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

Factual Questions
81

One blue-eyed islander, Day one. All the browns see the blue, assume the one blue is the “at least one” and their eyes are brown. The single blue sees no blue, and assumes the blue eyes he can’t see are his own.

Two Bs, Day one. All the browns see the two blues and assume their eyes are brown. The two blues see each other’s blue eyes and assume they other blue is the “at least one” and that their eyes are brown.
Day two. The blues see each other and realize the reason the other blue didn’t commit suicide is because because they saw one other blue… ergo, there must be “at least two” blue-eyes. Since they can see only one, the other set of blue eyes must be their own.

Three Bs, Day one. All the browns see the three blues and assume their eyes are brown. The blues see the other two blues, assume their eyes are brown, and think “won’t they be in for a surprise tomorrow.”
Day two. The blue see the other two blues, assume their eyes are brown, and think “well, now they know.”
Day three. The blues see the other two blues and realize that they still assume their eyes are brown because the other two blues can ALSO see two blues… ergo, there must be “at least three” blue eyes. Since they can see only two, the other set of blue eyes must be their own.

Four Bs, Day one. All the browns see the four blues and assume their eyes are brown. All the blues see the other three blues, assume their own eyes are brown, and think “Ha, those poor blues, they think the other blues are going to be in for a surprise tomorrow, but really they’ll be in for a surprise in two days.”
Day two. All the blues see the other three blues, assume their eyes are brown, and think "Ha, now those blues are each thinking the other two just figured out there are ‘at least two’’
Day three. All the blues see the other three blues, assume their eyes are brown, and think “There, now those blues know there are ‘at least three’”
Day four. All three blues see the other three blues, realize that all of the other blues must ALSO see three blues… ergo, there must be “at least four” blue eyes. Since they can see only three, the other set of blue eyes must be their own.

Et cet-er-a! Et cet-er-a! Et cet-er-a!

For N blues, each blue see N-1 blues, assumes their own eyes are brown, and every other blue sees N-2 blues. Since, by the above, everyone knows the value of N is figured out by day N, all the browns expect the blues to bump themselves off by day N. However, the blues expect the other blues to bump themselves by day N-1. On day N, when the blues see that the other blues haven’t bumped themselves off, they realize that it’s because all the other blues saw N-1 blues, not N-2. Since the other blues saw N-1 blues, each blue realizes there are really N blues, not N-1 blues. Since each blue can only see N-1 other blues, they other blue must be them, so they bump themselves off. On day N+1 all the browns see the blues are gone and rest secure in the knowledge that their eyes are brown.

82

Imagine if there is only one guy with blue eyes. And he’s so stupid he doesn’t get the point. Then everyone will kill themselves the second night!

It’s a big responsiblity to carry for the guy or group who has blue eyes!

83

Here’s what I’ve got.

Suppose N=1, BlueA.

BlueA thinks N=0. When he learns that N>=1, BlueA realizes that since he can see that no one else has blue eyes, he must be the blue-eyed one. He commits suicide on day 1.

Suppose N=2, BlueA and BlueB.

BlueA and BlueB both think N=1. BlueA thinks BlueB thinks N=0, BlueB thinks BlueA thinks N=0. BlueA thinks BlueB will commit suicide on the first day and BlueB thinks BlueA will commit suicide on the first day, given the scenario above. Neither commits suicide on the first day. After the first day, both will realize that N cannot equal 1, since otherwise that person would have killed themselves. Since they can see that there is already 1 set of blue eyes, there must be another set. Since they can see that no one else has blue eyes, they each realize that they must have the blue eyes. They both kill themselves the second day.

Suppose N=3, BlueA, BlueB and BlueC.

BlueA, BlueB, and BlueC think N=2. BlueA thinks BlueB and BlueC think N=1, BlueB thinks BlueA and BlueC think N=1, BlueC thinks BlueA and BlueB think N=1. On day 1 BlueA, BlueB, and BlueC realize that no one will kill themselves on day 1, since only if someone previously thought N=0 but now knows N=1 will anyone kill themselves. Since everyone thinks everyone thought N>=1, no suicides happen or are expected the first day. However, on the second day, BlueA will expect BlueB and BlueC to kill themselves, given the scenario above for N=2, but BlueA still doesn’t know he has blue eyes, so BlueA will not kill himself on day 2. Likewise, BlueB isn’t going to kill himself, instead he expects BlueA and BlueC to kill themselves, and likewise BlueC expects BlueA and BlueB to kill themselves. So, no one will kill themselves on the second day. Since each Blue’s expectation is not fulfilled, they each must assume there is some missing information. The only possible explanation is that there is another islander with blue eyes. Since each can see that every other islander has brown eyes, the only possible person with blue eyes is themselves. So…on the third day BlueA, BlueB and BlueC kill themselves.

Hmmm. I thought by working this out I would demonstrate that only if N=1 would there be any suicides. Then I realized that if N=2 there would also be suicides, but thought it would have to stop there. But I just showed that N=3 also works.

But the trouble is that for N=3, the oracle didn’t give any additional information. Every brown-eyed islander thought N=3 and thought the blues thought N=2. The blues thought N=2 and thought the other blues thought N=1. Since every islander already knows that N>=1, and knows that every other islander thinks N>=1, the fact that the oracle announced that N>=1 cannot be new information. So, where does the information come from?

84

There’s more layers. The blues thought that the other blues thought N=1. But the blues also thought that the other blues thought that the other blues thought that N=0. It’s at that deeper level that the new information is introduced. I would advise against trying to think about this in English, since one quickly gets bogged down in counting how many thoughts deep you are.

85

Wait Chronos, that can’t be true.

The browns think N=N, and think the blues think N=N-1
The blues think N=N-1, and think the other blues think N=N-2.

However, the blues DON’T think the other blues think the other blues think N=N-3, or N=N-N. They can see all the blues the other blues can see, except themselves. They should never think anyone thinks N=0. The least number for N that anyone can expect anyone to think is N=N-2. That is, personA accurately thinks a blue-eyed personB won’t count their own blue eyes, and (if personA has blue eyes) they mistakenly think that a blue-eyed personB won’t count personA’s blue eyes, although personB will.

The most anyone can actually be in error about N is by 1, and that is only if they are blue eyed. The most error anyone can have in their belief about someone else’s belief about N is 2. There is no daisy-chain where someone’s belief about someone’s belief about someone’s belief can be wrong by more than 2, since every islander can see every other islander’s eyes, and every islander knows that every islander can see every other islander’s eyes but not their own.

The only possible error islanderA can make about N is about their own eye color. The only possible error islanderA can have about islanderB’s belief about N is that that islanderA doesn’t know what islanderB’s belief is about islanderA’s eye color. And islanderA’s belief about islanderC’s belief is the same as islanderA’s belief about islanderB’s belief. IslanderA accurately believes islanderB does not count islanderB’s own blue eyes, but IslanderA mistakenly believes islanderB does not count islanderB’s blue eyes. But islanderA always believes accurately about islanderC’s belief about islanderB’s eyes, and vice versa.

The only mistakes possible are islanderA’s belief about islanderA’s own eyes, and islanderA’s belief about any other islander’s belief about islanderA’s eyes.

86

Three blue eyed people A, B, and C

Before the oracle speaks.

A thinks B sees one blue eyed person.
A thinks C sees one blue eyed person.
B thinks A sees one blue eyed person.
B thinks C sees one blue eyed person.
C thinks A sees one blue eyed person.
C thinks B sees one blue eyed person.

I’m sure we agree on that.

Addtionally:

A thinks B thinks C sees no blue eyed people.
A thinks C thinks B sees no blue eyed people.
B thinks A thinks C sees no blue eyed people.
B thinks C thinks A sees no blue eyed people.
C thinks A thinks B sees no blue eyed people.
C thinks B thinks A sees no blue eyed people.

It is important to note that no one thinks anyone sees no blue eyed people. This is where the new imformation comes in.

87

Each of the blues think N is N-1 and think all the other blues think N is N-2. Also, the blues think that the other blues that think N is N-2, think that all the other blues think than N is N-3. And the blues that the blues think are thinking that N-3? Well! They think the other blues think that N is N-4!

However, once you get past a certain number of blues, every blue realizes that the blue to which N>=1 is new information is entirely hypothetical. There is no such blue. No one expects any blue to commit suicide the first night, so the act of no blue commiting suicide provides no new information to any real blue.

88

I have one remaining question that has yet to be discussed. Does anyone know if this can be translated into first-order logic or if some other logic is needed?

The reason that first-order logic seems inadequate to me is that being able to express knowing what another person knows seems to require that predicates be able to take as arguments what other predicates return. This is not valid in first-order syntax, as far as I know.

89

The logic of knowledge is something else above and beyond sentential logic, and is a very active area of research right now. This book seems like a good place to start.

90

As I said previously, the starting of the clock is new information. Previous to the Oracle’s proclamation, there was no way to get inductive reasoning started, there were just some brown and blue-eyed people living together. You couldn’t reason out that since people didn’t kill themselves that must mean something.

The Oracle speaking provides a timepoint zero that is known by every villager, and known to be known by every villager. Therefore, logical reasoning can proceed from that point forward.

91

It says on the next sunrise , not during the night or any other time.

Why couldn’t the island be located at such a place on the Earth that the sun never fully rises or sets?

92

The new information is not “the starting of the clock.” The new information is that (everybody knows)[sup]n[/sup] that n is not 0.

What the hell does that mean?

For n=2 with A and B the two blue eyed islanders.

I’m going to explain A’s assumptions only, but B has a nearly identical set of assumptions.

A sees 1 blue eyed person, so A assumes n=1.
Since A assumes A has brown eyes, A assumes that B sees no blue eyed people.
Based on this assumption A assumes that B assumes n=0.

When the oracle speaks, A knows that B knows that n does not equal 0. This is the new information.

For n=3 with A, B and C the three blue eyed islanders.

I’m going to explain A’s assumptions only, but B and C have nearly identical sets of assumptions.

A sees 2 blue eyed people, so A assumes n=2.
Since A assumes A has brown eyes, A assumes that B sees 1 blue eyed person.
Based on this assumption A assumes that B assumes n=1.
Based on this assumption A assumes that B assumes that C sees no blue eyed people.
Based on this assumption A assumes that B assumes that C assumes n=0.

When the oracle speaks, A knows that B knows that C knows that n does not equal 0. This is the new information.

A similar argument can be made for any n.

If you disagree with this, please fill in the blank in this statement.

Before the oracle speaks, with n=3 and A, B and C the three blue eyed islanders, A assumes that B assumes C assumes n = ____ .

Be prepared to defend your answer.

93

You are correct. Just to illustrate this some more:
n = 4. A, B, C, and D have blue eyes.
A sees 3 people with blue eyes, so assumes n = 3.

A believes B sees 2 people with blue eyes and assumes n = 2.

A believes B assumes that C sees only 1 person with blue eyes and assumes n = 1.

A believes B assumes that C assumes that D sees nobody with blue eyes and n = 0.

This seems counterintuitive, yet it is true.
The reason is this: Every blue eyed islander believes that every blue eyed islander he sees is seeing one less blue eyed islander than he is. Namely, that they are missing their own blue eyes.

But every blue eyed islander also knows that every other blue eyed islander is reaching the same conclusion (that every blue eyed islander they see is seeing one less blue eyed islander than they are).
So islander n sees n-1 b.e.i. and believes that every b.e.i. sees n-2 b.e.i.

He further assumes that they will all do as he has done, and assume that every b.e.i. they see sees 1 less than them, and so think the others see n-3 b.e.i.

He further assumes that they will make that logical leap as well, and will assume that every b.e.i. they see assumes that every b.e.i. they see sees n-4 b.e.i.
And so on. Thus, the fact that everyone knows there is at least one b.e.i. is new information.

94

It works even if there are only 6 islanders, and ALL of them have blue eyes.
#6 sees 5 BE.

He thinks #5 sees 4 BE.

He thinks #5 thinks #4 sees 3 BE.

He thinks #5 thinks #4 thinks #3 sees 2 BE.

He thinks #5 thinks #4 thinks #3 thinks #2 sees 1 BE.

He thinks #5 thinks #4 thinks #3 thinks #2 thinks #1 sees 0 BE.

Even if every single islander has blue eyes, they don’t know that every islander knows that every islander sees at least 1 BE.
Thus they cannot determine their eye color without help from the oracle, and with that help they cannot fail.

95

This statement doesn’t quite compute. If #6 sees 5 BE, he knows that every one of those 5 sees at least 4 BE, doesn’t he?

96

Yes. But he doesn’t know that everyone knows that everyone else sees at least 1 BE.
The simplest way to look at (for me at least) is that every BE thinks every other BE sees 1 less BE than they do.
So #6, seeing 5 BE, thinks #5 sees 4 BE.

He then figures that #5, seeing 4 BE, thinks #4 sees 3 BE.

And that #5 thinks #4, seeing 3 BE, thinks #3 sees 2 BE.
And so on.

97

I understand that each islander might think that others estimates might not be the same as their own. However, I think that if I am one of 6 islanders and I see all others have blue eyes, I would know that every one of us knows that everyone sees at least 1 set of blue eyes.

Assuming I have brown eyes, I’d think that each and every one of the others sees 4 sets of blue eyes, and that they all assume that each of the 4 they see sees 3 sets. 5, 4 and 3 are all more than 1, so I know that everyone thinks that everyone sees more than 1 set.

98

Hard to believe that this thread lives on. The original logic problem has been answered very clearly in different ways many times already. FWIW, I think that some people are getting the argument too convoluted by stating what everyone on the island has to be thinking. You really only have to reason through each case from two perspectives: that of someone with blue eyes and that of someone with brown eyes.

99

I think there’s a flaw.

Assuming that there are, say, 10 blue eyed people, I understand the reasonning :

A thinks that B thinks that C thinks…etc…that X thinks there’s 0 blue eyed people.

But, A also knows for a fact that everybody can see at least 9 (if he’s brown-eyed) or 10 (if he’s blue eyed) blue eyed people. So, he knows that everybody knows there isn’t 0 blue eyed people, hence he knows that nobody can think someone else will think there’s 0 blue-eyed people.

To keep it simple, let’s assume there 4 b.e. people, A, B, C and D.

A, B, C and D all assume they have (or at least could have) brown eyes. Hence, they all think there are only 3 b.e. people.

They all assume the three others believe there are only 2 blue-eyed people (since they know each of the three others thinks ha has brown eyes).

But the inductive reasonning stops there. They all know nobody can believe there are less than 2 blue eyed people, since in any case, everybody can see at least 2 of them, and everybody knows that everybody knows that.

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.
So, I think the original argument is indeed flawed if there are at least 3 blue eyed people.
Is my reasonning flawed? If so, how?

100

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.