An unintuitive logic puzzle
I found this online many years ago. Don't look up the answer!
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
Comments (431)
I think I can justify all of them, but I think this is probably the best
Is this the crux of the puzzle?
I watched LOST a lot.
All the people except the guru have a pretty good idea what color their eyes are. After all they can count. So that night they go to the boat and get on. You havent said how they verify whether or not the person knows their eye color. Lets assume that the guy who is running the boat asks them. They tell him what they figured out. If theyre right, they get on the boat and go. If theyre wrong then they definitely know what color their eyes are so they get on the boat and go.
One possible problem with this is that each person making the decision might be wrong and might have red eyes or some other color. But they know that isnt true, because everybody showed up at the boat. They wouldnt have all showed up if one of the people had red eyes. If that happened, they couldnt be sure they didnt have red eyes too.
They can count, but... so what? What's the logic? From the point of view of any person showing up to the boat, how has he logically deduced the colour of his own eyes?
Youre right, their ability to count doesnt matter and I shouldnt have put it in there but it doesnt make any difference. The guy running the boat tells each person whether or not they got their eye color correctly. If they did they get on the boat. If they didnt on their first try, then they know and can get on the boat.
They know they dont have some color other than blue or brown, because if they did, no one wouldve showed up at the boat
I included this text from the author at the bottom:
QW55d2F5LCB5ZXMgdGhhdCdzIHRoZSBhbnN3ZXIhIFlvdSBmaWd1cmVkIHRoYXQgb3V0IHdpdGhvdXQgbG9va2luZyBpdCB1cD8gSW1wcmVzc2l2ZQ==
https://www.base64encode.org/
And send me the result that comes out below
There's a DECODE button at the top of the site to do the reverse process
You two guys are good, I actually couldn't figure this one out myself.
[hide]this is hidden[/hide]
[hide="Reveal"]It seems like no new information, because they could all see multiple blue eyes already, yet after 99 nights they knew something new that the guru hadn't told them, from the reactions of the others. So the new information was not what she told each one, but that she told them all at the same time. She set the clock ticking[/hide]
[hide]
I've heard it said that the new information is in iterative referential knowledge. If you look at the case with only one person with blue eyes, him saying "I see someone with blue eyes" doesn't give any new information to people without blue eyes, but to the person with blue eyes it does.
If you look at the case with two blue eyes, everyone knows "there's someone with blue eyes" before the guru speaks, but after he speaks, now you can say "everyone knows that everyone knows that there's someone with blue eyes" - two people didn't know that already.
And -- somehow, I forget how -- that everyone knows that everyone knows that ... keeps propagating upward, maybe after each passing day or something.
[/hide]
[hide]If they're perfect logicians then on the first day that they arrived on the island, even before the Guru speaks:
1. The Guru knows that 100 blue-eyed people each see either 99 or 100 blue-eyed people and either 100 or 101 brown-eyed people and that 100 brown-eyed people each see either 99 or 100 brown-eyed people and either 100 or 101 blue-eyed people.
2. Every blue-eyed person knows that 99 blue-eyed people each see either 98 or 99 blue-eyed people, either 100 or 101 brown-eyed people, and either 1 or 2 green-eyed people, that 100 brown-eyed people each see either 99 or 100 brown-eyed people, either 100 or 101 blue-eyed people, and either 1 or 2 green-eyed people, and that the Guru sees either 99 or 100 blue-eyed people, either 100 or 101 brown-eyed people, and either 0 or 1 green-eyed person.
3. Every brown-eyed person knows that 99 brown-eyed people each see either 98 or 99 brown-eyed people, either 100 or 101 blue-eyed people, and either 1 or 2 green-eyed people, that 100 blue-eyed people each see either 99 or 100 blue-eyed people, either 100 or 101 brown-eyed people, and either 1 or 2 green-eyed people, and that the Guru sees either 99 or 100 brown-eyed people, either 100 or 101 blue-eyed people, and either 0 or 1 green-eyed person.
Using the reasoning given by unenlightened, on day 100 every blue-eyed person would leave knowing that they have blue eyes, every brown-eyed person would leave knowing that they have brown eyes, and the Guru would stay knowing that they have neither blue nor brown eyes.
The Guru doesn't need to say anything; her saying either "I see at least one blue-eyed person" or "I see at least one brown-eyed person" provides no new information and is a red herring.[/hide]
What's the reasoning for brown eyed people? Unenlightened gave reasoning for blue-eyed people
The same, just change "blue" for "brown".
The Guru doesn't need to say it. Him saying it is a red herring. As perfect logicians, every blue-eyed person already knows that the Guru sees at least one blue-eyed person and every brown-eyed person already knows that the Guru sees at least one brown-eyed person.
[hide]If there was only 1 person w. blue eyes, that person would see no blue eyes and therefore know they had blue eyes and leave that night.[/hide]
This first step of reasoning doesn't work for brown eyed people, because that's not what the guru said.
So if his first step of reasnoning doesn't work, then you can't just swap out blue for brown.
Also I think the brown eyed people would not know their eye-colour for another 99 days after the blue eyes left, but only that they themselves didn't have blue eyes.
Why do you think that? Imagine the Guru were to have said "I see at least one blue-eyed person and at least one brown-eyed person".
But as I said to flannel, the Guru doesn't even need to say it because everyone already knows that she sees at least one blue-eyed person and at least one brown-eyed person, and so her saying it is a red herring.
You said you base your reasoning on unenlighteneds reasoning. Step 1 of his reasoning completely relies on the guru saying what he said. Can you see that?
So you must have different reasoning. Will you make it explicit?
That's the red herring; it doesn't. Everyone already knows that she sees at least one brown-eyed person, so her expressing this fact verbally provides no new information.
The first step in the reasoning is "the Guru sees at least one person with brown eyes". She doesn't need to say "I see at least one person with brown eyes" for this first step to be true. Her saying so is a red herring.
It's the same. Here are unenlightened's exact and complete words:
[hide]If there was only 1 person w. blue eyes, that person would see no blue eyes and therefore know they had blue eyes and leave that night.
If there were 2, they would both see one person w blue eyes and when they did not leave the first night, they would both know the second night that they must have blue eyes and leave.[/hide]
Notice that he doesn't mention the Guru or what she says at all.
It's implicit in the first sentence. The first sentence of his reasoning clearly depends on the guru saying what she said. Please look specifically at that first sentence, let's at least agree on that.
" there was only 1 person w. blue eyes, that person would see no blue eyes and therefore know they had blue eyes and leave that night."
If the guru doesn't say what she said, then no, the 1 blue eyed person wouldn't leave that night
No it doesn't. It only depends on "the Guru sees at least one blue-eyed person" being true. It doesn't depend on her saying so.
The reasoning is: if the Guru sees at least one person with blue eyes and if I don't see anybody with blue eyes then I have blue eyes.
Every person on the island already knows that the Guru sees at least one person with blue eyes and one person with brown eyes, whether or not she says so, as explained here. Given this fact:
If I don't see anyone with blue eyes then I have blue eyes, else if I don't see anyone with brown eyes then I have brown eyes.
This reasoning is valid even though I do in fact see others with blue eyes and brown eyes.
Not in the scenario with one blue eyed person they don't
I see 100 people with blue eyes and (unknown to me) I have brown eyes. The Guru says "I see at least one blue-eyed person". Now I imagine a scenario where these 100 blue-eyed people don't exist.
If I can still assume that the Guru says "I see at least one blue-eyed person" then I can still assume that the Guru sees at least one blue-eyed person, but doesn't say so.
If there's only one guy with blue eyes, he would only know that the guru sees blue eyes if the guru told him. There's no way around that.
In practice, perhaps, but the logic doesn't require that the Guru say anything. The logic only requires that I know that the Guru sees at least one blue-eyed person and one brown-eyed person.
Given that our reasoning stipulates (contrary to the facts) that I don't see any other blue-eyed person but that the Guru still says "I see at least one blue-eyed person", it can also stipulate (contrary to the facts) that I don't see any other blue-eyed person but that I know that the Guru still sees at least one blue-eyed person.
And I can stipulate all of this even if I in fact have brown eyes.
Im not stipulating random things, as shown by the fact that if the people on the island were to apply my reasoning then they would all correctly deduce the colour of their eyes. Thats not just some happy coincidence; its because the reasoning is sound.
If there's 2 blue 2 brown 1 guru and he doesn't say anything, no telepathy, nobody gets off the island.
Right?
No. It does depend on the guru saying so unless everyone already knows that everyone already knows at the same time, as I suggested above and you ignored. This is the extra information that the guru imparts: she doesn't inform them about what she sees, but she puts everyone in a synchronised state of knowing each other's knowing. That is what is required for the nested hypotheticals to begin.
I reason thus:
If there was only 1 person with blue eyes {PWBE} and that person knew that the guru sees blue eyes, then that person would know that they have blue eyes and would leave tonight.
Therefore:
If there were only 2 PWBE and the guru sees blue eyes {GSBE} then neither would leave tonight, and when they see that, they each know they have blue eyes and would leave on the second night.
And so on.
But the factual knowledge that I can see multiple blue eyes and thus already know that the guru can see blue eyes cannot be imported into the counterfactual hypothetical wherein the blue eyed person would know no such thing because he would not himself see blue eyes, and thus could not know therefore that the guru saw blue eyes ... wait for it ... UNLESS SHE SAID SO.
And if in this scenario I have brown eyes then the Guru wouldnt say I see someone with blue eyes, and yet we are allowed for the sake of argument to assume that she does.
Given that the Guru does in fact say it when there are 201 people we are allowed to assume that she still says it in a hypothetical scenario with 2 people.
And given that we all know what the Guru sees without her saying it when there are 201 people we are allowed to assume that we still do in some hypothetical scenario with 2 people.
The reasoning is sound in either case and gets us to the correct answer.
She doesnt need to say anything for perfect logicians to be in a synchronised state. At every moment they are in a synchronised state and will apply the same reasoning.
And if new people arrive or are killed by tigers during this process theyll adapt their premises (e.g the Guru sees either N or N+1 blue-eyed people) as needed.
Quoting unenlightened
She doesnt need to have actually said something in the real world in which I see 100 blue-eyed people for me to stipulate that she said something in this counterfactual hypothetical world in which I dont see any blue-eyed people.
In the real world I know that she sees at least one blue-eyed person even without her saying so, and so if it helps I can just assume that she says so even if she doesnt.
Oops.
I dont think thats a comparable scenario. I think a minimal example requires 3 blue, 3 brown, and 1 green.
Each blue reasons: green sees blue, and so if the two blue I see dont leave on the second day then I must be blue and green sees brown, and so if three brown I see dont leave on the third day then I must be brown.
Therefore on the third day each blue knows they are blue and leaves
Each brown reasons: green sees brown, and so if the two brown I see dont leave on the second day then I must be brown and green sees blue and so if the three blue I see dont leave on the third day then I must be blue.
Therefore on the third day each brown knows they are brown and leaves.
Each person (other than green) leaves knowing their eye colour, all without anyone saying anything. The proof is in the pudding, as it were.
Or is it just a coincidence?
Nope. Im going to keep trying. Here is some more non-canonical answers.
One guy, I dont know who, it might be @Baden, takes one of his eyes out, looks at it, and then leaves the island.
The guy who drives the ferry leaves the island every night.
If it doesn't matter what the green eyed person says, why is his presence required at all?
Im not sure, but my reasoning does allow all brown and all blue to leave knowing their eye colour, so either its sound or its a very lucky coincidence.
But as a question to you, why would it require green verbally expressing what everyone already knows?
I don't think it's sound or a coincidence. I don't think it's correct. I don't think there's any reason why the green eyed person being there, not saying anything, would allow anybody to decide their eye colour, and you haven't explained why there would be.
Quoting Michael
You have to follow the logic carefully one step at a time to find that out. It's very subtle and honestly strange - that what makes this such a good logic puzzle. It's completely counterintuitive, but also, once you fully grok it, undeniably true. That gives it this really unique flavour as a puzzle.
And yet every blue-eyed person leaves knowing they have blue eyes and every brown-eyed person leaves knowing they have brown eyes. So what do you mean by it not being correct?
Quoting flannel jesus
And I think that it goes even further: it may be counterintuitive, but one can get to the correct answer without green saying anything.
Clever. Obviously everyone could do that to their own eye. That's another loophole answer though - the real answer doesn't involve a loophole
You're saying "and yet" as if you've demonstrated that. You haven't
I demonstrated it in the example above.
The one with 3 brown, 3 blue, and 1 green
So why doesn't your example just include 3 blue 3 brown?
How can it be incomplete it it allows all brown and blue to leave knowing their eye colour?
I do not believe it allows that. Do you understand that? I don't think it does. I don't think it actually allows anyone to leave.
I explained the reasoning that each person performs and the conclusion they draw from it; a conclusion that is correct.
I dont understand what else youre looking for.
What does the green eyed person have to do with it? Why not just have 3 blue 3 brown?
As I said, Im 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 eyes and that if the brown-eyed people reason in such a way then they correctly deduce that they have brown eyes.
Therefore either the reasoning is sound or its a coincidence.
Just as you dont appear to be sure why green must verbally express what everyone already knows to be true.
"I don't know" isn't reasoning. Why is the green eyed person relevant in your scenario? If he's not saying anything? For every blue eyed person who relies on the green eyed person, couldn't they just as easily rely on a brown eyed person? For every brown eyed person who relies on the green eyed person, can't they just as easily rely on a blue eyed person? I don't think the green eyed person is doing anything.
Ive given up on being correct. Im working on being amusing.
And as I have repeatedly explained, it doesnt actually require the Guru to say anything. Its a red herring. It might appear to be necessary, but counterintuitively it isnt.
As evidenced by the fact that my reasoning allows all blues and browns to correctly deduce their eye colour and leave on the 100th day, answering the question in the OP.
I could justify unenlighteneds reasoning with actual complete syllogisms. I don't think you can do so for yours.
Its explained in the post.
You're still basing your reasoning off the words of a guy who explicitly has different premises from you. Until you stand on your own two feet with your reasoning, your explanations are on shaky ground
Like this for example. Why do you think this is true? If green eyed person says nothing, what reason would the two blue have to leave on the second day? Without resting on the coattails of unenlightened, why would this be the case?
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 stipulate some hypothetical in which green says "I see blue" (and so I can know that she sees blue), even though "in reality" she doesn't.
And we can do this because we know "in reality" that green sees blue even if she doesn't say so.
P1. Green sees blue
P2. Therefore, if I don't see blue then I must be blue
P3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
P4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
etc.
I know that P1 is true because I see 100 blue and I know that green can see them too, I know that the antecedent of P2 is false because I see 100 blue, I know that the antecedent of P3 is false because I see 100 blue, etc.
This reasoning doesn't require green to actually say "I see blue" and will allow me to correctly deduce my eye colour.
Therefore either the reasoning is sound or the correct "deduction" is a coincidence.
Same reasoning as unenlightened, who was assuming the green eyed person said something?
Make your reasoning explicit. You're still riding on coattails, you can't have your cake and eat it to. If you want to disagree with his premises, show your reasoning yourself.
This doesn't work if green doesn't say anything. If green doesn't say anything, p2 isn't the case. If green doesn't say anything, and you don't see blue, your eyes could literally be any colour.
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 coincidence?
https://xkcd.com/solution.html
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 eye colour and leave on the 100th day without green saying anything. I think the results speak for themselves.
I've explained it as clearly as I can, so there's nothing else to add.
What you're not understanding is that they could just add easily incorrectly deduce their eye colour. It's a coin flip at best, because the "deduction" isn't based on sound premises.
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 blue then I must be blue
A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue
B1. Green sees brown
B2. Therefore, if I don't see brown then I must be brown
B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown
Given that each blue sees 2 blue and 3 brown, they can rule out some of these premises:
A1. Green sees blue
[s]A2. Therefore, if I don't see blue then I must be blue[/s]
[s]A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue[/s]
[s]A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue[/s]
A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
[s]A3. Therefore, if I see three blue and they leave on the third day then I must not be blue[/s]
[s]A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue[/s]
B1. Green sees brown
[s]B2. Therefore, if I don't see brown then I must be brown[/s]
[s]B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown[/s]
[s]B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown[/s]
[s]B3. Therefore, if I see two brown and they leave on the second day then I must not be brown[/s]
[s]B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown[/s]
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown
Then, come the third day, they can rule out one more:
A1. Green sees blue
[s]A2. Therefore, if I don't see blue then I must be blue[/s]
[s]A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue[/s]
[s]A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue[/s]
[s]A3. Therefore, if I see two blue and they leave on the second day then I must not be blue[/s]
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
[s]A3. Therefore, if I see three blue and they leave on the third day then I must not be blue[/s]
[s]A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue[/s]
B1. Green sees brown
[s]B2. Therefore, if I don't see brown then I must be brown[/s]
[s]B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown[/s]
[s]B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown[/s]
[s]B3. Therefore, if I see two brown and they leave on the second day then I must not be brown[/s]
[s]B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown[/s]
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown
Given that one of the As and one of the Bs must obtain, and given that only A4 is left of the As, blue knows on the third day that A4 must obtain and so that they are blue.
The complimentary set of arguments will have all browns deducing that they are brown on the third day.
All without green having to say anything.
These pair of premises don't make sense together. If green hasn't said anything, then the only reason you could possibly know green sees blue is precisely because you see blue.
"Therefore if I don't see blue, I also don't know that green sees blue, so therefore A1 is no longer applicable"
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 some hypothetical world in which I don't see blue (even though in reality I do see blue) that she says "I see blue" (even though in reality she doesn't say "I see blue").
And again, the proof is in the pudding; as the above shows, all blues and all browns correctly deduce their eye colour on the third day.
I can't explain this any clearer than I already have. So if you disagree then we're just going to have to agree to disagree.
If there's only 2 blues, then each blue DOESN'T have justification to think that if the one blue they see wasn't blue, green would still see blue.
There's a canonical answer, and you're disagreeing with it. You have a bit of a burden of proof here. You can cop out if you want with "agree to disagree", but it is that, a cop out.
From the start, everyone knows there is not just one person with blue eyes, so why are these perfect logicians waiting the first day?
From the start, everyone knows there is not just two people with blue eyes, so why are these perfect logicians waiting the second day?
...
That's why you have to break it down into pieces. If everything were the same, (including what the guru says), except only one person has blue eyes, how and when could he figure out his eyes are blue? And then do the same question, but two blue eyed people.
What's tripping me up is this:
If only one person has blue eyes, the guru's statement is clearly informative: the person with blue eyes doesn't see any blue eyes.
If only two people have blue eyes, the guru's statemen is clearly informative, since no one leaving rules out the 1st case for each of the two blue eyed people.
But at three people, the Guru may as well not have spoken. Everyone knows that there is at least one blue person, and everyone knows that everyone knows that there is at least one blue person. Once you move beyond two blue people, the scenario shifts, yet you are relying on the one and two blue people cases to reason about it.
But here's the trick. You've agreed with the case of two blue eyed people. Which means, unambiguously, if there were two blue eyed people, they would leave on the second day, right?
Which implies, unambiguously, if there WEREN'T only two blue eyed people, they wouldn't leave on the second day.
Right?
I mean yeah, but... but...
Why should the step "If there were one blue, they would leave on the first day" appear in the brains of perfect logicians who already knew before the guru spoke that this was not the case?
If that is not an active possibility, which it is not when blue >2, the failure of anyone to leave on the first night also provides no information.
Whereas if blue = 2, blue = 1 is still an active possibility, so its disconfirmation on the first night provides new information.
You know everyone's eye colour except your own, and everyone else knows your eye colour but not in each case their own.
So to know the colour of your own eyes you need to compare what you see to what they see. Now as each day passes with no blue eyed people leaving, the minimum number of blue eyes each blue eyed person must be seeing increases by one. [Whereas a brown or green eyed person will see one more.] So if you are seeing 2 blue-eyes and they haven't left on the second night, they must also be seeing (at least) 2 blue-eyes, which means you must have blue eyes since they cannot see their own eyes and you can see everyone's but your own.
So when the days have passed that equal the number of blue-eyes that you see, that minimum requires that you have blue eyes too, otherwise the blue-eyes would have already gone. In which case you, and of course all the others remaining must have brown, grey, violet, green, or some other colour eyes, though as it happens you know as they each don't that they all have brown eyes except the guru.
Quoting hypericin
It's a counterfactual conditional from which valid deductions can be made thus:
If wishes were horses, then beggars would ride.
But beggars do not ride, but have to walk.
Therefore wishes are not horses.
It's a valid deduction, but we already know from the outset that wishes are not horses, it tells us nothing new. Similarly, the blue would have left if b=1, but we already know b>1, so their not leaving also tells us nothing new.
We agree that if b=1 or b=2, we MUST have the guru's statement to get the ball rolling. But if b>=3, then @Michael's reasoning seems to apply. We may as well just imagine the guru making the statement, which means we may as well just imagine the guru, and this imaginary guru can make the statement about blue or brown, and so everyone would have left long ago. But if this works with b>=3, surely it works with b=1 or b=2. But it does not.
At b=3, everyone can make the guaranteed true statement, "everyone must see at least 1 blue"
At b=2 or b=1, this is not a true statement.
So at b=3, but not b=1 or b=2, anyone can say of the guru, "she could truly say, 'I see a blue'", and so anyone could say "if there were a guru, she would say, 'I see a blue'".
[s]oh wait...[/s]
No, I was right, at b=2, a guru must see a blue, but it is not true that everyone else must also reach that conclusion. But at b=3, not only must everyone know that a guru must see a blue, everyone must arrive at the conclusion that everyone else knows that a guru must see a blue.
So someone says I see blue. And now we all know that if anyone did not see blue, they would be blue, and they would know they would be blue and be gone tonight. We know that we can see 99 blues, but that doesn't change the logic, because it's counterfactual conditional.
So tomorrow, we know that everyone can see at least one blue because no one left. But if anyone could only see 1 blue, they would know that, since that one blue did not leave, they themselves must be blue too. And in that case they would both leave that night. And so as each day passes, the counterfactual argument gets augmented by "but no one left therefore everyone must see one more blue", until it gets just exactly to the number of blues (which remember no one exactly knows, because they do do not know their own colour) So after 99 days you know that all the blues are seeing 99 blues, and you are seeing 99 blues and therefore you must be the extra blue that all the other blues must be seeing - because they cannot see themselves. You can see that no one else is.
And at this point I really cannot be arsed if anyone still doesn't get it. I done my bestest and thunked hard how to explain it - Over and out.
Don't edit posts, it confuses everything. If you made a mistake just admit the mistake in your next post. I know you edited a post, and that unfairly makes this whole conversation more confusing than it has to be.
I don't even think we need to do that.
It seems to be a simple mathematical fact that for all [math]n >= 3[/math], if I see [math]n - 1[/math] people with X-coloured eyes and if they don't leave on day [math]n - 1[/math] then I have X-coloured eyes.
So not only is the green person saying "I see blue" a red herring, but the green person being there at all is a red herring.
Anyone who applies the above reasoning will correctly deduce their eye colour without anyone having to say anything, or even imagined to have said anything.
I see 24 green, 36 blue, and 4 red.
Therefore if the 4 red leave on day 4 but the 24 green don't leave on day 24 then I have green eyes.
And why n >= 3, rather than n >= 2?
Maybe also when [math]n = 2[/math].
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 the 2 blues don't leave on day 2 then he is blue, and that if the 2 greens don't leave on day 2 then he is green.
So when the other brown doesn't leave on day 1 he correctly deduces that he is brown.
And then the same each for blue and green, all deducing the correct answer.
This is the part that logically fails. Why would 1 brown leave on day 1 anyway, if guru says nothing?
There's no reason whatsoever for 1 person, brown eyed or blue eyed, to leave on day one unless the guru says something. If the guru doesn't say anyhing, then all you know is there's X blue eyed people, Y brown eyed people, Z green eyed people and absolutely no way to know your own eye color. There's no mechanism in your logic Michael. You keep on riding the coattails of unenlightened's logic, but throwing out the fundamental premise of unenlightened's logic. You're trying to have your cake and eat it too.
Without the guru saying anything, there's no mechanism whatsoever short of magic for a single blue-eyed or brown-eyed person to know what color their own eye is. You can't skip past step 1.
This brings up a related question I had thought of before: if it wasn't given in the question, I would have said, no one leaves, end of story. Even after seeing the answer, I had a hard time accepting it.
Given that cases like this exist, how do we even trust our own reasoning? I think the answer is, we can't (except maybe unenlightened!)
You just have to accept that you aren't a perfect logician. Is that so bad?
The point is, usually when things feel logically certain, we think we at least know that much. That feeling of logical certainty amounts to a kind of psychological "proof". How else do we ultimately know anything logically follows?
Here, I was tripped up by the idea that the guru can't possibly be giving new information. But, amazingly, despite that feeling, she is, no matter how many blue eyed people there are.
Of course this forum, and philosophy in general, is a quagmire of mistakes. But it is probably much worse than we suspect. If our intuitions are that uncertain, even when they feel totally certain, it seems we are always on logical quicksand.
I guess I don't feel that way about this, because this is an especially contrived scenario, deliberately built to be counter intuitive. I don't think my failure here necessarily hints at a more wide, general failure at logic or thinking.
This is a stark example, but there have definitely been others, where it felt like something clearly was one way, when it turned out to be another. Surely you have experienced this as well, that the "clearly" feeling just isn't as reliable as it feels.
I actually think that's a good thing. I mean, we already have situations where two groups of people feel clearly that the other side is wrong - having examples where most people's "clearly" feelings are off base at least forces everyone to be a little more rigorous in their reasoning than just "it feels wrong".
I mean, maybe, if everyone went through this problem, or similar, and perfectly internalized that lesson. But, they won't, and frankly we will probably forget this too, sooner or later. But the deeper quandary to me is, how can we ever really be certain? No matter how rigorous we are, or think we are, there can always be some error.
Anyways, Smullyan - The Lady and the Tiger - or any of his logic puzzles are recommended to all.
Just to follow up on this.
In the case if 2 2 2 like you laid out, from the point of view of a brown, here's what he knows:
"There's 1 brown, 2 blue, 2 green and 1 unknown -me"
Now you're saying "if 1 brown doesn't leave on day 1..." But there's no reason for 1 brown to leave on day 1. If he was the only brown eyed person, he would see 5 people with non brown eyes, and nobody is saying anything in this scenario, so... what information is this hypothetical guy supposed to have that his eyes are brown? His eyes don't have to be brown. There's no rule that says "there's at least one brown eyed person on the island". For all he knows, there could be no brown eyed people. His eyes could be green, blue, yellow, grey - anything.
This is why the canonical answer does in fact rely on the guru saying something. You need that to get the logic rolling.
He wouldn't, but that's irrelevant. It can be demonstrated that if everyone just follows the rule: for all [math]n >= 3[/math], if I see [math]n - 1[/math] people with X-coloured eyes and if they don't leave on day [math]n - 1[/math] then I have X-coloured eyes, then they will correctly deduce their eye colour (unless they have a unique eye colour).
Knowing this fact is all it takes for everyone on the island to deduce their eye colour (except those with a unique eye colour). And perfect logicians would know this fact.
So say I see 4 blue, 5 brown, and 6 green
I reason:
1. If the 4 blues don't leave on day 4 then I am blue
2. If the 5 browns don't leave on day 5 then I am brown
3. If the 6 greens don't leave on day 6 then I am green
If I did have blue eyes then the others with blue eyes would reason:
1. If the 4 blues don't leave on day 4 then I am blue
2. If the 5 browns don't leave on day 5 then I am brown
3. If the 6 greens don't leave on day 6 then I am green
And the browns would reason:
4. If the 5 blues don't leave on day 5 then I am blue
2. If the 4 browns don't leave on day 4 then I am brown
3. If the 6 greens don't leave on day 6 then I am green
And the greens would reason:
4. If the 5 blues don't leave on day 5 then I am blue
2. If the 5 browns don't leave on day 5 then I am brown
3. If the 5 green don't leave on day 5 then I am green
All 5 of us with blue eyes would deduce after day 4 that we have blue eyes and leave on day 5.
And at the same time the 5 with brown eyes would deduce after day 4 that they have brown eyes and leave on day 5.
And then finally the 6 greens would deduce that they have green eyes after day 5 and leave on day 6.
Everyone deduced the correct answer without anyone having to say anything.
But since they do not know their eye colour they might all have unique eye colours and none of them can deduce their eye colour at all. Guess and hope is not deduction.
That's why I said: for all [math]n >= 3[/math] if I see [math]n ? 1[/math] people with X-coloured eyes...
So if there are at least 3 people with X-coloured eyes and at least 3 people with Y-coloured eyes (and at least 3 people with Z-coloured eyes, etc.) then everyone can deduce their eye colour without anyone saying anything.
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 eyes but the 100 blue and 100 brown can all determine their own eye colour by the 100th day, even without the Guru saying anything.
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 in the post here.
Everyone does in fact correctly deduce their eye colour.
Although I think the particular reasoning in that post only works if nobody has a unique eye colour. If somebody does have a unique eye colour then they can apply the reasoning in my original post.
No they don't because they could have a unique colour and being, unlike you, perfect logicians they know that, and therefore do not make the fallible guess that they do not have a unique eye colour, and so none of your predicted leavings happen and you will conclude that you must have eyes of every colour.
They dont assume that they dont have a unique eye colour. Rather, they infer it based on what the others dont do. Notice that each step is a conditional. The implicit final step is if everyone else has left then I have a unique eye colour.
Just as in your example; the blues and browns dont assume that their eyes arent green or red. They figure it out.
The point I am making is they that dont need to wait for green to say anything. They already know that she sees blue and brown. If it helps they could just imagine her saying I see blue and I see brown and apply the same reasoning.
It is bizzare to suggest that our perfect logicians all know this but must wait with bated breath for her to verbally express what they already know before they can start.
They can start the moment they arrive at the island.
But if 1 wouldn't leave, then you can't correctly deduce your eye colour at n=2. You already can't.
You said it applies to n=2 as well, that's what we're talking about. Here's the context:
https://thephilosophyforum.com/discussion/comment/1002914
So he wouldn't leave on day 1 if he was the only one, and he wouldn't leave on day 1 if he wasn't the only one.
So him not leaving on day one gives me no new information, because it would have happened regardless, and I can't use that non information to leave on day 2.
That's the state of play, that's how I and unenlightened analyze this situation.
You made that point before. but you are wrong. I have already also explained before why you are wrong.
It is bizarre but it is true because the puzzle was set up like that. The act of saying it changes the situation despite giving no new information in its content.
Furthermore, the reasoning cannot work for the brown eyed, because it begins:
If there was only one brown eyed person,and someone said "I see brown eyes" that person would know they had brown eyes.
But if no one said it, as no one did in this puzzle, then that unique brown eyed person could not have any idea of their eye colour, and therefore the whole chain of reasoning could not get started, and so no brown eyed people leave. Instead, they reason along with the blue eyed except that as they see 100 blue eyes, they will wait an extra day and learn that they do not have blue eyes because all the blue-eyed have gone. But they still won't know if their own eyes are brown, green or pink.
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 brown.
Were doing it right now. This island doesnt exist and there is no real Guru saying anything. And yet we can still say if there was one brown and if the Guru says I see brown then
And just as were allowed to say this, so too are our hypothetical islanders.
So if I was to be magically transported onto this island and see 99 blue, 100 brown, and 1 green, then even though I dont know if I am blue, brown, green, or other, knowing what I know I dont have to wait for green to say anything. I know that if the 99 blue dont leave on the 99th day then I am blue, else if the 100 brown dont leave on the 100th day then I am brown, else Im either green or other.
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.
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?
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?
If the 1 blue doesnt leave on the first day then I am blue, else if the 2 brown dont 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.
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 dont need to wait for some third party to actually say this. This assumption gets everyone to the correct answer.
I don't think it "helps", I think it's an entirely irrational thing for Tommy to imagine that.
I did here.
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.
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.
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.
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.
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 just randomly appear or disappear (whether before or after the Guru says anything).
we're not talkinfg about that case though. michael can't prove it for the case of 2.
That's why I explicitly said where [math]n >= 3[/math]. There are at least some occasions where it works where [math]n = 2[/math], but I haven't claimed that it will always work where [math]n = 2[/math], 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.
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.
Seems like you're not seriously considering the possibility that you're wrong. That's a mistake.
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?
Quoting Michael
Ît does say
Quoting flannel jesus
Maybe they were literally there forever.
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.
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.
No, because this is one of those [math]n = 2[/math] 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.
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 [math]n = 2[/math], so I don't know how showing that it doesn't work for some [math]n = 2[/math] proves that it doesn't work for [math]n = 3[/math].
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.
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.
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.
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 deduce their eye colour on the 100th day even if the Guru says nothing.
When? Everything seems so vague right now. When does it work? What does Tommy have to see for it to work?
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 knows) such that if the Guru were to say "I see X", nothing new would be learned.
In the OP, that green sees blue and that green sees brown is shared knowledge that everyone knows, and that shared knowledge allows all blues and all browns to deduce their eye colour, even without green saying anything.
In your example, that green sees brown is shared knowledge that everyone knows, and that shared knowledge allows all browns to deduce their eye colour, even without green saying anything.
Above doesn't include a specific scenario where it works. A specific scenario looks like "2 blue, 2 brown, 1 green". Does it work in that scenario, if the green eyed guy says nothing? If not that, what specific scenario does it work in?
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 green were to say "I see blue" and the 1 blue doesn't leave on the first day then they are blue, else that if green were to say "I see brown" and the 2 browns don't leave on the second day then they are brown.
The browns will reason that if green were to say "I see brown" and the 1 brown doesn't leave on the first day then they are brown, else if green were to say "I see blue" and the 2 blues don't leave on the second day then they are blue.
The blues leave on the second day knowing they are blue and the browns leave on the second day knowing they are brown, all without needing green to say anything.
It allows the blues and browns to deduce their eyes colour if there are 2 blue, 2 brown, and 1 green.
So it DOES work.
So if Tommy sees 1 blue, 2 brown, 1 green, Tommy can safely go to the boat on the second day knowing his eyes are blue. Right? That's your explicit position. It's not a "n=2 that doesn't always work" scenario, this is a scenario where you're saying explicitly, the logic DOES work, Tommy CAN deduce his eye color when he sees 1 blue, 2 brown, 1 green, and he sees that the blue eyed guy doesn't leave on day one.
It's clear, explicit, no "maybe sometimes doesn't work", you agree unambiguously that you think Tommy can deduce this.
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 blues.
It only works when there is shared knowledge, as I said. So in your scenario, only the browns can deduce their eye colour because only "green sees brown" is shared knowledge.
So Tommy can deduce it?
Tommy doesn't know that everyone knows that green sees blue because if Tommy doesn't have blue eyes then the 1 blue doesn't know that green sees blues.
The reasoning only works when a) everyone knows the same thing and b) everyone knows that everyone knows the same thing. That's the essence of the problem; that everyone knows that everyone knows that green sees blue, and so her saying "I see blue" cannot possibly provide anyone with new information.
So I actually think this requires [math]n >= 4[/math].
If I see 3 blue, 3 brown, and 1 green, then everyone knows that everyone knows that green sees blue and brown, and that allows the blues and browns to deduce their own eye colour (e.g. by hypothesising some counterfactual scenario in which green says "I see blue" and there is only 1 blue, etc.). It is rational for them to do so from the moment they lock eyes with one another, even if green says nothing. Them all "seeing everyone" is all the "synchronicity" one needs.
Quoting Michael
Is that n>=4? Are you talking from the perspective of a blue eyed person? Because that's only n=4 for blue, not for brown.
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 there are 2 blues then those blues would leave on the second day
etc.
The point I am making is that I can use this reasoning even if green hasn't actually said "I see blue". I can use this reasoning from the moment I lock eyes with everyone and come to know that everyone knows that everyone knows that green sees blue.
From the moment we lock eyes we can all just pretend that green has said "I see blue", and act as if she did, even if she doesn't.
Quoting flannel jesus
I might be blue, I might be brown, I might be green, or I might be other. What matters is that I see [math]n - 1[/math] (i.e. 3) of a particular colour.
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 runners on a track waiting for the gun to fire, is wrong. I can start right away, and just assume that she said so if it helps.
Okay, so... you wait for day 4, you think "the blues would have all left on day 3 if there were only 3", and then you leave on day 4 as a blue with all the other blues?
Doesn't that rely on the logic working for n=3?
Quoting Michael
If you think the logic works for n=4, but the logic relies on a premise that pretty much explicitly says "the logic also works for n=3", then... n=4 can only work if n=3 works, no? And n=3 doesn't work. We both agree n=3 doesn't work. So if n=3 doesn't work, can you rationally say "the blues would have all left on day 3 if there were only 3"? That IS what n=3 is saying. I don't think you can say "the blues would have all left on day 3 if there were only 3". We can't just freely accept that as a deductively valid premise to use in your logic.
True, perhaps its not as simple as defining some particular [math]n[/math].
Not that I think it matters to my argument. It is still the case that in the OP it is rational for all blues and all browns to counterfactually assume that green has said I see blue even if she hasnt, and counterfactually assume that there is only 1 blue, even though there are more, and then counterfactually assume that there are 2 blues, even though there are more, and so on, eventually deducing that if the blues I see havent left by a particular day then I must be blue, with comparable reasoning for brown. This will allow all blues and all browns to leave on the 100th day, knowing their eye colour.
If you need some kind of synchronicity then locking eyes with everyone else will suffice (a premise in the experiment). We dont need anyone to say anything, whether that be a blue saying begin! or green saying I see blue.
There's a pattern here. We proved that it doesn't work for n=2, and because of that, you immediately accepted that it doesn't work for n=3.
For some reason, you didn't apply that to n=4 - for the exact same reason you can reject n=3 if n=2 doesn't work, you can reject n=4 if n=3 doesn't work. And if you can reject n=4, you can reject n=5. And so on. Up to 100.
Your confidence should be shaken. You were so confident about n=2, to the point of even saying "I've already explained it, I can't make it any more simple" - and then we proved you wrong. And you agree that you were wrong about n=2. So... don't you think maybe you should be a little less confident about n=100?
I'm not saying you should immediately reject n=100, but maybe accept that you have a serious burden of proof there, because if n=2 doesn't work, neither does 3, and so on, right?
So don't be so sure. Do the logic. Work it out. Don't just state a conclusion and call it a day, this is a deduction puzzle. I want to see you deduce.
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 there were 1 blue then that blue would leave on day 1.
P2. If green says "I see blue" and if there were 2 blues then those blues would leave on day 2.
P3. If green says "I see blue" and if there were 3 blues then those blues would leave on day 3.
...
PX. If green says "I see blue" and if there were X blues then those blues would leave on day X.
PX+1. Therefore, if the X blues I see don't leave on day X then I am blue.
What you don't seem to understand is that we don't need to wait for green to say "I see blue" to start this reasoning. We can start this reasoning as soon as we lock eyes with each other, or as soon as someone says "Begin!" or as soon as green says "I see brown".
Notice that you allow for everyone to assume P1 as part of their reasoning even though everyone knows that there is more than 1 blue. It is a counterfactual. And for the exact same reason we are allowed to assume P1 as part of our reasoning even if green hasn't yet said "I see blue".
Shouldn't it work at n=3 then? But we've both agreed it doesn't.
At this point, I'm just answering the question in the OP. All the blues and all the browns can, and will, correctly deduce their eye colour 100 days after they lock eyes even if green says nothing, as they can use the reasoning above and counterfactually assume that green has said "I see blue" and "I see brown" even if she hasn't.
I said before that I wasn't going to try to explain this again, as this is as simple as I can explain it. I'm actually going to commit to that promise now. There's nothing more I can add to what I've already said.
You telling me something doesn't make it true. If the logic works, there's got to be a point at which it works. If there's no n at which the logic works, then... it doesn't work.
The last time you said it was simple, you said that about something you were wrong about. Your overconfidence is...weird. I mean it was weird to begin with, but to use the same overconfident line again, after you know the last time you said it you were wrong, is like... double weird. Why are you doing that? Don't you think that's weird? You've been consistently wrong, I don't know where your arrogance comes from.
If you're right, you should be able to prove it. This is a case of pure deductive reasoning, and I'm quite frankly enjoying it. If you want to throw in the towel, fine, I just... don't even know why you bothered to say anything if you don't want to even try to prove it.
It doesn't allow any such deduction. Knowing is not the same as saying, and I think we agree that if someone has a unique colour, they cannot deduce it.
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.
If the argument begins with "everyone can see that there are multiple blue and brown but no one says anything." What is the next step?
For him the next step is just imagining someone says something.
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.
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.
Quoting Michael
You can know that too. but you cannot apply it to your situation because no one has said anything.
So you can only get to "if 99 days have passed and no one has left and someone had said I see blue then I would know my eyes are blue."
But no one spoke so you don't know.
Ok, so one person knows all the other people on the island, and knows their eye color. The only one they don't know is themselves.
Quoting flannel jesus
Ok. So I'm assuming that this is taking place when there is this number of people on the island. This is VERY important as if the elder can speak at any time and the number of people would be different, then this is a different story. In fact, if they can change from this number, this initial number never should be mentioned and this needs to be restated more carefully.
Quoting flannel jesus
Ok, so no one knows what color eyes another has, and apparently we can make up colors of eyes that people don't have like purple. Are you sure this is just not limited to green, blue, or brown?
Quoting flannel jesus
What is "The islanders?" Because you mentioned earlier that there were 100 with blue eyes and 100 with green eyes. "Islanders" implies "All of them". Is this poorly worded? Is it "At least one islander?"
As the initial premise is written, this is the set up.
A. There are 100 blue eyes, 100 brown eyes, and one green eyed elder.
B. However, the islanders do not know that this is the limit of eye color, and their eye colors could be any color under the rainbow. They also don't know the actual number. So even if they see 100 blue eyed individuals, they're own eye color could be blue or anything else.
C. The elder is speaking to all 200 other people on the island, and we're assuming he sees all 200 people, and says, "I see someone with blue eyes".
The only uncertainty that isn't listed here is how many people the elder saw while speaking to everyone. If its more than one person, there is no one who could know their eye color, as they don't know the total of blue eyes on the island. Therefore the only logical conclusion I can think of if we know someone does figure out their eye color, is if the elder is speaking to everyone, but one person is the only one being seen, and understands they are the only one being seen. That person could figure it out then and leave that night.
I'm not wrong because I didn't say "only the 100 of us".
Quoting 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.
Out of all the replies I didn't expect this one. Any good logic problem needs to be broken down and poked at carefully. If I appear confused while breaking down your logic problem, perhaps your logic problem isn't very straight forward and needs some refinement? Feel free to correct my points if I have misread anything.
Your paragraph here shows you were pretty adept at getting over most of your self-inflicted confusions. You're right about the setup.
The elder saw all of them and was looking at everyone when she said it. Not any one person. Even while saying it, she knew and could see 100 blue eyed people.
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.
What you are doing is inserting 'we all know we can all see blue' in to substitute for "x says 'I see blue'"
It doesn't work, precisely because this is the counterfactual situation in which the speaking is absolutely necessary because the hypothetical solitary blue does not see blue and has to be told in order to deduce their eye colour. This produces a contradiction that the hypothetical solitary blue cannot but does see blue, and cannot but does know their own eye colour.
If there was only one blue, that blue would not know there were multiple blues or any blues.
Therefore?
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.
Quoting flannel jesus
And when you say, "All of them" do you mean the 100 blue eyes and 100 brown eyed individuals on the island? Or could this be a variable number like there only being 1 besides the elder at the point the elder speaks?
This puzzle isn't trying to trick you with wording. The most natural interpretation of that bit of the text is that ALL of the islanders are there, and that's how you should interpret it.
"Standing before the islanders" - no need to try to think of clever alternate ways of interpreting it, at face value "the islanders" means all of them.
Thank you. Then logically taking only the information given, no one would be able to leave the island. All the elder has confirmed is that blue is a color of eye that at least someone has. Of course, everyone already knew that. It didn't need to be the elder that stated it, it could be anyone. "Someone" in the logical sense means "at least one".
At least, this is assuming there is no other outside information that is needed to know about eye color etc. Taking the problem verbatim with no outside knowledge needed, its impossible for anyone to determine the necessary logical conclusion about their own eye color. If you're concerned about posting the answer here, feel free to give me a direct message. I promise I won't reveal the answer.
I'm not sure about this.
If we take as a premise that "everyone sees at least one blue", then the counterfactual still works: If there is one blue, he would leave on day one. As you pointed out, that the counterfactual is false is irrelevant.
What if the sage had said instead, "I see at least two blues"?
Imagine instead that of the 200 people the guru was speaking to, 199 of them had brown eyes and 1 had blue eyes. The guru says "I see someone with blue eyes". What happens next? Can anybody leave then?
If there were only one blue, then it WOULDN'T be true that everyone sees at least one blue.
You agreed with me earlier here:
Quoting Philosophim
Are you now saying this was incorrect and that the number of people with different eye color could be different when the elder finally speaks?
A = Only one Blue
B = Everyone sees one blue
C = Blue leaves on first night
B
A -> ~B
A -> C
Still valid.
If there wer eonly one blue, it wouldn't be true that everyone sees one blue. Right? Do you understand why that is?
I'm asking you to clarify the rules. Imagining something that isn't in the rules is pointless if I'm unsure of the rules. Please clarify the rules as I noted, then I will gladly imagine it.
I said in my summary that there are at the time of the elder speaking, 100 blue eyed, 100 brown eyed, and 1 green eyed elder. You said that summary was correct. You are now presenting a scenario in which there could be a different balance of eye color. Are you saying that the balance of eye color could be any variety at the time the elder speaks?
Ok, so I'm going to assume that YES, its always 201 people, but that the eye color can vary at any one time. In that case, the solution is trivial. Obviously if the eye color can vary, then in the case where one person could see everyone else did not have blue eyes, they would know they have blue eyes. Flannel, your logic puzzle needs another pass on clarity. People not understanding the rules of your puzzle isn't a puzzle, that's just confusing.
I didn't invent this logic puzzle. This isn't mine. This was invented by a smart guy, and many smart people did the puzzle and liked it. People aren't confused, you're confused.
So, now imagine this:
2 blue eyed people, 198 brown eyes. Guru says "I see someone with blue eyes". What do you think happens then?
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.
If this answer is something like, "Everyone will turn and stare at the two blue eyed people, I'm going to be angry. That's not a logic puzzle, that's a riddle. Logically it is not definite that people will turn around and all stare at the blue-eyed people at the same time, as the blue eyed people would need to be looking at both blue eyed and green eyed people to see who's staring at who.
Logic puzzles leave no room for human error or uncertainty.
so you're inventing nonsense to be confused about, and now you're inventing stuff to be angry at.
Try to use logic and think about it. Let me know if you want the answer to this scenario with 2 blue eyed people.
No, I'm done if you won't confirm that I had the rules right at this point. I just feel like you're trolling. If you want me to keep playing, please confirm my understanding that there are at the time the elder is speaking, 100 blue eyes, 100 brown eyes, and 1 green eyed elder.
There were no "rules" about how many people can be on the island. It's an island. However many people you want are on the island. I laid out the scenario and told you how many people of each eye color ARE on the island. That's not a rule, that's just a fact. I don't know why you want it to be a "rule" - seems like something you're just actively confusing yourself about.
There are 100 blue eyed, 100 brown eyed, 1 green eyed. That's not a "rule", that's just the scenario.
Those people can't see their own eye color, so they don't know that's the scenario. A blue eyed person thinks it could be the case that there are 99 blue eyed, 101 brown eyed, 1 green eyed, or maybe his eyes are green instead of brown, or maybe his eyes are amber.
This is a logic puzzle, every detail is an important rule. If you misunderstand or don't think critically about everything, you're going to miss out. Thank you, I'll think about your scenario again. I find it odd that you mention the day they would leave.
"They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island"
So, in the case of 2 blue eyed people, let's get into one of their heads. I'm going to be thinking as the blue-eyed person right now:
I see 1 blue eyed person and 198 brown eyed people and 1 green eyed guru. The guru just said she sees someone with blue eyes.
either (a) the blue eyed person I see is the ONLY blue eyed person,
or (b) my eyes are blue as well, and we both have blue eyes..
If (a) is true, the blue eyed guy I see will leave the first night.
So that means (a) isn't true, and (b) must be true, so I can catch the ferry on the second night.
Ok, I concede. You are unteachable.
Alright, let me tackle 3 now as I think I see what you're getting at. From my perspective, I would see 2 blue eyed people. I don't know if I have blue or green eyes. But if I had green eyes, I would see one more blue-eyed person than someone with blue eyes. So after the second day, if no one left, from the blue eyes perspective people, they would know that the other two blue eyes see one more blue eyed person. So by day 3, all blue eyed people will leave.
I'm assuming this pattern continues up to day 100. The reason why blue can do this is because green eyed people will be doing the same calculus, but one day behind blue. Ok, that's pretty cool!
Yes!
Quoting Philosophim
Will they?
Yeah, I follow, there is definitely a case to be made. This puzzle has been confusing the fuck out of me. The core problem is, I think you understand, at what point is (1)?
n=3: no, every blue thinks it could be 2
n=4: no, every blue thinks it could be 3
n=5: no, every blue thinks it could be 4
...
Hey no editing.
For (1) to be false, blue A must see blue B , and know that B sees blue C, but not know that B knows that C sees a blue.
This doesn't seem possible when n=100.
Actually Michael still keeps green:
Quoting Michael
So for this to be false, we must find some blue that can find some blue that they aren't sure knows green sees a blue.
How will you do this when n=100?
It's gotta be something like 3 or 4 right?
Ok, and I might be wrong on this, but I'll put my logic out.
Once all the blue eyed people leave, then everyone else sees just brown eyed people and the elder. Meaning that essentially the elder just said, "I see someone with brown eyes". And at this point, I think I was wrong on it taking another 100 days. If you see 99 people with the same eye color and they don't leave, they all are uncertain of their own eye color. But, since all 99 don't leave the next day after blue eyes leave, that's because they each brown eyed person realizes 'I must have brown eyes, otherwise they all would have left'. So its day 101, that none of the brown eyes people leave, then they leave on day 102. Finally day 103 the elder would leave as no one is left on the island.
Why would they have?
Let me think through it again. The elder has green, first person view doesn't know if they have green, brown, or some other color eyes.
After all the blue eyes leave, everyone sees that there are only brown and one green eyed person. No one would leave the first day. But what do we learn from that? The elder learns they do not have brown eyes, nor blue eyes, but, something I missed, they'll never know they have green eyes so can't ever leave the island.
Since the elder doesn't leave, the brown eyes who don't know their own color know they don't have blue or green eyes. So I guess this leaves the idea that they could have some unknown color like red. At this point it means there is only one eye color that is uncertain, each brown eyed person doesn't know if they have brown eyes. So perhaps I was right the first time and they would simply follow the logic that the blue eyes people did. I would think further, but I have to go to work. Again, fun puzzle. :)
How do they learn that? The elder could easily have brown eyes, as far as she's concerned.
Quoting Philosophim
I think so too. I wanted to spark some debates.
This is the beating heart of the puzzle, if you can't answer this you don't understand the puzzle. It is not to synchronize, not to make the counterfactual work.
It is to make sure, not that everybody knows everybody sees a blue, but that everybody knows everybody knows everybody knows..., n-1 times, that everybody sees a blue.
N=0 nobody is a blue.
N=1 not everybody sees a blue.
N=2 everybody sees a blue, everybody does not know everybody sees a blue.
N=3 everybody knows everybody sees a blue, everybody does not know everybody knows everybody sees a blue.
N=4 everybody knows everybody knows everybody sees a blue, everybody does not know everybody knows everybody knows everybody sees a blue.
And so on.
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.
at three blue, everyone knows everyone sees a blue
at four blue, everyone knows everyone knows everyone sees a blue. But, at this stage, you can add as many "everyone knows" as you want, I think. Can't you? At four blue, everyone knows * infinity that everyone knows that everyone sees a blue.
I think
Or maybe at 5?
Idk I'm lost
Because the elder sees that everyone left has brown eyes. Its the same as one person who doesn't know if they have blue eyes looking at other blue eyes, though perhaps it would take 101 days to figure it out.
Number blues A B C...
At n=3, A doesn't know B knows C sees a blue.
At n=4, A doesn't know B knows C knows D sees a blue.
And so on.
Of course, every permutation of these are true as well.
Like imagine I know something. Maybe you know I know it, maybe you don't - that doesn't explode.
Now imagine I know something and you know I know it. Now, that also doesn't explode - maybe you know I know, but I don't know you know I know.
Now imagine I know you know I know.
And then imagine you know I know you know I know.
If I know the fact, and you know I know, and I know you know I know, and you know I know that, then... at that point, can't we realistically add as many "I know you know"s as we want and it still remain deductively true, assuming we're perfect logicians and both know each other are perfect logicians?
That's my intuition. I could be wrong.
Even in the case of two people, they are all distinct facts, all the way up. It's just that we lose our ability to mentally grasp them pretty fast.
Yeah I think that's probably true. I changed my mind
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.
.You are missing the recursion.
Once your list is fully expanded, #1 knows a fact, call it A:
A: everybody knows that everybody knows guru can see blue.
Ok, #1 knows A. But then #1 realizes, everyone has to know A to proceed, not just me. Otherwise we cannot act in concert.
So, really #1 must establish a meta-fact, B:
B: everyone knows A.
So #1 goes though a longer expansion, and proves B. But then #1 realizes, wait, if I have to know B to proceed, everyone has to know it. So now really he has to establish
C: everyone knows B.
...
And so on.
Im not. Im explicitly saying that I dont think it needs to be recursive.
If you are missing the need, you are missing it.
His logic for 100 relies on the assumption 99 would leave on day 99.
And that in turn relies on the assumption that 98 would leave on day 98.
And you can continue to trace that back, all the way down to:
Logic for 6 relies on 5.
Logic for 5 relies on 4.
Logic for 4 relies on 3.
Logic for 3 relies on 2.
And we KNOW 2 doesn't work if the guru says nothing. Even Michael agrees with that.
If 2 doesn't work, 3 doesn't work. If 3 doesn't work, 4 doesn't work. Trace that all the way back up to 99, then 100.
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 ...
If there were two people with blue eyes, and nothing was said, each would know that there was at least one person with blue eyes but would have no idea of their own eye colour. So neither would leave on any day. This is because they would know that the person with blue eyes that they could see could not know their eye colour any more than they knew their own.
If there were 3 people with blue eyes, and nothing was said, each would see 2 people with blue eyes, but they would have no idea what their own eye colour was, even though they knew everyone knew that at least 1 person had blue eyes, but the would still also know that no one had any way to know the colour of their own eyes, and so no one would leave.
If there were 4 people with blue eyes, and nothing was said, each would see 3 people with blue eyes, but they would have no idea what their own eye colour was, even though they knew everyone knew that at least 2 people had blue eyes, but they would still also know that no one had any way to know the colour of their own eyes, and so no one would leave.
If there were 5 people with blue eyes, and nothing was said, each would see 4 people with blue eyes, but they would have no idea what their own eye colour was, even though they knew everyone knew that at least 3 people had blue eyes, but they would still also know that no one had any way to know the colour of their own eyes, and so no one would leave.
Can anyone see a pattern emerging? The non leaving of the counterfactual solitary person entails the non leaving of any number of people, because nothing ever tells anyone their own eye colour
My last substantive reply he only replied to the first sentence, I suspected he didn't really read it
In his defense, that's a pretty good reason to ignore someone
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 ...
It's worse than your amended 2. It recurses endlessly.
There is a fundamental problem. Whatever condition you think is sufficient ((1) in your case), , everybody must know it, not just you. But not in the omniscient sense, you, the islander, must know everybody knows it. But if you prove that, that new thing you now know is an additional fact that everyone must know, and you have to know they know it, and everybody must know that, and ..
This is the logic being discussed, right?
But we already have a simple, straight-forward case that this logic doesn't work. We know, because he's already acknolwedged, that 2-blue-eyed doesn't work. 2 blue-eyed people cannot leave on the second day.
If it's true that 2 blue-eyed people cannot leave on the second day, then it must also be true that 3 blue-eyed people cannot deduce that there's more than 2 blue-eyed people just because they don't leave on the second day. So 3 blue-eyed people cannot leave on the third day.
But premise 1, "everyone has come to know, through some means or another, that everyone knows that #101 sees blue", is true in the case of 3 -- and yet it still doesn't work.
So we have a tangible, specific case where Michael should be able to apply this logic, and yet can't.
It genuinely feels like these simple cases, for low numbers of blue-eyed people, are being ignored because it's easier to hide the reasoning behind the obscurity and confusion of very large numbers. The beauty of unenlightend's logic is that it clearly unambiguously works for small numbers, and so we can work our way up to large numbers. In contrast, Michael's logic, we know for sure doesn't work for small numbers, so instead of working his way up to large numbers, he just kinda ignores the problems at small numbers and hopes nobody notices the gaps in logic once there's 100 people to talk about. It's easier to hide the cracks with so many blue-eyed people to think about.
If Michael wasn't so worried about getting tortured for eternity, I'd be encouraging him to find the lowest number of blue-eyed people that it works for. Michael it's only a fictional torturing for eternity.
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?
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. It's hard to explain, it's unintuitive, but if you work through the logic from the ground up, it's nevertheless true. For some reason, it makes a difference.
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.)
I explain it in the first part of the post above:
"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.
Just to recap, We've already agreed that it does make a difference for the case of one blue eyed person, and two, and three. There must be some number where it starts making a difference. I'm very interested in that number. You want me to accept that it starts making a difference some time before 100 - if I'm going to accept that, I'm gonna need you to show me when.
For every number of blue eyed people x, your reasoning seems to rely on the premise that if there were x-1 blue eyed people, they leave in x-1 days. You're obscuring your logic by jumping to 100 blue eyed people. I'm trying to explore with you the numbers that aren't obscured.
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.
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.
That's not my premise.
So if the 99 you see leave on the 99th day, on the 100th day you'll conclude you have blue eyes anyway?
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.
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 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.}
I was imagining myself as one of the blues though, putting myself in the place of BL1. That's what I was going for
No, I'll conclude that I don't have blue eyes.
Quoting 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.
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.
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.
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.
No, nobody is going to leave on day 98 because nobody sees only 97 blue.
Otherwise, how could it be true that "if there were only 99 they would leave on day 99"?
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.
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.
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.
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.
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.
Quoting unenlightened
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.
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).
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.
(1) doesn't say "nobody has told me anything".
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.
Then it should say '...and someone has said "I see blue"' because otherwise it is contradictory.
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.
You are flailing. If you are magic and a mind reader then bla bla blah, anything you like. 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. So piss or get off the pot, you can't have it both ways.
You seemed to understand why that means it can't work for 3 people, so if there are only 3 eyed blue people, we also know they won't leave on the 3rd day.
Do you see why that means it can't work for 4 blue eyed people, why they can't leave on the 4th day?
If you are patient and take this seriously, I'm pretty sure you'll find what I have to say compelling. But we gotta start small.
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 wed be surprised and dumbfounded if anyone left earlier than that.
Given that we can dismiss the first counterfactual situation outright, it doesnt 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.
1. I know x
2. Everyone knows x
3. I know everyone knows x
4. Everyone knows everyone knows x.
5. I know everyone knows everyone knows x
6. Everyone knows everyone knows everyone knows x
...
You claim that at some number in this series, they stop being distinct facts, so that the next number is the same fact as the previous?
Everyone can see 99 blues in the n=100 case. But this is not the same as the guru speaking, or everyone seeing on a piece of paper "there is one blue". With only visual evidence there can always be cases where 1 believes 2 believes 3 believes... 99 believes there is no blue. Only communication can collapse this chain.
It is not purely performative. If it was I'm sure the perfect logicians could find a way to skip it.
After the guru speaks, everybody knows everybody knows ... there is at least one blue
After the first day, everybody knows everybody knows... There is at least two blues.
And so on
Nobody will ever come any closer to agreement as long as we focus on numbers we can't even completely imagine.
That's what I think, and I have given a fairly strong argument for it, which you have ignored. I have seen no argument from you to show otherwise, and no reference to such an argument, whereas I have given a reference to a supporting argument and widely accepted solution. But carry on incorrigible.
Probably. But the interesting part to me is exploring different aspects and arguments. it's pretty rich, for a logic puzzle!
Like I tried to meet him where he is, at 100, and it took him a long time to come around to the idea that his logic for leaving on day 100 relies on it being true that if there were only 99, they'd leave on day 99. But eventually, I got him to see that, I think.
And so then I said, so surely in turn it's true that "if there were only 99, they'd leave on day 99" relies on it also being true that "if there were only 98, they'd leave on day 98". For some reason that just doesn't compute for him. Applying the same logic he's applying to n100, to n99... that's where I lose him.
He's so ultra focused in on 100 that he refuses to look at any of the surrounding logic.
Seems like he just wants to conclude that his logic works, not look at it, not have it be questioned, end of story. Which is fine but like... keep it to yourself then lol. That's not much fun for the rest of us.
What do you think it explains about the forums?
Yes, I still believe that.
Well the initial state problem is true because as the opening post describes that they would not be able to count the amount of individuals in groups of different eye colors.
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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] 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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] 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.
You're skipping steps again. Usually you skip up - you go from some low number, get tired of thinking about that, and skip all the way up to 100. Now you're doing the opposite - you're going from 100 straight down to 1.
Don't. Skip.
Be patient, take it one step at a time
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.
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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] 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.
Surely your logic involves the following statements at some level, implicitly or explicitly:
If there are only 99, they'll leave on day 99.
If there are 100, the 99 I see won't leave on day 99.
No?
No it doesn't.
If you can't accept that Argument A is valid then we can't continue.
It's not my conclusion. It's one of my premises. And it's a premise that I demonstrated to be true here.
if the n
people I see with X
eyes don't leave on day n
so that means, surely, that it's completely agreeable when I point out that your logic relies on this also being true:
If there are only 99, they'll leave on day 99.
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 [math]m[/math] people with brown eyes and [math]m[/math] people with blue eyes then it is valid when there are [math]m - 1[/math] people with brown eyes and [math]m - 1[/math] people with blue eyes.
The number of each colour makes a difference.
Because that's what this is about at root. There are only 2 possibilities from the perspective of a blue eyed person: either there are m-1 blue eyed people, or m. He's trying to deduce which world has in.
If he's waiting to see if m-1 people maybe don't leave in m-1 days, but it turns out to be FALSE that m-1 people would leave in m-1 days, then waiting for that doesn't tell him what world he's in. He could be in a world where m-1 people have blue eyes, or m people have blue eyes.
These guys don't want to get tortured for eternity. They can't rely on iffy reasoning. They have to be SURE. No guessing allowed. Only deductions.
So if it's at all possible that m-1 people WOULDN'T leave in m-1 days, then we absolutely cannot then say, "oh well I didn't see m-1 people leave in m-1 days, so therefore there must be m blue eyed little"
So if m is 100, each blue eyed person sees 99 blue eyed people and they, as perfect logicians (not perfect planners, not perfect committers to rules), have to ask themselves, can I really be sure 99 people would leave in 99 days? If they're anything less than deductively sure, they can't leave in 100 days.
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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] 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
It's not even an assumption. This blue eyed person doesn't just immediately start assuming everyone has committed to the rule.
What is it?
It's a premise in the argument.
I'm not saying that they are. I'm simply saying that the argument is valid.
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.
I'm not assuming anything about anyone. I am simply saying that Argument A is valid.
It is.
Quoting flannel jesus
The premises aren't arbitrary. [math]n[/math] is the number of people seen with [math]X[/math] eyes. Adding or subtracting some arbitrary number to or from [math]n[/math] 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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] 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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] eyes" then everyone will leave the island having correctly declared their eye colour.
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"
Yes, but adding or subtracting some arbitrary number to or from [math]n[/math] is arbitrary, whereas [math]n[/math] isn't arbitrary.
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
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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] eyes" then everyone will leave the island having correctly declared their eye colour.
Yes but you don't know that every person will do that. Therein lies the problem
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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] 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.
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 [math]n[/math] is arbitrary, and so perfect logicians wouldn't do it. But [math]n[/math] isn't arbitrary.
They could subtract 95, but that would be arbitrary and so they wouldn't do it. Perfect logicians would stick to the non-arbitrary [math]n[/math].
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 [math]n[/math] people I see with [math]X[/math] eyes don't leave on day [math]n[/math] then I will leave on day [math]n + 1[/math] and declare that I have [math]X[/math] 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.
I'm leaning towards thinking it's not correct but it's only a lean.
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.
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.
But this aspect is less important than the previous one. It really matters when it starts working
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.
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.
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.
What's the logical deduction?
99+A=100
A=1
Apparently, you can't comprehend what you wrote, so I've bolded the relevant information below. Everything else is a red herring.
So, if you have blue eyes, you'll only be able to see 99 people with blue eyes, 100 people with brown eyes, and 1 person with green eyes. This means that you must have blue eyes in order for there to be 100 people with blue eyes.
But, if you have brown eyes, you'll only be able to see 99 people with brown eyes, 100 people with blue eyes, and 1 person with green eyes. This means that you must have brown eyes in order for there to be 100 people with brown eyes.
Since everyone on the island is logical, everyone is able to come up with the same conclusion. Therefore, everyone knows what color eyes they have.
The answer to this riddle isn't hard.
Just because it's true that there's 100 blue eyed people doesn't mean any individual blue eyed person knows there's 100 blue eyed people. They don't have that fact available to them. There could be 101 brown eyed people for all they know.
You can't use information available to us, from outside the island, as if it's necessarily available to them.
A blue-eyed person knows there are either 99 or 100 blue-eyed people.
A brown-eyed person knows there are either 100 or 101 blue-eyed people.
No-one knows the number of eye-clolours, but they do know it's either 2 (the ones they can see), or 3 (if their own isn't among the ones they can see). Therefore:
A blue-eyed person knows there's:
(a) 99 blue-eyed people and 101 brown-eyed people (and their eye-colour is brown)
(b) 100 blue-eyed people and 100 brown-eyed people (and their eye-colour is blue)
(c) 99 blue-eyed people, 100 brown-eyed people, and themselves (with a unique, unknown eye-colour)
If (a) and (b) were the only options, Michael's rule would work from a logical point of view, but (c) messes up things, here. We know that [# of blue]+[# of brown]=200; they don't: [# of blue]+[# of brown] could be 199.
So what changes when the Guru tells them something they already know?
That's the problem. There's a wedge between a logical sequence and an empirical reality I find hard to reconcile:
See, the catch is this: If an islander sees no blue-eyed person, then all other islanders see exactly one person with blue eyes. So all of the logic here is counterfactual: you don't really have to go see if someone leaves; you know nobody will. But at the same time, this logic spins forward until Day 99 or Day 100, depending on wether the person is blue-eyed or brown-eyed (unbeknowest to themselves), and then it's supposed to work in the world we live in. How?
It's the event on Day 100 of all blue-eyed people leaving that tips the brown-eyed people off that they're not blue-eyed (but not that they're brown-eyed). All that hinges on the fact that systematically brown-eyed people see more blue-eyed people than blue-eyed people do, and thus their set-off point is later. And their set-off point is only later, because the Guru talked about blue-eyed people.
The important fact seems to me this:
Every Islander knows that blue-eyed people see one fewer blue eyed person than non-blue eyed people.
If there had been 100 blue-eyed people, 90 brown-eyed people, and 10 green-eyed people, they'd still have had to wait 99 days before making the decision, because both brown-eyed and green-eyed people form the relevant group of people whose eyes aren't blue.
If there had been 20 blue-eyed people and 180, the game would be over much sooner if blue-eyed people were "seen" by the guru, or much later if brow-eyed people were "seen".
It's obvious to me, matemathically, that the announcement matters. I just don't know how to interpret this in pragmatical terms. It's baffling.
That's not the catch, it's the hook on which the whole thing hangs. If the guru says he sees blue eyes but I see no blue eyes then I must be the blue eyed one and I leave that night.
But if I see 1 and only 1 person with blue eyes and they do not leave that night, then they too must see blue eyes and since I only see him, I must be the other that he can see with blue eyes, So the next night we will both know we are blue eyed and leave.
But if I see 2 and only 2 people with blue eyes and they do not leave the 2nd night, again there must be another blue eyed person that is me, and they will be reasoning the same way and so we will all leave together on the 3rd night.
etc.
And so everyone, whatever colour their eyes (because no one knows their own eye colour), is waiting to see if after n nights (where n is the number of blue eyed people they see) the blue eyed people leave, and if they don't, they can conclude they also have blue eyes, and if they do then they conclude they have eyes of some other colour.
Yes, that's all perfectly clear to me. What's not clear to me, for example, is why they can't skip forward to day n. I know they can't, but it makes no sense other than in purely logical terms. That's what I find so nuts about this riddle. There's a rift between logic and experience here I don't know how to bridge.
n is not the same for everyone. So there's no skipping to n possible because we have to all reach every step of the argument together. I'm waiting for my nth day, and you're waiting for yours and we don't know yet if they are the same day or not. If one of us has blue eyes and the other doesn't, our n is different and we find out on the nth day of the person with blue eyes when that one of us leaves, along with all the other blue eyed folk. And the other is no longer waiting because there are no blue eyes left and no more argument to be made and their n was never reached.
Ah, yes, of course. I missed that (thought of it in another context, but somehow didn't make the connection on the practical front). Thanks.
If we follow it through, then if I'm an islander with red eyes, I will still erroneously conclude on day 100 that I have blue eyes and get thrown off the ferry.
based on what?
I meant the bit about when all the brown eyed people realize they have brown eyes...but on checking, that wasn't from the OP, that was from someone's solution.
You're right that in the OP as stated, the blue eyed people all leave on day 100 or whatever, and no-one else.
1. The guru spoke "I can see someone who has blue eyes" when speaking in front of the islanders. The crowd I imagine has a mixture of blue and brown eyes. So, what is the point of the guru's comment? The guru spoke only once in many years and this is the sentence?
2. Can the islanders not know by counting how many islanders present and how many blue eyes and brown eyes? ( I get it that each one of them will end up counting 99 and 100) But is it just us who know this fact?
3. In what context is the "on what night" the islanders leave? Do we respond, the first, the second, the third, and so on?
That's half the puzzle.
Quoting L'éléphant
They don't know their own eye colour. They can count everyone's colours except their own, but counting 99 and 100 doesn't tell them their own. They can't just assume it's an even split.
Quoting L'éléphant
You could say, "the islanders don't rely on the guru saying anything, everyone leaves the island on the third night" or "the islanders leave the island on the second night after the guru speaks." The thing that happens every night, once a night, is that the ferry comes. Maybe they can't figure it out the first night, but somehow waiting a day gives them extra information
From the point of view of any given blue-eyed person on the island they must be either the 101st brown-eyed person, the 100th blue-eyed person, or neither blue nor brown-eyed. From the point of view of any given brown-eyed person, they must be either the 100th blue-eyed or 101st brown-eyed person, or neither blue nor brown-eyed.
If any given islander realizes that it is actually a 100/100 split brown/blue (the guru not being included in that count) they will deduce that they must be either the 100th blue-eyed person or 100th brown-eyed person because they see 99 people with the eye color that corresponds to their own; they must be the hundredth for everything to add up. Therefore, everyone but the guru would leave the island on the first night.
I will now show why this is the ultimate outcome:
We can assume that everyone deduces everything in the first paragraph of this solution and thus they can check their own possible deductions/considerations against everything the other islanders could deduce. Any given brown-eyed person must consider that:
- They could be the 101st blue-eyed person
- They could have neither blue nor brown eyes
- They are the hundredth brown-eyed person
Any given blue-eyed person must consider that:
- They could be the hundredth blue-eyed person
- They are neither blue nor brown-eyed
- They are the 101st brown-eyed person
From here we check each possible deduction/consideration against the other: a given brown-eyed person cannot correctly reason themselves to be the 101st blue-eyed person if a blue-eyed person reasons that they are the hundredth blue-eyed person because, given the guru is not blue-eyed, that would add up to 202 people on the island. The same goes for the reverse. So those possibilities can be thrown out.
Next: if there were two or more islanders that had neither blue nor brown eyes, then there would have to be 98 or less people with either brown or blue eyes instead of 99 (other than the guru), and any islander could see that that is not the case.
We are then left with the possibility of a brown eyed-person reasoning that they are the 101st blue-eyed person or a blue-eyed person reasoning they are the 101st brown-eyed person while the other has neither blue nor brown eyes. Any given islander can see that this is clearly not the case because they are not seeing anyone with eyes that are not brown or blue (other than the guru).
Thus, we are left only with the possibility of it being a 100/100 split between brown and blue, and, deducing this, the islanders all leave on the first night and the guru stays behind. I guess forever.
Is it really that crappy of a solution?
Like what? Maybe I can explain it. If you are confused about the discussion of possible considerations/deductions being measured against each other, it comes from this:
Quoting ToothyMaw
I don't get this paragraph. There's a green eyed person, and everyone who doesn't have green eyes sees her.
I'm only considering the reasoning of brown or blue-eyed people about potentially blue-eyed or brown-eyed people.
As such, what I'm saying there is that there would be a group of islanders that would be part of the whole but would also not have brown or blue eyes, which would mean there being less than 99 of either blue or brown (disregarding the guru). Any given brown-eyed person or blue-eyed person would see this is not true and rule out the corresponding possibility that there are at least two (relevant) islanders with non-brown or blue eyes. The guru doesn't really have to factor into this part, although I understand your concern. I could have phrased it better.
Alright. If a brown-eyed islander reasons that it is true that they have neither brown nor blue eyes, and a blue-eyed islander also reasons in parallel that they have neither brown nor blue eyes, then from the point of view of a brown-eyed islander, there would be 98 brown-eyed islanders and one with non-blue or brown eyes and from the point of view of a blue-eyed islander there would be 98 blue-eyed islanders and one with non-blue or brown eyes. We know that this cannot be the case, however, because in the problem it is stipulated that both blue-eyed and brown-eyed islanders know that there are at least 99 islanders of each eye color.
It appears that that is the one possibility I left out, of course. And I doubt I could account for it with the approach I took. Whatever.
[hide]
Let's call the the islanders W, X, Y, Z, and they'll all be male (for grammar simplicity).
[/hide]
If I counted 100 brown eyes, and I guessed that I am brown-eyed, then the probability of me being a brown-eyed is 101/200, which is roughly the same as the probability of blue-eyed.
But if I guessed that I was neither blue-eyed or brown-eyed, then the probability is 1/200. Which amounts to a very small chance that I was neither blue-eyed or brown-eyed.
In conclusion, I would pick that my eyes were blue if I counted 99 blue eyes.
Because if my eyes were red, then that would put me in a unique position as the guru.
The guru is the only one with a unique eye color.
Though your explanation of the rules of this puzzle.
The islanders can know the number (sans himself) the number of blue, brown, and green eyes. Isn't it?
I mean hell, your probabilities don't even make basic sense. They add up to more than 100%. There's a more than 100% probability that this person's eyes are blue, brown or green by your given probabilities. How is that possible?
You said that there are 100 blue eyed and 100 brown eyed yet the green eyed sees 101 brown and 99 blue? Are there 200 or 201 people?
All the shaman has to say is "if you have counted 100 people that have blue eyes leave, you have brown eyes".
"If you have counted 99 people you have blue eyes, stay for the tonight after party."
How I solved it:
This is basically X=C-A where C is brown eyed and A is blue eyed, so to solve it we require just C and A. Knowing that logisticians counted each other except themselves we get X=C-A where X= colour of your own eyes and C-A is the other people. So its two simple equations(or one?).
Nobody asked you what the shaman has to say though. I told you what the shaman says. You've solved a question that isn't being asked.
Quoting flannel jesus
Guru=Shaman, you are being overly pendantic, yes a shaman is not a guru.
Right at the end:
Quoting flannel jesus
I googled the puzzle to understand it so now I know the answer and the puzzle is ruined for me, I cant answer.
I think there are two factors:
1. The level of indirection. In daily life, you might sometimes think on a level of "I know, that you know, that he knows" but a couple of levels of indirection like that is usually sufficient for most things. It feels weird to reason that "He will reason that, he will reason that, he will reason that...x99"...we aren't used to it
2. The premise of everyone knowing everyone else is perfectly logical. Pretty easy to say, but pretty alien in practice. We rarely have the privilege of knowing how others will reason.
Some would say that the key thing is that it seems like pointing out someone has blue eyes isn't adding new information of course. And that's true, but I think that flows from (1) and (2) above. Most people can figure out that the logic works in the 2 or 3 villagers scenarios, even though for those cases everyone already sees at least one person with blue eyes.
So I'd say it's more a symptom of the confusion, rather than the cause.