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...
Here's my best attempt to prove this: 1. As of right now, everyone has come to know that everyone knows that green sees blue through some means or ano...
That someone speaking is required in the counterfactual scenario isn't that someone speaking is required in the actual scenario. You're making a false...
I'm not wrong because I didn't say "only the 100 of us". We don't need someone to say something to apply it to our current situation. We all just need...
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 s...
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...
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 ther...
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 ...
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 ru...
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...
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 ...
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 ...
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 gre...
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 know...
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 deduc...
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 ...
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...
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...
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 th...
That's why I explicitly said where n >= 3. There are at least some occasions where it works where n = 2, but I haven't claimed that it will always wor...
From the OP: 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 ju...
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 th...
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 sou...
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 thi...
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...
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 scena...
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 ...
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 condit...
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...
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 e...
That's why I said: for all n >= 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 ...
He wouldn't, but that's irrelevant. It can be demonstrated that if everyone just follows the rule: for all n >= 3, if I see n - 1 people with X-colour...
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...
I don't even think we need to do that. It seems to be a simple mathematical fact that for all n >= 3, if I see n - 1 people with X-coloured eyes and i...
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...
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 b...
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 e...
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 coincidenc...
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...
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 ...
I already answered. I don’t know. But the reasoning nonetheless allows all blues and browns to correctly deduce their eye colour and leave on the 100t...
And as I have repeatedly explained, it doesn’t actually require the Guru to say anything. It’s a red herring. It might appear to be necessary, but cou...
As I said, I’m not sure. But it appears to be a fact that if the blue-eyed people reason in such a way then they correctly deduce that they have blue ...
I explained the reasoning that each person performs and the conclusion they draw from it; a conclusion that is correct. I don’t understand what else y...
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