The elder learns they do not have brown eyes — Philosophim

How do they learn that? The elder could easily have brown eyes, as far as she's concerned.

Again, fun puzzle. :) — Philosophim

I think so too. I wanted to spark some debates.

here's the more tricky part - what new information did the Guru give them that they didn't already have? — flannel jesus

This is the beating heart of the puzzle, if you can't answer this you don't understand the puzzle. It is not to synchronize, not to make the counterfactual work.

It is to make sure, not that everybody knows everybody sees a blue, but that everybody knows everybody knows everybody knows..., n-1 times, that everybody sees a blue.

N=0 nobody is a blue.
N=1 not everybody sees a blue.
N=2 everybody sees a blue, everybody does not know everybody sees a blue.
N=3 everybody knows everybody sees a blue, everybody does not know everybody knows everybody sees a blue.
N=4 everybody knows everybody knows everybody sees a blue, everybody does not know everybody knows everybody knows everybody sees a blue.

And so on.


In order to get started, so that the failure of anybody to leave is meaningful, all this must be known. And, for @Michael solution to work, all this must be known too. Only a truth telling guru communicating to everyone that indeed there is a blue can cut through this recursive epistemic conundrum.

at two blue, everyone sees a blue

at three blue, everyone knows everyone sees a blue

at four blue, everyone knows everyone knows everyone sees a blue. But, at this stage, you can add as many "everyone knows" as you want, I think. Can't you? At four blue, everyone knows * infinity that everyone knows that everyone sees a blue.

I think

Or maybe at 5?

Idk I'm lost

How do they learn that? The elder could easily have brown eyes, as far as she's concerned. — flannel jesus

Because the elder sees that everyone left has brown eyes. Its the same as one person who doesn't know if they have blue eyes looking at other blue eyes, though perhaps it would take 101 days to figure it out.

Number blues A B C...

At n=3, A doesn't know B knows C sees a blue.
At n=4, A doesn't know B knows C knows D sees a blue.
And so on.
Of course, every permutation of these are true as well.

I don't know why, I can't justify it right now, I feel as though it explodes to infinity at some threshold.

Like imagine I know something. Maybe you know I know it, maybe you don't - that doesn't explode.

Now imagine I know something and you know I know it. Now, that also doesn't explode - maybe you know I know, but I don't know you know I know.

Now imagine I know you know I know.

And then imagine you know I know you know I know.

If I know the fact, and you know I know, and I know you know I know, and you know I know that, then... at that point, can't we realistically add as many "I know you know"s as we want and it still remain deductively true, assuming we're perfect logicians and both know each other are perfect logicians?

That's my intuition. I could be wrong.

Thinking about it more, I probably am wrong. Maybe it never explodes to infinity

Even in the case of two people, they are all distinct facts, all the way up. It's just that we lose our ability to mentally grasp them pretty fast.

they are all distinct facts, all the way up — hypericin

Yeah I think that's probably true. I changed my mind

In order to get started, so that the failure of anybody to leave is meaningful, all this must be known. And, for Michael solution to work, all this must be known too. Only a truth telling guru communicating to everyone that indeed there is a blue can cut through this recursive epistemic conundrum. — hypericin

I'm not convinced that it need be recursive. It's sufficient that each person knows that each person knows that green sees blue.

If we assume that the participants are numbered, each participant asks himself "is there some X and Y such that #X does not know that #Y knows that #101 sees blue?".

Given that there are 201 participants, there are 40,401 possible combinations, so it's unfeasible for us to list them all, although our perfect logicians will be able to.

But we can make a start.

I'm #1 and I see 99 blue (#2-100), 1 green (#101), and 100 brown (#102-#201).

I ask myself:
Does #2 know that #101 sees blue? Yes; #2 knows that #101 can see blue #3.
Does #2 know that #1 knows that #101 sees blue? Yes; #2 knows that #1 and #101 both see blue #3.
Does #2 know that #2 knows that #101 sees blue? Yes; #2 knowing what #2 knows is a tautology.
Does #2 know that #3 knows that #101 sees blue? Yes; #2 knows that #3 and #101 both see blue #4.
Does #2 know that #4 knows that #101 sees blue? Yes; #2 knows that #4 and #101 both see blue #3.
Does #2 know that #5 knows that #101 sees blue? Yes; #2 knows that #5 and #101 both see blue #3.
Does #2 know that #6 knows that #101 sees blue? Yes; #2 knows that #6 and #101 both see blue #3.
...

I'm fairly certain the answer is always going to be "Yes", and so that there is no X and Y such that #X does not know that #Y knows that #101 sees blue.

So, returning back to the previous steps, I deduce that (1) is true, and can make use of the subsequent counterfactuals to determine whether or not I am blue, which I will deduce on either the 99th day (if I'm not) or the 100th day (if I am):

1. As of right now everyone has come to know, through some means or another, that everyone knows that #101 sees blue
2. If (1) is true and if I do not see blue then I am blue and will leave this evening
3. If (1) is true and if I see 1 blue then if he does not leave this evening then I am blue and will leave tomorrow evening
4. If (1) is true and if I see 2 blue then ...
...

And it bears repeating (if any reader missed the previous comment), that even though as a practical matter (1) is true in counterfactual scenarios (2) and (3) only if someone says "I see blue" isn't that someone must say "I see blue" in every counterfactual and actual scenario for (1) to be true and for this reasoning to be usable.

.You are missing the recursion.

Once your list is fully expanded, #1 knows a fact, call it A:

A: everybody knows that everybody knows guru can see blue.

Ok, #1 knows A. But then #1 realizes, everyone has to know A to proceed, not just me. Otherwise we cannot act in concert.

So, really #1 must establish a meta-fact, B:

B: everyone knows A.

So #1 goes though a longer expansion, and proves B. But then #1 realizes, wait, if I have to know B to proceed, everyone has to know it. So now really he has to establish

C: everyone knows B.

...

And so on.

You are missing the recursion. — hypericin

I’m not. I’m explicitly saying that I don’t think it needs to be recursive.

I’m not. I’m explicitly saying that I don’t think it needs to be recursive. — Michael

If you are missing the need, you are missing it.

the reason I know Michael's answer doesn't work... is related to the recursion you're speaking of, I think, I would phrase it like this:

His logic for 100 relies on the assumption 99 would leave on day 99.

And that in turn relies on the assumption that 98 would leave on day 98.

And you can continue to trace that back, all the way down to:

Logic for 6 relies on 5.
Logic for 5 relies on 4.
Logic for 4 relies on 3.
Logic for 3 relies on 2.

And we KNOW 2 doesn't work if the guru says nothing. Even Michael agrees with that.

If 2 doesn't work, 3 doesn't work. If 3 doesn't work, 4 doesn't work. Trace that all the way back up to 99, then 100.

If you are missing the need, you are missing it. — hypericin

So given these:

1. As of right now everyone has come to know that everyone knows that #101 sees blue
2. If (1) is true and if ...

Are you saying that none of the participants can deduce (1) or are you saying that (2) is false and should instead say:

2. If everyone knows that (1) is true and if ...

If there was only 1 person with blue eyes, and nothing was said, that person would be unique in the group and could not know their colour. So they would not leave.

If there were two people with blue eyes, and nothing was said, each would know that there was at least one person with blue eyes but would have no idea of their own eye colour. So neither would leave on any day. This is because they would know that the person with blue eyes that they could see could not know their eye colour any more than they knew their own.

If there were 3 people with blue eyes, and nothing was said, each would see 2 people with blue eyes, but they would have no idea what their own eye colour was, even though they knew everyone knew that at least 1 person had blue eyes, but the would still also know that no one had any way to know the colour of their own eyes, and so no one would leave.

If there were 4 people with blue eyes, and nothing was said, each would see 3 people with blue eyes, but they would have no idea what their own eye colour was, even though they knew everyone knew that at least 2 people had blue eyes, but they would still also know that no one had any way to know the colour of their own eyes, and so no one would leave.

If there were 5 people with blue eyes, and nothing was said, each would see 4 people with blue eyes, but they would have no idea what their own eye colour was, even though they knew everyone knew that at least 3 people had blue eyes, but they would still also know that no one had any way to know the colour of their own eyes, and so no one would leave.

Can anyone see a pattern emerging? The non leaving of the counterfactual solitary person entails the non leaving of any number of people, because nothing ever tells anyone their own eye colour

Did you read my last post? I don't want to repeat of you didn't.

he replied to your last reply my brother

My last substantive reply he only replied to the first sentence, I suspected he didn't really read it

He's been ignoring me since I tortured him for eternity.

In his defense, that's a pretty good reason to ignore someone

I did and I don't see that it clearly answers my question.

So I'll ask again; given these:

1. As of right now everyone has come to know that everyone knows that #101 sees blue
2. If (1) is true and if ...

Are you saying that none of the participants can deduce (1) or are you saying that (2) is false and should instead say:

2. If everyone knows that (1) is true and if ...

are you saying that (2) is false and should instead say:

2. If everyone knows that (1) is true and if ... — Michael

It's worse than your amended 2. It recurses endlessly.



There is a fundamental problem. Whatever condition you think is sufficient ((1) in your case), , everybody must know it, not just you. But not in the omniscient sense, you, the islander, must know everybody knows it. But if you prove that, that new thing you now know is an additional fact that everyone must know, and you have to know they know it, and everybody must know that, and ..

This is the logic being discussed, right?

1. As of right now everyone has come to know, through some means or another, that everyone knows that #101 sees blue
2. If (1) is true and if I do not see blue then I am blue and will leave this evening
3. If (1) is true and if I see 1 blue then if he does not leave this evening then I am blue and will leave tomorrow evening
4. If (1) is true and if I see 2 blue then ...
...

And it bears repeating (if any reader missed the previous comment), that even though as a practical matter (1) is true in counterfactual scenarios (2) and (3) only if someone says "I see blue" isn't that someone must say "I see blue" in every counterfactual and actual scenario for (1) to be true and for this reasoning to be usable.

But we already have a simple, straight-forward case that this logic doesn't work. We know, because he's already acknolwedged, that 2-blue-eyed doesn't work. 2 blue-eyed people cannot leave on the second day.

If it's true that 2 blue-eyed people cannot leave on the second day, then it must also be true that 3 blue-eyed people cannot deduce that there's more than 2 blue-eyed people just because they don't leave on the second day. So 3 blue-eyed people cannot leave on the third day.

But premise 1, "everyone has come to know, through some means or another, that everyone knows that #101 sees blue", is true in the case of 3 -- and yet it still doesn't work.

So we have a tangible, specific case where Michael should be able to apply this logic, and yet can't.

It genuinely feels like these simple cases, for low numbers of blue-eyed people, are being ignored because it's easier to hide the reasoning behind the obscurity and confusion of very large numbers. The beauty of unenlightend's logic is that it clearly unambiguously works for small numbers, and so we can work our way up to large numbers. In contrast, Michael's logic, we know for sure doesn't work for small numbers, so instead of working his way up to large numbers, he just kinda ignores the problems at small numbers and hopes nobody notices the gaps in logic once there's 100 people to talk about. It's easier to hide the cracks with so many blue-eyed people to think about.

If Michael wasn't so worried about getting tortured for eternity, I'd be encouraging him to find the lowest number of blue-eyed people that it works for. Michael it's only a fictional torturing for eternity.

It's worse than your amended 2. It recurses endlessly. — hypericin

I don't know what it would mean for (1) to be true but for "everyone knows that (1) is true" to be false, much like I don't know what it would mean for "I know that Paris is the capital of France" to be true but for "I know that I know that Paris is the capital of France" to be false.

It seems to me that if (1) is true then everyone knows that (1) is true and everyone knows that everyone knows that (1) is true, etc. So you get your recursion.

And you said before that the Guru saying "I see a blue" can "cut through this recursive epistemic conundrum", but it's not the only thing that can. Another thing that can is seeing a piece of paper with the words "there is at least one blue" written on it.

But what's the relevant difference between seeing a piece of paper with the words "there is at least one blue" written on it and seeing 99 blue? How and why is it that the former can "cut through this recursive epistemic conundrum" but the latter can't?

But what's the relevant difference between seeing a piece of paper with the words "there is at least one blue" written on it and seeing 99 blue? How and why is it that the former can "cut through this recursive epistemic conundrum" but the latter can't? — Michael

That's what makes this puzzle so interesting. Truly, that's one of the biggest points, and why people find it fascinating. It's weird. It's hard to explain, it's unintuitive, but if you work through the logic from the ground up, it's nevertheless true. For some reason, it makes a difference.

That's what makes this puzzle so interesting. Truly, that's one of the biggest points, and why people find it fascinating. It's weird. — flannel jesus

Okay, well I think the answer is that there isn't a difference. Seeing 99 blue does exactly what seeing a piece of paper with the words "there is at least one blue" does; it makes (1) true (which makes "everyone knows (1)" true, which makes "everyone knows that everyone knows (1)" true, etc.)

okay so I dare you to not leap to thinking about 100, and think about smaller numbers. We've talked about 2, we've talked about 3, I think we agreed 2 can't leave on the second day, I think we agree 3 can't leave on third. Can 4 leave on the 4th?

I explain it in the first part of the post above:

If we assume that the participants are numbered, each participant asks himself "is there some X and Y such that #X does not know that #Y knows that #101 sees blue?".

Given that there are 201 participants, there are 40,401 possible combinations, so it's unfeasible for us to list them all, although our perfect logicians will be able to.

But we can make a start.

I'm #1 and I see 99 blue (#2-100), 1 green (#101), and 100 brown (#102-#201).

I ask myself:
Does #2 know that #101 sees blue? Yes; #2 knows that #101 can see blue #3.
Does #2 know that #1 knows that #101 sees blue? Yes; #2 knows that #1 and #101 both see blue #3.
Does #2 know that #2 knows that #101 sees blue? Yes; #2 knowing what #2 knows is a tautology.
Does #2 know that #3 knows that #101 sees blue? Yes; #2 knows that #3 and #101 both see blue #4.
Does #2 know that #4 knows that #101 sees blue? Yes; #2 knows that #4 and #101 both see blue #3.
Does #2 know that #5 knows that #101 sees blue? Yes; #2 knows that #5 and #101 both see blue #3.
Does #2 know that #6 knows that #101 sees blue? Yes; #2 knows that #6 and #101 both see blue #3.
...

I'm fairly certain the answer is always going to be "Yes", and so that there is no X and Y such that #X does not know that #Y knows that #101 sees blue.

So, returning back to the previous steps, I deduce that (1) is true

"is there some X and Y such that #X does not know that #Y knows that #1 sees blue?"

I don't know what the minimum number of participants must be for the answer to this question to be "no", but by induction it appears that the answer to the question is "no" when applied to the problem in the OP, which is the only thing I'm addressing.

Okay, well I think the answer is that there isn't a difference — Michael

Just to recap, We've already agreed that it does make a difference for the case of one blue eyed person, and two, and three. There must be some number where it starts making a difference. I'm very interested in that number. You want me to accept that it starts making a difference some time before 100 - if I'm going to accept that, I'm gonna need you to show me when.

For every number of blue eyed people x, your reasoning seems to rely on the premise that if there were x-1 blue eyed people, they leave in x-1 days. You're obscuring your logic by jumping to 100 blue eyed people. I'm trying to explore with you the numbers that aren't obscured.

I'm very interested in that number. — flannel jesus

Then go through all the numbers and for each number imagine the participants asking themselves "is there some X and Y such that #X does not know that #Y knows that #1 sees blue/brown/green?"

I'm just addressing the problem in the OP, and I think that what I say in that post above shows that the blues and browns can and will leave on the 100th day having deduced their eye colour even without the Guru having said anything.