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?" — Michael

X knows that everyone knows that guru sees blue at 3 blue. But we've already established that 3 can't leave on the third day.

You're trying to address the problem, but this is a deduction puzzle, and your deduction has a false premise. The premise that's false is 99 blue eyed people would leave on the 99th day.

But for me to show you that's false, I would have to show you that it's false that 98 people would leave on the 98th day.

And to prove that's false, I would have to prove to you that it's false that 97 people would leave on the 97th day.

And so on.

That's a lot.

But here's the deal- you keep counting down, 99 98 97... eventually you get to 3. And we know 3 don't leave on the third day.

It's easier to talk about small numbers than big numbers.

The premise that's false is 99 blue eyed people would leave on the 99th day. — flannel jesus

That's not my premise.

That's not my premise. — Michael

So if the 99 you see leave on the 99th day, on the 100th day you'll conclude you have blue eyes anyway?

It might not be the explicit premise you're trying to focus on, is what I'm saying, but it's still a direct consequence of the reasoning you're trying to apply. If your reasoning works, then it must be true that 99 leave on the 99th day. Right?

Let's assume that the Guru says "I see blue" or "I see brown".

Despite all the counterfactuals, every person on the island knows for a fact that nobody will leave on the first day, or the second day, or the third day, etc.

Them waiting is purely performative (up to the 99th/100th day), albeit a necessary performance. Everyone who can see 99 blue knows that none of them can leave before the 99th day and everyone who can see 99 brown knows that none of them can leave before the 99th day and everyone who can see 100 blue knows that none of them can leave before the 100th day and everyone who can see 100 brown knows that none of them can leave before the 100th day and everyone who can see the Guru knows that she cannot leave.

The moment that the Guru says "I see blue" everyone just commits themselves to the rule:

1. If the 99/100 blue I see don't leave on the 99th/100th day then I am blue and will leave on the 100th/101st day, else I am not blue

And the moment the Guru says "I see brown" everyone just commits themselves to the rule:

2. If the 99/100 brown I see don't leave on the 99th/100th day then I am brown and will leave on the 100th/101st day, else I am not brown

And it is a mathematical fact that if they do commit themselves to these rules then every blue will leave having deduced that they are blue, that every brown will leave having deduced that they are brown, and that the Guru will remain having deduced that she is neither blue nor brown.

I simply believe that the participants do not need to wait for the Guru to say "I see blue" or "I see brown" to commit themselves to these rules. I believe, and I believe that I have shown, that seeing 99/100 blue and 99/100 brown (and possibly 1 green) is all the evidence that perfect logicians need to deduce that committing themselves to these rules can, and will, allow every brown and every blue to leave knowing their eye colour, and so that they will commit themselves to these rules from the moment they lock eyes, without having to wait for the Guru to say anything.

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?".

And just to be clear, we can apply this to 3 blues.

Imagine 3 blues and 5 browns and 1 green.

BL1(#X) sees 2 blues, and looks at one of them (#Y) and knows that he sees at least 1 blue, and because #Y sees at least one blue, #X can reason that #Y also knows that guru sees at least one blue.

So if this is truly the basis of the reasoning, it has to work at 3 blues.

Imagine 3 blues and 5 browns and 1 green. — flannel jesus

Imagine rather, that there are 3 blues, 5 browns, 1 green, and you. You know thus that everyone can see at least 2 blues if they are blue, and at least 4 browns if they are brown and so on.

I think this is the source of a lot of the confusion. In order for you to know your colour you have to know that other people can reason their way to knowing their colour from what they can see. So what is that reasoning? No one has begun to show it for any numbers, but because from outside the situation we know the complete numbers, we are told in advance. We can reason from that to what we think they all should be able to reason. But they don't know the very thing we start with, how many blues, browns and greens there are. If they all knew that, everyone would leave immediately, assuming logicians can count.

But it ought to be obvious, really, that for any person looking at any number of other people with eyes of this that and the other colour, and with no other information, no one can deduce their own eye colour so no one can leave, until someone actually says something.

So in the above situation, the person with green eyes says, "I see black eyes", and that night you leave.
And now the situation is exactly what you proposed above. How does everyone else deduce their eye colour? {Hint: obviously they only know extra, that they don't have black eyes.}

Imagine rather, that there are 3 blues, 5 browns, 1 green, and you. You know thus that everyone can see at least 2 blues if they are blue, and at least 4 browns if they are brown and so on. — unenlightened

I was imagining myself as one of the blues though, putting myself in the place of BL1. That's what I was going for

So if the 99 you see leave on the 99th day, on the 100th day you'll conclude you have blue eyes anyway? — flannel jesus

No, I'll conclude that I don't have blue eyes.

If your reasoning works, then it must be true that 99 leave on the 99th day. Right? — flannel jesus

No.

My reasoning is: if the 99 blue leave on the 99th day then I am not blue, else I am blue and will leave with the other blues on the 100th day.

It is the exact same reasoning that I would make were someone to write "there is at least one blue". I just don't need to wait to see this written down. Seeing 99 blue does exactly what seeing "there is at least one blue" written on a piece of paper does.

No.

My reasoning is: if the 99 blue leave on the 99th day then I am not blue, else I am blue — Michael

I'm really not trying to be sense here but, doesn't that make the answer to the question "yes"? Yes, if there's only 99, they leave on the 99th day.

I'm really not trying to be sense here but, doesn't that make the answer to the question "yes"? — flannel jesus

Your question was:

"If your reasoning works, then it must be true that 99 leave on the 99th day. Right?"

And the answer is "no", because if I have blue eyes then the 99 blue I see won't leave on the 99th day.

So it is possible that the 99 blue I see leave on the 99th day without me and possible that the 99 blue I see leave on the 100th day with me.

okay let me rephrase, I thought you would understand my more casual phrasing:

Your logic relies on it being true that if there were only 99, they would leave on day 99. That's what I meant by "If your reasoning works, then it must be true that 99 leave on the 99th day. Right?" Forgive my sloppy wording.

Do you agree with the new wording?

Yes. If there are 99 blue then every blue will commit to the rule:

1. If the 98 blue I see don't leave on the 98th day then I am blue and will leave on the 99th day, else I am not blue

And in committing to this rule, every blue will leave on the 99th day having deduced that they are blue.

right, and in order for that to be true, that only 99 would leave on day 99, then it must also be true that only 98 would leave on day 98, right?

right, and in order for that to be true, that only 99 would leave on day 99, then it must also be true that only 98 would leave on day 98, right? — flannel jesus

No, nobody is going to leave on day 98 because nobody sees only 97 blue.

I'm not saying anybody is going to leave on day 98. I'm saying the statement, "if there were only 99, they would leave on day 99" can only be true if it's also true that "if there were only 98, they would leave on day 98"

Otherwise, how could it be true that "if there were only 99 they would leave on day 99"?

I'm saying the statement, "if there were only 99, they would leave on day 99" can only be true if it's also true that "if there were only 98, they would leave on day 98" — flannel jesus

No, that's false. Although both statements are true, neither depends on the other.

This is a standalone, deductive argument:

1. There are 100 blue
2. Therefore, every blue sees 99 blue
3. Every blue commits to the rule: if the 99 blue I see don't leave on the 99th day then I am blue and will leave on the 100th day, else I am not blue
4. Therefore, every blue will leave on the 100th day, declaring themselves to be blue

It is not possible for (1) to be true but (2) false, and it is not possible for (1) and (3) to be true but (4) false.

This is the exact principle that applies to the canonical answer to the problem.

Our disagreement stems only over what it would take for (3) to be true.

You say that (3) is true only after someone says "I see blue", i.e., that our logicians will only commit to this rule after hearing someone say "I see blue".

I say that (2) is reason enough for our logicians to commit to this rule, and so for (3) to be true.

No, that's false. Although both statements are true, neither depends on the other. — Michael

Why would 99 leave on day 99 if they didn't reason that only 98 would leave on day 98?

You're saying 100 would be able to leave on day 100 because they're reasoning that if there were only 99 they would leave on day 99. Why do you think it's different for 99? Surely the proof for 99 leaving on day 99 is the same - surely it relies on it being true that only 98 would leave on day 98, just as much as 100 relies on it being true that only 99 would leave on day 99.

Why would 99 leave on day 99 if they didn't reason that only 98 would leave on day 98? — flannel jesus

Because they have committed to the rule: if the 99 blue I see don't leave on the 99th day then I am blue and will leave on the 100th day, else I am not blue

They already know from the start that it is not possible for any blue to leave on the 98th day, so they don't need to consider it all.

if there were only 99, then no they wouldn't think it's not possible for blues to leave on day 98. That's what we're reasoning about. We're reasoning about "if there were only 99". If there were only 99, they WOULD think it's possible for the 98 they see to leave on day 98, if your logic holds. They would have to

if there were only 99, then no they wouldn't think it's not possible for blues to leave on day 98. That's what we're reasoning about. We're reasoning about "if there were only 99" — flannel jesus

Which is irrelevant.

Again, this is a valid argument:

1. There are 100 blue
2. Therefore, every blue sees 99 blue
3. Every blue commits to the rule: if the 99 blue I see don't leave on the 99th day then I am blue and will leave on the 100th day, else I am not blue
4. Therefore, every blue will leave on the 100th day, declaring themselves to be blue

If (1) and (3) are true then (4) is true. This cannot be avoided.

The only thing we need to ask is: what does it take for (3) to be true, i.e. what does it take for our logicians to commit to following this rule?

You say that they will only commit to this rule after hearing someone say "I see blue". I say that they will commit to this rule after seeing 99 blue.

Why would they commit to 3?

Again, this is a valid argument:

1. There are 100 blue
2. Therefore, every blue sees 99 blue
3. Every blue commits to the rule: if the 99 blue I see don't leave on the 99th day then I am blue and will leave on the 100th day, else I am not blue
4. Therefore, every blue will leave on the 100th day, declaring themselves to be blue — Michael

Perfectly valid.

As is this:

There are 100 blue.
I see 99 blue.
Therefore I know I have blue eyes and leave immediately.

Unfortunately, no one within the puzzle knows premise 1.

No one has begun to show it for any numbers, but because from outside the situation we know the complete numbers, we are told in advance. We can reason from that to what we think they all should be able to reason. But they don't know the very thing we start with, how many blues, browns and greens there are. If they all knew that, everyone would leave immediately, assuming logicians can count. — unenlightened

Why would they commit to 3? — flannel jesus

For the exact same reason that they would commit to it after hearing someone say "I see blue" or write "there is at least one blue".

None of them need to hear someone say "I see blue" to know that the following counterfactuals are true, or to know that everyone else knows that they are true:

1. If I know that there is at least one blue and if I do not see a blue then I am blue and will leave tonight
2. If everyone knows that there is at least one blue and if I see only one blue and if he doesn't leave tonight then I am blue and will leave tomorrow night
...

Being perfect logicians, this is just background knowledge.

And as I said in the post above, counterfactual scenarios (1) and (2) can be ruled out from the start. Given what they know of the actual scenario, it's not possible that a blue will leave on the first or second night.

Unfortunately, no one within the puzzle knows premise 1. — unenlightened

Which isn't relevant to what I am saying.

Given this argument:

1. There are 100 blue
2. Therefore, every blue sees 99 blue
3. Every blue commits to the rule: if the 99 blue I see don't leave on the 99th day then I am blue and will leave on the 100th day, else I am not blue
4. Therefore, every blue will leave on the 100th day, declaring themselves to be blue

I am saying:

a) if (1) and (3) are true then (4) is true, and
b) seeing 99 blue is reason enough for our logicians to commit to the rule defined in (3).

1. If I know that there is at least one blue and if I do not see a blue then I am blue and will leave tonight — Michael

This is an impossible condition, because if you do not see a blue, and no one has told you anything you cannot know that there is at least 1 blue.

This is an impossible condition, because if you do not see a blue, and no one has told you anything you cannot know that there is at least 1 blue. — unenlightened

(1) doesn't say "nobody has told me anything".

well then that logic should work when there are just 3 blue eyed people. But it doesn't.

I really want you to consider the lowest possible number this can work at, so we can actually analyse it without being confounded by big numbers.

(1) doesn't say "nobody has told me anything". — Michael

Then it should say '...and someone has said "I see blue"' because otherwise it is contradictory.

Then it should say '...and someone has said "I see blue"' because otherwise it is contradictory. — unenlightened

It doesn't need to say that.

1. If I know that there is at least one blue and if I do not see a blue then I am blue and will leave tonight

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.

Much like:

2. If I kill myself then my parents will have only 1 living son

This is true even without specifying the practical mechanism by which I kill myself.