why would they imagine someone saying that?

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.

I'm trying to work with you on making your case, but you don't want to take n= 2 to is logical conclusion. Try it. Try reasoning about the case where there's 2 blue, and the guru says nothing. What's the actual logic? We'll move on when it's settled, but I think we can actually come to an agreement on n=2 if you commit to actually working it to its conclusion.

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?

let's name the blue-eyed people. We're going to talk about Timmy and Tommy. Timmy and Tommy both have blue eyes so let's look at this from Tommy's perspective

Tommy sees that Timmy has blue eyes and nobody else that he sees has blue eyes. So from Tommy's perspective there's really two possibilities: either Timmy is the only one with blue eyes or Timmy and Tommy both have blue eyes.

So you're saying that Tommy would expect Timmy to leave on day one if Timmy was the only one with blue eyes, but why would Tommy expect him to do that? Think about it

If Timmy was the only one with blue eyes, which is the scenario we're thinking about as Tommy, then Timmy wouldn't see anyone with blue eyes. If Timmy didn't see anyone with blue eyes he wouldn't know anybody on the island had blue eyes and therefore he wouldn't know that he has blue eyes and therefore he wouldn't have a reason to leave the island on day one.

This is why I think your n=2 logic doesn't work. Is there anything wrong with my reasoning?

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.

that's not deductive though. 1 blue wouldn't leave on the first day anyway, right? Why would he?

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.

spell out the reasoning then.

I don't think it "helps", I think it's an entirely irrational thing for Tommy to imagine that.

spell out the reasoning then. — flannel jesus

I did here.

A1 and A2 don't make sense as two separate premises

A1. Green sees blue
A2. Therefore, if I don't see blue then I must be blue

The only reaason you know Green sees blue is BECAUSE you see blue. You see blue, and so you know green sees blue. This step in the reasoning doesn't ever get off the ground.

A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue

This doesn't work either, because the one blue you see, you don't necessarily know that he knows that green sees blue. Right? How would you know that he knows green sees blue?

That's really what this logic puzzle is about - what can you know other people know? A3 only works if you know that the blue eyed person you see knows green sees blue. But you don't know that he knows that.

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.

sure, it's a "coincidence". You're using non-deduction and incorrect reasoning, as an outside observer, to get to the correct conclusion.

If I asked you, what's 2+2, you might say 4. If I asked, how do you know that? You might say, because rhinosceroses have horns.

You're using nonsensical reasoning to arrive at an answer.

Now YOU as an outside observer can do that in this case because YOU have the privilege of knowing Timmy and Tommy both have blue eyes. But Timmy and Tommy don't know that. THEY only have the information available to them. For all Timmy and Tommy know, the other person could be the only one with blue eyes. Tommy thinks, I couild easily have brown eyes, I have no way of knowing. Even on day 2.

The logic puzzle requires correct deductive reasoning, not just using your magic powers as an outside observer to use wrong reasoning to guess at the right answer.

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.

A3 only works if you know that the blue eyed person you see knows green sees blue. But you don't know that he knows that. — flannel jesus

When b>=3, you absolutely DO know that. You can prove that everyone (including a real or hypothetical green) sees blue. The problem I see with @Michael reasoning is the use of "days". Days from what? There is a hidden assumption that everyone arrives at the island at the same time, and can all see each other at that time.

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).

When b>=3, you absolutely DO know that. — hypericin

we're not talkinfg about that case though. michael can't prove it for the case of 2.

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.

Tommy genuinely thinks Timmy might be the only one with blue eyes, right?

If that's the case, Tommy has NO REASON whatsoever to think Timmy will leave on day 1. You aren't even trying to make a case for it. You're giving up without even trying.

From Tommy's perspective, Timmy doesn't have any reason at all to imagine green eyed guy saying "I see someone with blue eyes". From tommy's perspective, timmy doesn't know that green eyed guy DOES see someone with blue eyes. So from Tommy's perspective, Timmy could be imagining the blue eyed guy saying "I see someone with green eyes" -- Tommy thinks Tommy could have green eyes, right?

So Timmy might be imagining the guru saying "I see someone with green eyes" as far as Tommy is concerned, and Timmy might see Tommy with green eyes, and Timmy might be waiting day 1 to see what Tommy does, to see if Tommy leaves.

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

Seems like you're not seriously considering the possibility that you're wrong. That's a mistake.

How about we do a real experiment:

YOU'RE Tommy.

I'm not going to tell you your eye color.

You see Timmy with blue eyes, you see George and Jack with brown eyes, and you see Guru with green eyes. Nobody says anything on day 1, nobody leaves on night 1. It's day 2. It's approaching time to board the boat. Do you board the boat? Have you deduced your own eye color? What is it?

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 — Michael

Ît does say

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. — flannel jesus

Maybe they were literally there forever.

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.

I don't know if you're going to play the game from my previous post, so I'll play it on your behalf using the logic you've given already.

You see Timmy with blue eyes, you see George and Jack with brown eyes, and you see Guru with green eyes. Nobody says anything on day 1, nobody leaves on night 1. It's day 2.

Using all your reasoning that you've said so far, you've "deduced" that because nobody left on night 1, you have blue eyes.

You go to the boat, you ask to board. You tell Charon that you've deduced that you have blue eyes.

He says I'm sorry Tommy (Michael), that's incorrect, now I'm going to torture you and keep you alive in agony for as long as I can, thems the rules.

Unfortunately, you didn't have blue eyes. You had green eyes. Timmy had blue eyes, but he didn't leave on day one because he DIDN'T imagine the Guru saying "I see someone with blue eyes" like you expected him to. Because why would he imagine that? He didn't see anyone with blue eyes. Of course he didn't imagine that.

Your reasoning skills got you tortured. Try again next life.

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.

No, because this is one of those n=2

=
2
scenarios that I explicitly accept doesn't always work. — Michael

This is the exact scenario that you said did work above.

https://thephilosophyforum.com/discussion/comment/1003049

You said the reasoning worked. You're just randomly changing your mind now.

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 $n = 3$.

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. — Michael

But you're Tommy. You don't know ahead of time if there are 2 brown 2 blue or 3 brown 1 blue. That's the point. That's LOGIC. You don't know. Your reasoning has to apply in any case where you see the things you see. You're not magic, you're just Tommy.

Tommy sees the same thing in the 2 brown 2 blue scenario as he does in the 3 brown 1 blue scenario, doesn't he? Your reasoning, as Tommy, has to be based on what he sees. You said it works if he sees 1 blue and 2 brown - so I put that to the test, and your reasoning got you tortured.

You now accept that it doesn't work for n=2 - that's why it doesn't work for n=3. It would only work for n=3 if it did work for n=2. And it doesn't.

For the longest time, you've been riding unenlightend's logical coattails. The thing is, his logic works for n=1 and because it works for n=1, it works for n=2 and because it works for n=2 it works for n=3.

That's not the case with your logic, as we can see. That's why it's so important to work through n=2 before we even begin with n=3. If it doesn't work for n=2, it can't work for n=3.

if they were perfect logicians then they wouldn't have been there for endless years; — Michael

Since they are perfect logicians, anything that would have allowed them to synchronize and leave before the guru spoke can be ruled out, since they are still there.

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.

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