but you also know it's a valid argument if you replace a2 and B2 with this premise: — flannel jesus

Yes, but adding or subtracting some arbitrary number to or from $n$ is arbitrary, whereas $n$ isn't arbitrary.

it might be — flannel jesus

It is.

it exists in a sea of equally valid and arbitrary premises. Suppose they replace 2 with committing to leave on X + 5 days. Or even X - 10 days. — flannel jesus

The premises aren't arbitrary. $n$ is the number of people seen with $X$ eyes. Adding or subtracting some arbitrary number to or from $n$ would be arbitrary though, which is why perfect logicians wouldn't do it.

Next we consider Argument B:

B1. There are 99 people with blue eyes and 101 people 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"

B4. Therefore, from (B1) and (B2), every person with blues eyes commits to the rule: "if the 98 people I see with blue eyes don't leave on day 98 then I will leave on day 99 and declare that I have blue eyes"

B5. Therefore, from (B1) and (B2), every person with blues eyes commits to the rule: "if the 101 people I see with brown eyes don't leave on day 101 then I will leave on day 102 and declare that I have blue eyes"

B6. Therefore, from (B1) and (B2), every person with brown 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"

B7. Therefore, from (B1) and (B2), every person with brown 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"

B8. Therefore, from (B4), every person with blues eyes leaves on day 99 and declares that they have brown eyes

B9. Therefore, from (B7), every person with brown eyes leaves on day 101 and declares that they have brown eyes

This argument is also valid.

Now we get to the more interesting part. If we know that these arguments are valid then so too do our islanders. They might not yet know if any of the premises are true, but they do know that the arguments are valid.

For the next step let's start by considering a simplified version of the argument in the OP. The islanders arrive on the island together and are told that everyone has either blue or brown eyes — which is not the same as being told that there is at least one person with blue eyes and one person with brown eyes (it could be that everyone has blue eyes or everyone has brown eyes); it is only meant to dismiss the possibility that one's own eyes are green or red or pink or whatever.

I am an islander.

I know that Arguments A and B are valid.

I see 99 people with blue eyes and 100 people with brown eyes. Therefore I know that either A1 or B1 is true.

Therefore, I know that if 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" then everyone will leave the island having correctly declared their eye colour.

Seems like it requires mind reading to me for them to assume that about everyone else.

If they all could assume that about everyone else, sure, they get off the island. But they have no idea what everyone is committing to. — flannel jesus

I'm not assuming anything about anyone. I am simply saying that Argument A is valid.

Why would one of these blue eyed people think of that particular premise? — flannel jesus

I'm not saying that they are. I'm simply saying that the argument is valid.

What is it? — flannel jesus

It's a premise in the argument.

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

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.

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.

sure it follows — flannel jesus

No it doesn't.

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

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.

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.

Well, I am my biology, my brain activity, my thoughts and so on, so to me this is another instance of everything being willed by yours truly.

Good times. — NOS4A2

Not everything our body does is voluntary.

Just as “one’s heartbeat” refers to a particular thing in the body, not the body as a whole, so too does “one’s will”. If eliminative materialism is correct then one’s will is a particular kind of neurological phenomena, and only bodily behaviour caused by that particular neurological phenomena is “being willed by yours truly”.

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.

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.

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".

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).

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.

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 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.

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.

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.

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.

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.

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.

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.

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

That's not my premise.

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.

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.

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.)

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?

Organisms operate by different principles to non-organic matter. — Wayfarer

Perhaps, but organic matter is still a collection of material components. So if we have a reason to believe that organic matter can be conscious then we have a reason to believe that a collection of material components can be conscious.

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 ...

Perhaps you’re something other than a collection of material components. — Wayfarer

Okay, but you said: "there’s no reason to believe that any collection of material components has ever been conscious".

So are you saying that there's no reason to believe that I am a collection of material components?

You possess something that instruments don’t, namely, organic unity. — Wayfarer

Is "organic unity" not a collection of material components? Because as far as I'm aware, organic matter is matter.

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 ...

You are missing the recursion. — hypericin

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

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.

My disagreement is that you need the guru to say something just to make the counterfactual work. — hypericin

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 another
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 ...
...

[repeat for brown]

That 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.

The only requirement is that (1) be true, and in the actual scenario in which there are 100 brown, 100 blue, and 1 green, (1) is true even if nobody says anything. Our perfect logicians know (1) from the moment they lock eyes, and so immediately apply the above reasoning and start their daily counting, allowing the blues and browns to leave on the 100th day knowing their eye colour without anyone having to say anything.

Yes you do need someone to say it because the first counterfactual needs someone to say it and every iteration thereafter rests on that necessity; you cannot discharge that assumption along the way. — unenlightened

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 logical step.

Edit
See here which I think explains it best.

Original
It is a fact that if 100 browns, 100 blues, and 1 green lock eyes then if everyone immediately starts applying the counterfactuals and says "if the 98/99/100 blues I see don't leave on the 98/99/100th day then I am blue" and "if the 98/99/100 browns I see don't leave on the 98/99/100th day then I am brown" then the 100 browns and 100 blues will leave on the 100th day knowing their eye colour, and the 1 green will remain knowing that they are neither blue nor brown.

You might think that they shouldn't reason this way, but nonetheless if they do reason this way then (other than green) they will leave knowing their eye colour — and each of them knows from the start that reasoning this way will allow either 199 or 200 people to leave by the 101st day knowing their eye colour.

You've gone wrong already.You see 99 blues. The blues that you see, all see 98 or 99 blues. The 200 of you are all thinking that. — unenlightened

I'm not wrong because I didn't say "only the 100 of us".

You can know that too. but you cannot apply it to your situation because no one has said anything. — unenlightened

We don't need someone to say something to apply it to our current situation. We all just need to know when we will all start counting, which will be the first possible "synchronisation" point — when everyone first locks eyes. At that moment, if it helps, they can pretend that someone says "I see blue and I see brown", even though nobody does.

And if everyone starts counting from the moment they first lock eyes, the 100 blue will leave on the 100th day knowing that their eyes are blue, the 100 brown will leave on the 100th day knowing that their eyes are brown, and the green will remain knowing that they are neither blue nor brown.

So I think the initial intuitive assumption that green saying "I see blue" shouldn't make a difference is correct, even though the initial intuitive assumption that nobody will ever leave is false. 200 people will leave on the 100th day knowing their eye colour without anyone having to say anything.

When the deduction begins, it has to begin with: 'if there is only one blue, and someone says "I see blue" then they will know that they have blue eyes', and someone has to say it out loud, because in this case they have no idea that anyone sees blue because they are the only blue. And that is why the argument only runs when it is said out loud, not when everyone just knows from their own experience that in fact everyone can see blue. — unenlightened

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 someone says "I see blue" then that blue will know that they are blue and leave

The 100 of us do not need to wait for someone to say "I see blue" for us to think and know that (1) is true.

And given that there are at least 99 blue, everyone knows that everyone knows that green sees blue, and so can make use of (1) to try to deduce their eye colour.

You said at the start that there must be a "synchronisation" point, and that's right; but that "synchronisation" point doesn't need to be green saying "I see blue". It can be anything that everyone recognises as being the signal to start our deduction. It can be green saying "I see blue", it can be green saying "I see brown", it can be green saying "Begin!", or it can be everyone locking eyes with each other for the first time.