Logicians: What (if anything) is wrong with this argument? Blue-eyed islanders

[clairobscur]

@muttrox

It’s inductive reasoning. As nivlac said, there’ve been dozens of good explanations posted.

If there were less than 2 blue eyed people, then they would have killed themselves, since they could have reasoned it through. Since that didn’t happen, there are 2 or more blue-eyed people. And so on up the chain… Each day that goes by without suicides eliminates possibilities until the only remaining possibility is that it’s time for you to end your miserable existence on the Blue-Eyed Island of The Damned.

I understood that. I’m just saying (like another poster above. I had not noticed his post previously) that the reasonning followed in this thread is flawed. It works for 1 or 2 blue eyed people, but I think your “and so up the chain” is not correct. For some reason , I can’t wrap my mind about the case of 3 blue eyed people. But if there are at least 4 of them, the reasonning doesn’t work.
Because if there are say, 5 blue eyed people, any islander knows for sure there is at least 4 blue eyed people (excluding him, if he has blue eyes, since he doesn’t know his own eyes color) , and also knows for sure that every other islander knows there are at least 3 blue eyed people, because every one of them can see at least three of the 4 b.e. islander he himself can see.

And since all islanders know that all other islanders know there are at least 3 blue eyed people, nobody can assume that someone else will believe that someone else believe there’s zero (or 1 or 2) blue eyed islander.

So, the beginning of the reasonning is correct :

A thinks there might be only 4 b.e. islander (5 minus him, since he doesn’t know he’s blue-eyed)

A thinks that B might think there are only 3 of them (5 minus him because he doesn’t know he’s blue eyed, and B because B doesn’t know his own eye color, either).
But going further is not possible :

A and B both know that everybody can see at least 3 blue eyed people. So, they know that nobody can believe there are less than 3.

Therefore, A cannot think that B might think that C might think there are only 2 of them.
As a result, the oracle statement doesn’t add any information about what other islanders could think about what yet other islanders could think. The “information” provided by the oracle (“there are at least one blue eyed person here”) not only is already known by everybody, but also everybody knows that everybody else already knows it.

Hence, the oracle statement won’t change anything. Nobody is going to jump in the volcano.

[davenportavenger]

I’ve only skimmed the thread, so I apologize if this has been answered already, but how would the problem work out if everyone on the island had blue eyes? I would think that in that case, there would be no suicides, since it would be implicit that there was at least one brown-eyed person on the island, and everyone (looking around and seeing that every other person was blue-eyed) would incorrectly assume they were the only safe one. Or would it work out in some other fashion?

[Nightime]

More generally, assuming N blue eyed people, stating that A believes that B believes that C believes …etc…that X believes that there’s 0 blue eyed people, the mistake is to ignore the fact that all know that nobody can believe that there are less than N-2 blue eyed people…

Is my reasonning flawed? If so, how?

Actual   #1 thinks  #1 thinks   #1 thinks     #1 thinks
                             #2 thinks    #2 thinks     #2 thinks
                                                #3 thinks     #3 thinks
                                                                    #4 thinks

1 be s s s s

2 be be s s s

3 be be be s s

4 be be be be s

As someone else said, the best way to be convinced of this is to try it yourself. Say there are 4 BE islanders.
Fill in the blank: #1 thinks #2 thinks #3 thinks #4 thinks there are ____ BE.

[Nightime]

Ignore that failed attempt at a table.

[Chronos]

And since all islanders know that all other islanders know there are at least 3 blue eyed people, nobody can assume that someone else will believe that someone else believe there’s zero (or 1 or 2) blue eyed islander.

OK, using the case of five blue-eyed folks, myself one of them. The following statements are true:
A: There are at least five blues.
B: I know there are at least four blues.
C: I know that everybody else knows that there are at least three blues.
D: I know that everybody knows that everybody else knows there are at least two blues.
E: I know that everybody knows that everybody knows that everybody knows there is at least one blue.
F: I know that everybody knows that everybody knows that everybody knows that everybody knows there are at least zero blues.

Now, the Oracle speaks. Statements A-E remain unchanged. But Statement F (which doesn’t really say much) is now strengthened: I now have F prime:
F’: I know that everybody knows that everybody knows that everybody knows that everybody knows there is at least one blue. I also gain statements G’, H’, etc., all of which end with “… there is at least one blue.”, but I don’t need those.

Now comes the first designated killing time, and nobody dies. Now, it’s universal knowledge that there are at least two blues. It’s just as if, on the second day, the Oracle said to all “There are at least two blues”. We now can replace E with E prime:
E’: I know that everybody knows that everybody knows that everybody knows there are at least two blues.
We can also replace F’, G’, H’, etc. with double-primed statements, ending with “… there are at least two blues.”, but we don’t need those any more.

Comes the second killing-time, and it’s now universal knowledge that there are at least 3. We replace statement D with D’:
D’: I know that everybody knows that everybody else knows there are at least three blues.
Again, we could also ammend all of the statements after D, but we don’t need to.

3rd night, it’s universal knowledge that there are at least 4, and we have C’:
C’: I know that everybody else knows that there are at least four blues.
And finally, after the fourth designated killing time, we have B’:
B’: I know there are at least five blues.
And therefore I know I’m one of them.
Ack.

[clairobscur]

@Nightime

As someone else said, the best way to be convinced of this is to try it yourself. Say there are 4 BE islanders.
Fill in the blank: #1 thinks #2 thinks #3 thinks #4 thinks there are ____ BE.

But once again, you’re ignoring a critical piece of information known by all islanders : the fact that if there are N blue eyed islanders, everybody :

-can see at least N-1 bei (blue eyed islanders)

-knows that everybody else can see at least N-2 bei

-knows that everybody else knows all the others can see at least N-3 bei (I was mistaken above when I stated N-2).
Let’s take again the example with 5 bei

A is a bei. he assume he isn’t, and think there are (or at least there could be) only 4 bei.

He then wonders what the beis might think. He assume they themselves believe they aren’t bei. So, he assumes that one of them, B, think there are only 3 beis.

No, he wonders what B thinks C could think. He assume B is following the same reasonning as him, hence that B thinks C thinks there are only 2 beis.

Can be go further, following the same reasonning, and reducing the number of potential bei by one at each step?

No. Because A can see there are at least 4 beis . Hence he knows that B, C D and E (the other beis) can see at least 3 beis. Hence he knows that B, C, D and E knows that the three others can see at least 2 beis.

As a consequence, A knows that B can’t think that any of the other beis believe there are less than 2 beis. A knows that B knows that all the others can see at least 2.
The mistake is to believe that the reasonning, which is correct as the beginning, can be followed until you reach 0. It’s only correct until you reach N-3.
If it’s unclear, assume you’re A. You happen to be with all the 4 others beis. You can assume you’re not a filthy bei yourself, hence that there are only 4 beis. You can also assume that each of them believe the same secretly. So, in your mind, they believe there are only 3 beis (actually, they believe like you that there are 4, but you don’t know that). Can one of them follow the same reasonning and believe that someone else inthe group believe there are only two? Sure.

But can you go on. Nope. Because you know each of them see 3 beis (according to you…actually, they see 4). So he knows that each of the others can see 2.
Get a mental picture of the situation, and you should realize that everybody knows all the others can see at least 3 beis, hence everybody knows that the others can see at least 2. So, nobody can believe that any other could assume there are less than 2. A knows that B knows that C, D and E can see 2 bei.

I essentially repeated twice the same thing, but I asssumed that a mental image of a group of 5 blue eyed people could help understand why it’s impossible for A to believe that B believes that C…etc…When you reach N-3 the reasonning isn’t true any more.
I maintain that nobody will ever jump in the volcano.

[Nightime]

You didn’t answer the question though.
Let’s make n = 5 this time. Fill in the blank:
#5 thinks #4 thinks #3 thinks #2 thinks #1 thinks there are ____ BE.

You should be able to fill in the blank even if you think the logic is flawed - you would just have a number other than zero.

[clairobscur]

@Chronos

OK, using the case of five blue-eyed folks, myself one of them. The following statements are true:

Let’s see :

A: There are at least five blues.

True. We’re assuming there are five.

B: I know there are at least four blues.

True. You can see four of them

C: I know that everybody else knows that there are at least three blues.

True. You know that they can see at least three.

D: I know that everybody knows that everybody else knows there are at least two blues.

True. You know that everybody else knows that everybody else can see at least two.

E: I know that everybody knows that everybody knows that everybody knows there is at least one blue.

BUZZZ!!! False. As you yourself stated in (D) everybody knows that everybody else knows there are at least two.

(Well…technically true, since if they know there are at least two, it means also that there is at least 1, 1/2 or 0. But you know what I mean.)

Since, you established in (D) that everybody knows that everybody knows there are at least two blues, everybody knows that nobody can believe there are less than two.

F: I know that everybody knows that everybody knows that everybody knows that everybody knows there are at least zero blues.

False, as above (or more exactly, as mentionned, pointless).

Now, the Oracle speaks. Statements A-E remain unchanged. But Statement F (which doesn’t really say much) is now strengthened: I now have F prime:
F’: I know that everybody knows that everybody knows that everybody knows that everybody knows there is at least one blue. I also gain statements G’, H’, etc., all of which end with “… there is at least one blue.”, but I don’t need those.

Once again, pointless. Everybody already knew that everybody knew there were at least two blues. The new “information” doesn’t add anything. Adding others “everybody knows” besides the two first ones is meaningless. Knowing that everybody knows that everybody knows there are two blues is exactly the same as knowing that everybody knows that everybody knows that everybody knows that everybody knows, etc…(a dozen times) that there are two blues.

What is important is that everyknow knows (you’re one of everybody) that everybody else knows there are at least two blues. Hence that everybody knows that nobody can believe there are less than two.

Now comes the first designated killing time, and nobody dies. Now, it’s universal knowledge that there are at least two blues.

It was already universal knowledge before the oracle came, as you established in (C). And more importantly, everybody knew it was universal knowledge, as you established in (D).

It’s just as if, on the second day, the Oracle said to all “There are at least two blues”. We now can replace E with E prime:
E’: I know that everybody knows that everybody knows that everybody knows there are at least two blues.

The oracle wouldn’t need to say that. It was already known. Once again, adding several “everybody knows” in a row doesn’t mean anything. Once it’s established that everybody knows that everybody knows, adding another “every knows” doesn’t change the meaning of the sentence.

Comes the second killing-time, and it’s now universal knowledge that there are at least 3. We replace statement D with D’:
D’: I know that everybody knows that everybody else knows there are at least three blues.
Again, we could also ammend all of the statements after D, but we don’t need to.

3rd night, it’s universal knowledge that there are at least 4, and we have C’:
C’: I know that everybody else knows that there are at least four blues.
And finally, after the fourth designated killing time, we have B’:
B’: I know there are at least five blues.
And therefore I know I’m one of them.
Ack.

Nope. You don’t. You don’t because since everybody knew that everybody knew that there was at least two blues since the beginning, the reasonning you’re following is false.

You’re reasonning about all blue jumping in the volcano on day N (N= number of blues) only hold true if you can believe that an islander could believe that another islander could believe that another…etc… believe that there were no blue.

It’s true for instance if there’s only one islander. On the first day, he jumps in the volcano. He believed there were no blue, he learnt that there was one, it can only be him, so he jumps.

It’s true also if there are two blues. Noticing that the other one didn’t jump, both understand the other isn’t the only blue, as they believed , hence both jump on the second day.
But it’s ** not true ** for 5 blues, since, as once again you established yourself, with five blues, everybody knows that everybody knows there are several blues since the beginning. Nobody can believe that someone else can believe there are no blue, or only one blue.

As a result, when the oracle speaks, nobody jump because everybody assume the oracle is refering to one of the four other blue they know about (as you established in B). And nobody wonders on the following day, either, why a blue didn’t jump, because they all knew that each blue knew there were at least two other blues (as you established in D) . So, everybody assume (correctly) that each blue believed the oracle was refering to the other blues.
The oracle brought no new information allowing anybody to determine his status as blue, and as a result nobody jump in the volcano ** except ** if there are only one, two, or three blues, in which case, they indeed jump all together on the first, second or third day.

With at least four blue, everybody is safe.

[clairobscur]

@Nightime

You didn’t answer the question though.
Let’s make n = 5 this time. Fill in the blank:
#5 thinks #4 thinks #3 thinks #2 thinks #1 thinks there are ____ BE.

You should be able to fill in the blank even if you think the logic is flawed - you would just have a number other than zero.

Yes I can :
#5 thinks #4 thinks #3 thinks #2 thinks #1 thinks there are ** at least 2 ** BE
Because, as I (and ** Chronos **, actually) explained, everybody (including #5) knows that everybody else (including #4, #3, #2, #1, they are interchangeable, anyway, there are no difference between their individual reasonnings) knows there are at least two blues.

You could actually more simply write that 5 thinks that 4, 3, 2 and 1 all think the other 3 think there are at least two blues. Because there’s no particular reason to add that many X thinks that Y thinks. Actually, I believe that putting it that way is the reason why you reach an erroneous conclusion.

[Nightime]

#5 thinks #4 thinks #3 thinks #2 thinks #1 thinks there are at least 2 BE

Ahh… but which 2 BE?
Which BE might 5 think 4 thinks 3 thinks 2 thinks 1 could see?

[clairobscur]

While I’m at it, I’m going to explain what happens when the oracle speaks. I’m going to pretend I’m A, a blue, but anyway, all the blues (B, C, etc…) will follow the same reasonning. I’m also going to assume that someone not knowing he’s blue assume he’s not (actually, being perfectly logic, he doesn’t know, but it allow me to write “he assumes”, it will be simpler and doesn’t change a thing).

1st scenario) There’s only one blue, A. I believe there’s no blue on the island. When the oracle speaks, I understand that since I can’t see any other other blue, I must a blue. So I jump in the volcano on this very day.
2nd scenario) There are two blues, A and B. Not knowing I’m blue, I believe B is the only one. So, when the oracle speaks, I expect him to follow the reasonning above and jump in the volcano. But he doesn’t. The only possible explanation is that he believed the oracle was refering to someone else besides him (or to several other people) . Since I can’t see any other blue besides him, this “someone else” is necessarily me. B follows the same reasonning. So, on the second day, we both jump together in the volcano.
3rd scenario) There are three blues, A, B and C. I believe there are only two, B and C. So, I expect them to follow the reasonning above. They’re are both going to assume there’s only one blue, the other one, and upon realizing their mistake, they’ll jump in the volcano on the second day. But they don’t. The only possible explanation is that B (for instance) could see two other blues (at least) and was expecting these two blues to jump. Like me. One of these two other blues is C, but who is the other? Since besides B and C, I can’t notice any other blue, it must be me. B and C follow the exact same reasonning, and on the third day we all jump in the volcano together.
4th scenario) There are 4 blues, A, B, C and D. I believe there are only three : B, C and D. I know that B, C and D, seeing only the two other ones, each assume there are only two. B believe (In my opinion, actually, he can see that i’m also blue) only C and D are blues, C believe only B and D are and D believes only B and C are. Further, I assume that for instance B (not knowing he’s blue) is going to believe that for instance C believes D is the only blue. But also that D believes C is the only blue.

And similarily, in my opinion, C believing there are only two blues, B and D, is going to believe that B believes D is the only blue and D believes that B is the only blue.

It’s becoming a little more complicated than in the previous example, but the point is that :

-all of us believe there are only three blues, the three others.

-all of us believe that the three others only see two blues.

-all of us believe that the three others assume that the two blues they can see both think there’s only one blue, the other one. Only one blue, but there’s one, and it’s the important point.

Then the oracle comes and speaks. He’s not bringing any new information to me, in this case. But also, I know that he’s not bringing any information to the others, either. I already knew that everybody knew that everybody knew there was at least one blue in the tribe. Not only I’ve no reason to jump (as in the first scenario), not only I’ve no reason to expect somebody else to jump (as in the seond scenario), but also I’ve no reason to believe that somebody else will expect someone to jump (as in the third scenario). I’ve no way to deduce, from the statement of the oracle, or from the behavior of the other tribesmen (not jumping) that I’m myself a blue. The other blues can’t, either.

Since all of us blues know that all the others know there’s at least one blue who is not himself, nobody expect anybody to jump or to believe that somebody else will jump. So, nobody does, ever, and we revel in our blissful ignorance.

The oracle didn’t bring any new information in the system, in this case, so it stays unchanged.

[Chronos]

BUZZZ!!! False. As you yourself stated in (D) everybody knows that everybody else knows there are at least two.

Statement D is true, and everybody knows it’s true. However, not everybody knows that everybody knows that statement D is true.

Do you see what’s happening with your argument, incidentally? First, you said that it only worked down to N-2. Now, you’re seeing another layer, and saying it works to N-3. Exactly the same sort of layering argument continues to work indefinitely.

[nivlac]

@continuity eror

I’ve only skimmed the thread, so I apologize if this has been answered already, but how would the problem work out if everyone on the island had blue eyes? I would think that in that case, there would be no suicides, since it would be implicit that there was at least one brown-eyed person on the island, and everyone (looking around and seeing that every other person was blue-eyed) would incorrectly assume they were the only safe one. Or would it work out in some other fashion?

If there were N>1 people on the island and they ALL have blue eyes, then the mass suicide will occur right after the dawn of day N+1. On that day, each blue eye person will first notice that the other N-1 people did not off themselves as dawn came. Then they will all realize that they all have blue eyes and it’s bye bye time. Note that this problem assumes that everyone can reason logically and presumably at the same rate.

[Nightime]

And similarily, in my opinion, C believing there are only two blues, B and D, is going to believe that B believes D is the only blue and D believes that B is the only blue.

But what follows from “B believes D is the only blue”?
It means that “B believes D can see zero blues”.
They are actually equivalent - if you believe someone is the only blue, then you believe they can see zero blues.

[clairobscur]

@Nightime

But what follows from “B believes D is the only blue”?

It means that “B believes D can see zero blues”.

They are actually equivalent - if you believe someone is the only blue, then you believe they can see zero blues.

It’s 3 a.m. here, and my brain melted. I think I’m going to sleep on this and I’ll come back tomorrow.

My issue with your last posts is that indeed I can’t wrap my mind about what A believes that B believes that C believes…etc… without coming at the same conclusion as you :

I (A, for instance) believe that someone believe…etc…that someone believe there’s zero blue.

That would be fine and dandy, however, there a contradiction, here.

Despite whatever A could believe B believes about the beliefs of C about the beliefs of D etc…, nevertheless, A knows for a fact that B, C and D all knows for a fact that there are two blues (C and D for B, B and D for C, B and D for C). Hence, A also knows for a fact that each of them also know for a fact that the others know for a fact there’s at least one blue.

If it’s unclear, A knows that B can see C and D are blue. Hence that B knows that both C and D know there’s at least one blue (C for D and D for C).

And it works for any combination of A, B, C and D. A also knows that C knows both B and D know there’s at least one blue. C knows that B knows that both A and D know there is at least one blue, etc…

IOW, everybody necessarily knows that everybody knows there’s at least one blue.

Which is quite obvious if you imagine the situation. Four blue eyed people are put together. Each of them can assume he’s not blue. He can also assume that the others, individually, don’t know they are blue, either. But still, he knows that everybody can see at least two blues, hence that everybody knows with certainty that everybody else can see at least one blue.

This part, I assume, is not disputable.

So, there must be a flaw in the reasonning “A believes that B believes that C believes…etc…” because nobody can at the same time :

-believes that someone could construct a “chain of belief” resulting in zero blue.

-knows for a fact that everybody knows everybody sees at least one blue.

A for instance knows that B knows both C and D see one blue, that C knows both B and D see one blue, that D knows both B and C sees one blue. And of course, B knows the same about what C knows about A and D, and so on…

There’s a contradiction here. There are two things I assume I should think about :

-The flaw being related to the difference between the “beliefs about beliefs” and the actually known facts. For instance, when A believes that B believes that C believes…A also knows that these beliefs are erroneous. He knows that, regardless what B could think, C actually see two blues, B and D. And all of them similarily ascribe to the others beliefs that they know are erroneous. A mistakenly believes that B mistakenly believes that C…etc…

-The flaw has something to do with the “at least” part.

As for the comment about me adding a layer, from N-2 to N-3, no, I don’t believe it can on indefinitely. Just that I was mistaken the first time. Because, once again, once you reach N-3, everybody knows that everybody knows there’s at least one blue eyed islander. In other word, at this point, the oracle doesn’t bring any new information, nor directly (like in “I didn’t know there is at least a blue”) nor indirectly (like in “someone else could not have known that everybody knows there is at least one blue”).

Now, I might be mistaken. But if one of you think so, I would want you to point out exactly where is my mistake.

[Chronos]

Despite whatever A could believe B believes about the beliefs of C about the beliefs of D etc…, nevertheless, A knows for a fact that B, C and D all knows for a fact that there are two blues (C and D for B, B and D for C, B and D for C). Hence, A also knows for a fact that each of them also know for a fact that the others know for a fact there’s at least one blue.

Aha! I think we have the mistake, here! It is, in fact, the case that A knows all the same information that B has. But, A doesn’t know that that’s a fact, unless he knows that he and B both have the same eye color. But if A and B have different eye color (as A so fervently hopes), then B won’t have the same information as A. In other words, B has the same information as A (since they both have the blues), but A thinks that B has different information.

[wilsta]

Not sure if this has already been stated but for my money the answer is that there must be only one person with blue eyes and on the next sunrise after the announcement they will commit suicide.

If there were more than one person they would have already known and already committed suicide.

[nivlac]

@wilsta

Not sure if this has already been stated but for my money the answer is that there must be only one person with blue eyes and on the next sunrise after the announcement they will commit suicide.

If there were more than one person they would have already known and already committed suicide.

Not true at all. Just read the posts that discussed the case where there are only two blue eyes.

[muttrox]

@clairobscur

But going further is not possible :

A and B both know that everybody can see at least 3 blue eyed people. So, they know that nobody can believe there are less than 3.

Therefore, A cannot think that B might think that C might think there are only 2 of them.

Consider what A thinks that B thinks about A, rather than what A thinks that B thinks about C. More interesting stuff there.

---

You said you accept the two person case. Let’s start there. As of day two, no one has killed themselves. There would have been suicides if there were exactly two blue-eyed people, therefore, any remaining villagers know there must be at least a third blue.

Next question A would ask is: are there exactly three blue-eyed people, or are there more? Well, I can see 4 others. So I know there is more. Question answered.

At the third day though, something else happens. Or rather fails to happen, no one kills themselves. Had there been only three blue-eyed people, they would have killed themselves. After all, if you were a blue-eyed villager and only saw two blues, and knew there more than two blues, you could absolutely know that you yourself had blue eyes. So when A sees that no one has killed themselves on day 3, he knows that everyone knows there is at least 4 blue-eyed people.

A doesn’t know whether there are 4 or 5 blue-eyed people. And he knows that everyone on the island knows there are 4 or 5 blue-eyed people. So let’s take the two cases.

There are 4: In this case, A must be brown-eyed, and everyone else is blue-eyed. So those blue-eyed people only see 3 blue-eyed people. They can all reason out that there are 4 blue-eyed people, the exact same way that A did. So if they see 3 blue-eyeds, but know there must be more than 3 blue-eyeds, then obviously the missing blue-eyed person is themselves. Therefore, on the Day 4, they will each commit suicide, and dawn will find Person A sitting by themselves.

There are 5 (the actual case you asked about):
Each villager goes through the identical reason as above. However, on Day 4, nothing happens. Therefore, each villager can additionally conclude that there are more than 4 blue-eyed people. If there were 4 or less, there would have been suicides by now. Since there weren’t, there must be a fifth person with blue eyes. Our villager A has no choice but to commit suicide.

---

OK, you say you understand all that. But, you feel the Oracle has added no information. I claim the Oracle has added two pieces of information.

It provides the knowledge that you know that they know that you know that they know bla bla bla. It’s not what A and B know about C, it’s what A knows about himself based on B’s actions or lack of actions.

It also provides the clock, so when Day 4 hits (for example), you can confidently make the deduction that there are at least 4 blue-eyed (not everyone seems to agree with me on that last info, though it seems self-evident to me)