The practical mechanism by which I have come to know that there is at least one blue does not need to be specified for this conditional to be true. It is true even when left unspecified. — Michael

You are flailing. If you are magic and a mind reader then bla bla blah, anything you like. But in the scenario there is no magic, no one knows their eye colour and yet you think everyone can logically deduce their own eye colour without anyone saying anything. So piss or get off the pot, you can't have it both ways.

we agreed that it can't work for 2 people. 2 people don't leave on the 2nd day.

You seemed to understand why that means it can't work for 3 people, so if there are only 3 eyed blue people, we also know they won't leave on the 3rd day.

Do you see why that means it can't work for 4 blue eyed people, why they can't leave on the 4th day?

If you are patient and take this seriously, I'm pretty sure you'll find what I have to say compelling. But we gotta start small.

But in the scenario there is no magic, no one knows their eye colour and yet you think everyone can logically deduce their own eye colour without anyone saying anything. — unenlightened

I am saying both of these:

1. If I do not see anyone with blue eyes then I cannot deduce that I have blue eyes unless someone says "there is at least one person with blue eyes"

2. If I see 99 people with blue eyes then I can deduce whether or not I have blue eyes even if no-one says "there is at least one person with blue eyes"

You seem to think that because (1) is true then (2) is false? I don't think that follows at all.

As I mentioned in an earlier comment, even if we wait for the Guru to say "there is at least one person with blue eyes" it's not as if anyone is actually waiting to see if someone will leave on the first day. We already know that no-one will. Our waiting those first 98/99 days is purely performative, not informative, and we’d be surprised and dumbfounded if anyone left earlier than that.

Given that we can dismiss the first counterfactual situation outright, it doesn’t matter what would be required in that situation to know that there is at least one person with blue eyes; that requirement is not a requirement in our actual situation in which we see 99/100 people with blue eyes.

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. — Michael

1. I know x
2. Everyone knows x
3. I know everyone knows x
4. Everyone knows everyone knows x.
5. I know everyone knows everyone knows x
6. Everyone knows everyone knows everyone knows x
...

You claim that at some number in this series, they stop being distinct facts, so that the next number is the same fact as the previous?

Everyone can see 99 blues in the n=100 case. But this is not the same as the guru speaking, or everyone seeing on a piece of paper "there is one blue". With only visual evidence there can always be cases where 1 believes 2 believes 3 believes... 99 believes there is no blue. Only communication can collapse this chain.

We already know that no-one will. Our waiting those first 98/99 days is purely performative, not informative, — Michael

It is not purely performative. If it was I'm sure the perfect logicians could find a way to skip it.

After the guru speaks, everybody knows everybody knows ... there is at least one blue
After the first day, everybody knows everybody knows... There is at least two blues.
And so on

I swear to god it will be easy to convince him if you just convince him to start with small numbers. He's allowing himself to get confused by the number 100 and 99. There's at lot less room to get confused at 2, 3, 4. He's already admitted it's impossible at 2. He's half way admitted it's impossible at 3.

Nobody will ever come any closer to agreement as long as we focus on numbers we can't even completely imagine.

2. If I see 99 people with blue eyes then I can deduce whether or not I have blue eyes even if no-one says "there is at least one person with blue eyes"

You seem to think that because (1) is true then (2) is false? I don't think that follows at all. — Michael

That's what I think, and I have given a fairly strong argument for it, which you have ignored. I have seen no argument from you to show otherwise, and no reference to such an argument, whereas I have given a reference to a supporting argument and widely accepted solution. But carry on incorrigible.

Probably. But the interesting part to me is exploring different aspects and arguments. it's pretty rich, for a logic puzzle!

I'd be willing to explore his angle too but he doesn't bite on anything!

Like I tried to meet him where he is, at 100, and it took him a long time to come around to the idea that his logic for leaving on day 100 relies on it being true that if there were only 99, they'd leave on day 99. But eventually, I got him to see that, I think.

And so then I said, so surely in turn it's true that "if there were only 99, they'd leave on day 99" relies on it also being true that "if there were only 98, they'd leave on day 98". For some reason that just doesn't compute for him. Applying the same logic he's applying to n100, to n99... that's where I lose him.

He's so ultra focused in on 100 that he refuses to look at any of the surrounding logic.

Seems like he just wants to conclude that his logic works, not look at it, not have it be questioned, end of story. Which is fine but like... keep it to yourself then lol. That's not much fun for the rest of us.

That a simple puzzle such as this can go for twelve pages explains so much about the forums.

This is actually kinda usual for this particular puzzle. I brought this up on another forum 12 years ago and the same thing happened - someone with more or less the same position as Michael went on for pages and pages about why everyone else was wrong and he was right. He did come around in the end.

What do you think it explains about the forums?

I thought this puzzle has an initial state problem that makes it impossible to solve.

do you still think that, or you just used to think it?

Yes, I still believe that.

can you describe what the problem is?

Well the initial state problem is true because as the opening post describes that they would not be able to count the amount of individuals in groups of different eye colors.

they can't count themselves. They can count everyone else

Your insistence that if my reasoning works for 100 then it must work for 1, and so that if it doesn't work for 1 then it doesn't work for 100, is false.

Take these two arguments:

A1. There are 100 people with blue eyes and 100 people with brown eyes
A2. Every person commits to the rule: if the $n$ people I see with $X$ eyes don't leave on day $n$ then I will leave on day $n + 1$ and declare that I have $X$ eyes
A3. Therefore, every person will leave and correctly declare their eye colour

B1. There is 1 person with blue eyes and 1 person with brown eyes
B2. Every person commits to the rule: if the $n$ people I see with $X$ eyes don't leave on day $n$ then I will leave on day $n + 1$ and declare that I have $X$ eyes
B3. Therefore, every person will leave and correctly declare their eye colour

Argument A is valid even though argument B is invalid.

To show this:

If there are 100 people with brown eyes and 100 people with blue eyes then:

1. Every person with brown eyes commits to the rule: if the 99 people I see with brown eyes don't leave on day 99 then I will leave on day 100 and declare that I have brown eyes
2. Every person with brown eyes commits to the rule: if the 100 people I see with blue eyes don't leave on day 100 then I will leave on day 101 and declare that I have blue eyes
3. Every person with blue eyes commits to the rule: if the 99 people I see with blue eyes don't leave on day 99 then I will leave on day 100 and declare that I have blue eyes
4. Every person with blue eyes commits to the rule: if the 100 people I see with brown eyes don't leave on day 100 then I will leave on day 101 and declare that I have brown eyes
5. Therefore, every person with brown eyes will leave on day 100 and correctly declare that they have brown eyes and every person with blue eyes will leave on day 100 and correctly declare that they have blue eyes

If there is 1 person with brown eyes and 1 person with blue eyes then:

1. The person with brown eyes commits to the rule: if the person I see with blue eyes doesn't leave on day 1 then I will leave on day 2 and declare that I have blue eyes
2. The person with blue eyes commits to the rule: if the person I see with brown eyes doesn't leave on day 1 then I will leave on day 2 and declare that I have brown eyes
3. Therefore, the person with brown eyes will leave on day 2 and incorrectly declare that they have blue eyes and the person with blue eyes will leave on day 2 and incorrectly declare that they have brown eyes

So it doesn't matter how many "counterarguments" you come up with where the reasoning doesn't work with lower numbers or different combinations of eye colours; argument A is valid.

Your insistence that if my reasoning works for 100 then it must work for 1, and so that if it doesn't work for 1 then it doesn't work for 100, is false. — Michael

You're skipping steps again. Usually you skip up - you go from some low number, get tired of thinking about that, and skip all the way up to 100. Now you're doing the opposite - you're going from 100 straight down to 1.

Don't. Skip.

Be patient, take it one step at a time

I didn't say if it works for 100, it must work for 1. I said if it works for 100, it works for 99. If it doesn't work for 99, it can't work for 100.

I didn't say if it works for 100, it must work for 1. I said if it works for 100, it works for 99. If it doesn't work for 99, it can't work for 100. — flannel jesus

That doesn't follow.

This is valid, regardless of whether or not a comparable argument is valid for some other number:

A1. There are 100 people with blue eyes and 100 people with brown eyes
A2. Every person commits to the rule: if the $n$ people I see with $X$ eyes don't leave on day $n$ then I will leave on day $n + 1$ and declare that I have $X$ eyes
A3. Therefore, every person will leave and correctly declare their eye colour

It is impossible for A1 and A2 to be true but for A3 to be false.

sure it follows. This is a deduction puzzle. You see 99 people with blue eyes, you have two possibilities: either you're on an island with 99 blue eyed people and you don't have blue eyes, or 100 and you do.

Surely your logic involves the following statements at some level, implicitly or explicitly:

If there are only 99, they'll leave on day 99.

If there are 100, the 99 I see won't leave on day 99.

No?

sure it follows — flannel jesus

No it doesn't.

If you can't accept that Argument A is valid then we can't continue.

if I don't agree with your conclusion we can't continue. Yeah okay buddy. I don't know why you want to talk to anybody lol. This is a philosophy forum. We can disagree with you, don't be weird about it.

if I don't agree with your conclusion we can't continue. Yeah okay buddy. I don't know why you want to talk to anybody lol. This is a philosophy forum. We can disagree with you, don't be weird about it. — flannel jesus

It's not my conclusion. It's one of my premises. And it's a premise that I demonstrated to be true here.

and I'm trying to talk to you about that. You have this here:

if the n
people I see with X
eyes don't leave on day n

so that means, surely, that it's completely agreeable when I point out that your logic relies on this also being true:

If there are only 99, they'll leave on day 99.

As shown above, the argument is valid when there are 100 people with brown eyes and 100 people with blue eyes but invalid when there is 1 person with brown eyes and 1 person with blue eyes.

Therefore, it's not the case that if the argument is valid when there are $m$ people with brown eyes and $m$ people with blue eyes then it is valid when there are $m - 1$ people with brown eyes and $m - 1$ people with blue eyes.

The number of each colour makes a difference.

okay so if that's not valid, then when you start out unsure if you're on an island with m blue eyed people or m-1 blue eyed people, you can't rely on it being true that "if there were m-1 blue eyed people, they would have left in m-1 days - therefore there are more than m-1 people with blue eyes, therefore I can leave on night m"

Because that's what this is about at root. There are only 2 possibilities from the perspective of a blue eyed person: either there are m-1 blue eyed people, or m. He's trying to deduce which world has in.

If he's waiting to see if m-1 people maybe don't leave in m-1 days, but it turns out to be FALSE that m-1 people would leave in m-1 days, then waiting for that doesn't tell him what world he's in. He could be in a world where m-1 people have blue eyes, or m people have blue eyes.

These guys don't want to get tortured for eternity. They can't rely on iffy reasoning. They have to be SURE. No guessing allowed. Only deductions.

So if it's at all possible that m-1 people WOULDN'T leave in m-1 days, then we absolutely cannot then say, "oh well I didn't see m-1 people leave in m-1 days, so therefore there must be m blue eyed little"

So if m is 100, each blue eyed person sees 99 blue eyed people and they, as perfect logicians (not perfect planners, not perfect committers to rules), have to ask themselves, can I really be sure 99 people would leave in 99 days? If they're anything less than deductively sure, they can't leave in 100 days.

You're getting ahead of yourself. I'm not yet talking about what the people on the island see or know. I am simply saying that Argument A is valid.

I'll break it down even further if it helps:

A1. There are 100 people with blue eyes and 100 people with brown eyes

A2. Every person commits to the rule: "if the $n$ people I see with $X$ eyes don't leave on day $n$ then I will leave on day $n + 1$ and declare that I have $X$ eyes"

A4. Therefore, from (A1) and (A2), every person with brown eyes commits to the rule: "if the 99 people I see with brown eyes don't leave on day 99 then I will leave on day 100 and declare that I have brown eyes"

A5. Therefore, from (A1) and (A2), every person with brown eyes commits to the rule: "if the 100 people I see with blue eyes don't leave on day 100 then I will leave on day 101 and declare that I have blue eyes"

A6. Therefore, from (A1) and (A2), every person with blue eyes commits to the rule: "if the 99 people I see with blue eyes don't leave on day 99 then I will leave on day 100 and declare that I have blue eyes"

A7. Therefore, from (A1) and (A2), every person with blue eyes commits to the rule: "if the 100 people I see with brown eyes don't leave on day 100 then I will leave on day 101 and declare that I have brown eyes"

A8. Therefore, from (A4), every person with brown eyes leaves on day 100 and declares that they have brown eyes

A9. Therefore, from (A6), every person with blue eyes leaves on day 100 and declares that they have blue eyes

what is 2? What do you call that? It's not part of the setup. It's not a known fact about the scenario. It's also not a necessary consequence of the scenario setup.

It's not even an assumption. This blue eyed person doesn't just immediately start assuming everyone has committed to the rule.

What is it?

What is it? — flannel jesus

It's a premise in the argument.