a premise that comes from where? Why would one of these blue eyed people think of that particular premise?

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.

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. This isn't a commitment puzzle.

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.

it might be, but unfortunately it exists in a sea of equally valid arguments with equally arbitrary premises. Suppose they replace 2 with committing to leave on X + 5 days. Or even X - 10 days. Hell maybe even X - 95 days, why not?

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.

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

Every person commits to the rule: "if the n
people I see with X
eyes don't leave on day n-95
then I will leave on day n-94
and declare that I have X
eyes"

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.

the point you were focusing on is it's validity.

You enter some premise - some premise that isn't derived from the problem statement - and if you can use that premise with the rest of the problem statement to get everyone to leave with the correct eye colours, then it's valid. That's the way you've been arguing.

You've inserted a "commitment" and once inserted it allows you to get everyone to leave. I did the same with n-95

the point you were focusing on is it's validity. — flannel jesus

Yes, that's the first step: arguments A and B are valid.

The next step is: premises A2 and B2 are not arbitrary.

And the next step is: 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.

Therefore, I know that if every person commits to the rule: — Michael

Yes but you don't know that every person will do that. Therein lies the problem

Yes but you don't know that every person will do that. Therein lies the problem — flannel jesus

Then we move on:

1. If I have blue eyes then every person with blue eyes knows exactly what I know
2. If I have brown eyes then every person with brown eyes knows exactly what I know

Either way, I know that everyone with my eye colour knows exactly what I know, and so knows 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.

I put it to you that if perfect logicians know that everyone with their eye colour knows that committing to this rule will work then they will commit to this rule, and so they will leave the island having correctly declared their eye colour.

Either way, I know that everyone with my eye colour knows exactly what I know, and so knows that if every person commits to the rule: "if the n
people I see with X
eyes don't leave on day n-95
then I will leave on day n-94
and declare that I have X
eyes" then everyone will leave the island having correctly declared their eye colour.

Again, subtracting an arbitrary number from $n$ is arbitrary, and so perfect logicians wouldn't do it. But $n$ isn't arbitrary.

but they all know they could, and they all, according to you, know exactly the same thing, so they all know they could subtract 95 and it would still work. That'll save them some waiting. Why not?

but they all know they could, and they all, according to you, know exactly the same thing, so they all know they should subtract 95 and it would still work. — flannel jesus

They could subtract 95, but that would be arbitrary and so they wouldn't do it. Perfect logicians would stick to the non-arbitrary $n$.

I think adding "we all know the same thing" is something unnatural you added tbh. It's not in the problem statement. It's basically cheating yourself into a false solution.

I think adding "we all know the same thing" is something unnatural you added tbh. — flannel jesus

It's not.

Everyone knows that Arguments A and B are valid, because they are and everyone is a perfect logician.

I see 99 blue and 100 brown.

If I have blue eyes then every person with blue eyes sees 99 blue and 100 brown.
If I have brown eyes then every person with brown eyes sees 99 blue and 100 brown.

Everyone with my eye colour knows that either A1 or B1 is true.

Therefore everyone with my eye colour has come to the same conclusion: that if we all commit 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.

So I commit to this rule, as will they — and we leave the island having correctly declared our eye colour without anyone having to say anything.

too much mind reading for me personally. Not deductive.

I will say that it's food for thought for me. I might seem dismissive and like I'm refusing to accept it, but I'm running it around in my mind and there are moments where I think, maybe... maybe Michael actually does have it right and the guru doesn't need to say anything.

I'm leaning towards thinking it's not correct but it's only a lean.

I might seem dismissive and like I'm refusing to accept it — flannel jesus

I know that I've come across this way too and I don't mean to be.

We all think this never works. You know this doesn't work at low n, but think it does at high n. Therefore it is incumbent on you to find the special n where it starts working.


I still think you are missing the recursion. To act in concert, everyone must model everyone's mental state. And that model must include the modeling of everyones mental state. Every time you move from x's mental state to their model of y's mental state, x can no longer be counted as blue; whether or not x is blue, x doesn't know what y sees when y looks at x's eyes.

Humans wouldn't think this way, but these are perfect logicians, not humans.

We all think this never works. You know this doesn't work at low n, but think it does at high n. Therefore it is incumbent on you to find the special n where it starts working — hypericin

Yeah this is definitely an aspect that still bothers me. And it will endlessly make the "guru says nothing" solution distasteful unless it's figured out.

Another aspect is that because it relies on commitment rather than deduction, the easy counter to it is, it's assuming they're committing because they WANT to leave. I don't think anything in the problem statement explicitly indicates that they want to leave, just that they do leave when they've correctly deduced their eye colour.

But this aspect is less important than the previous one. It really matters when it starts working

Yeah this is definitely an aspect that still bothers me. And it will endlessly make the "guru says nothing" solution distasteful unless it's figured out. — flannel jesus

It is not a solution whatsoever until @Michael can prove it.

In the "official" formulation I saw on Popular Mechanics, it says something to the effect that the islanders do not do anything unless they are logically certain of the outcome. I think this is key. It is impossible to be logically certain that in everyone's mental modelling of everyone's mental modelling, everyone can see blue. In reality, no matter how logical the islanders are, they would see 99 other blues and just say fuck it and act as Michael suggests.

Either way, I know that everyone with my eye colour knows exactly what I know, and so knows 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. — Michael

Except that there's a possibility that one islander is a red-eye.
So that islander can count that there are 201 islanders total, assuming no one has left yet. They can deduce between the blue and the brown, but not if there's a third/fourth color.

I'll present my answer without searching for the correct answer or before reading any of the comments.

Everyone leaves on the first night because the Guru can see that 100 people with blue eyes and 100 people with brown eyes, making her the only one who has the green eyes. Everyone with blue eyes can only see 99 people with blue eyes. So, through logical deduction, that person must have blue eyes since there are only 100 people in total with blue eyes, that person must conclude that he/she is the 100th person with blue eyes. This same logic goes the same for the people with brown eyes. That's how everyone can leave the island on the first night.

So, through logical deduction, that person must have blue eyes since there are only 100 people in total with blue eyes, that person must conclude that he/she is the 100th person with blue eyes. — night912

What's the logical deduction?

99+A=100
A=1

how does he know there's 100? Nobody said that.