. I don't think you can say "the blues would have all left on day 3 if there were only 3". We can't just freely accept that as a deductively valid premise to use in your logic.

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”.

why are you so eager to skip to the end solution without building up working premises first?

There's a pattern here. We proved that it doesn't work for n=2, and because of that, you immediately accepted that it doesn't work for n=3.

For some reason, you didn't apply that to n=4 - for the exact same reason you can reject n=3 if n=2 doesn't work, you can reject n=4 if n=3 doesn't work. And if you can reject n=4, you can reject n=5. And so on. Up to 100.

Your confidence should be shaken. You were so confident about n=2, to the point of even saying "I've already explained it, I can't make it any more simple" - and then we proved you wrong. And you agree that you were wrong about n=2. So... don't you think maybe you should be a little less confident about n=100?

I'm not saying you should immediately reject n=100, but maybe accept that you have a serious burden of proof there, because if n=2 doesn't work, neither does 3, and so on, right?

So don't be so sure. Do the logic. Work it out. Don't just state a conclusion and call it a day, this is a deduction puzzle. I want to see you deduce.

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".

so what number n does it start working at? Not 2. Not 3. Which one?

It is sufficient that all blues know that all blues know that green sees blue.

Shouldn't it work at n=3 then? But we've both agreed it doesn't.

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've told you, it's probably not as simple as there being some specific n — Michael

You telling me something doesn't make it true. If the logic works, there's got to be a point at which it works. If there's no n at which the logic works, then... it doesn't work.

The last time you said it was simple, you said that about something you were wrong about. Your overconfidence is...weird. I mean it was weird to begin with, but to use the same overconfident line again, after you know the last time you said it you were wrong, is like... double weird. Why are you doing that? Don't you think that's weird? You've been consistently wrong, I don't know where your arrogance comes from.

If you're right, you should be able to prove it. This is a case of pure deductive reasoning, and I'm quite frankly enjoying it. If you want to throw in the towel, fine, I just... don't even know why you bothered to say anything if you don't want to even try to prove it.

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 greens to deduce their eye colour, even without green saying anything. — Michael

It doesn't allow any such deduction. Knowing is not the same as saying, and I think we agree that if someone has a unique colour, they cannot deduce it.

When the deduction begins, it has to begin with: 'if there is only one blue, and someone says "I see blue" then they will know that they have blue eyes', and someone has to say it out loud, because in this case they have no idea that anyone sees blue because they are the only blue. And that is why the argument only runs when it is said out loud, not when everyone just knows from their own experience that in fact everyone can see blue.

If the argument begins with "everyone can see that there are multiple blue and brown but no one says anything." What is the next step?

If the argument begins with "everyone can see that there are multiple blue and brown but no one says anything." What is the next step? — unenlightened

For him the next step is just imagining someone says something.

This is just what I was talking about. It seems so damn reasonable that at some n, they could skip the stupid guru, lock eyes, and start from there. Like, suppose the universe was packed tight with 10^100 blues, you need a guru to tell you that... she sees a blue??? Yet, afaict, you do.

yeah, totally, i get it. I get why it seems like you ought not to need the guru. But when none of the proofs work, and you can't even bootstrap the deductive process on n=1 or n=2, then there's nowhere to go from there.

When the deduction begins, it has to begin with: 'if there is only one blue, and someone says "I see blue" then they will know that they have blue eyes', and someone has to say it out loud, because in this case they have no idea that anyone sees blue because they are the only blue. And that is why the argument only runs when it is said out loud, not when everyone just knows from their own experience that in fact everyone can see blue. — unenlightened

I see 99 blue. These 99 blue see either 98 or 99 blue. The 100 of us are all capable of thinking and knowing that:

1. If there is only 1 blue and if someone says "I see blue" then that blue will know that they are blue and leave

The 100 of us do not need to wait for someone to say "I see blue" for us to think and know that (1) is true.

And given that there are at least 99 blue, everyone knows that everyone knows that green sees blue, and so can make use of (1) to try to deduce their eye colour.

You said at the start that there must be a "synchronisation" point, and that's right; but that "synchronisation" point doesn't need to be green saying "I see blue". It can be anything that everyone recognises as being the signal to start our deduction. It can be green saying "I see blue", it can be green saying "I see brown", it can be green saying "Begin!", or it can be everyone locking eyes with each other for the first time.