1801

Well that’s just the nuts & bolts. If you broke down the formal proofs of the probability in the other one, you’d have to develop an entire axiomatic system like Bayesianism, or develop the whole of real analysis if there are continuous probability distributions involved.

You can also spend about 100 pages proving 1+1=2 if you get foundational enough.

The intuitive explanation of the blue-eyed islander problem is just a simple induction proof.

But yeah, I find it preferable anyway!

1802

I think the island is suicide-free for any n in the absence of the announcement. If not, why not? The announcement “n>=1” reveals information for any n, even though at first glance it doesn’t seem to:

On island n=1, before the announcement, the blue person reasons that either n=0 or n=1. After the announcment, the blue knows we are not on island n=0, and kills himself on day 1.

On island n=2, call the islanders A and B. Before the announcement, A reasons as follows: “I know that n=1 or n=2, but B either knows that n=0 or n=1 or B knows that n=1 or n=2, depending on which world we are in. Neither of us can determine n, so no one need kill themselves.” Then it is announced “n>=1”, which seems to provide no information to A, but it does: it gives A information about what B knows. A now reasons. “I already knew that n=1 or n=2, but I now know that B knows that n=1, that is, I know that B knows that we are not in world n=0. B will off himself on day 1 if n=1, if B does not kill himself on day 1, we must be on island n=2 and we must kill ourselves on day 2.”

On island n=3, A knows that n=2 or n=3, and A knows that B and C each know either that (n=2 or n=3) or that (n=1 or n=2), depending on A’s eye color. A knows that if n=2, then B and C each see the world as islander A did on island n=2 in the previous paragraph. If n=2, then, A will observe B and C kill themselves on day 2. If day 2 passes with no suicides, then A realizes that n=3, and everyone dies on day 3.

And so on for any n.

It’s a “I knew that you knew that I knew that you knew that I knew, but I did not know that you knew that…” kind of thing. Making it common knowledge that n>=1 means that everyone knows this “to infinity,” so speak, I know that you know that I know that you know… I don’t claim to understand the formal logic in Tao’s explanation, but I think that’s what he’s getting at.

1803

I agree that with 1 or 2 villagers, the explorer’s observation is fatally informative. But the existence of three or more blue eyed villagers has the identical effect, and at that point the explorer does not have any additional effect.

Before the person talks, they know of themselves that they have one of many eye colors. Once the person notes a specific eye color then he creates two groupings, people that see all X of the existing people with that color and people that see X-1 of that eye color.

But how does he accomplish that merely by noting an eye color? With 100 blue eyed natives, they can create those logical groupings without any outside help.

I think the real trick here is that the explorer creates a narrative reference frame, where the reader can assume that the use of logic truly starts. You also have to assume that those perfect logicians never actually used logic before his arrival, otherwise there would be no story. And that’s why what he says is ultimately insignificant.

1804

That seems right to me. I’m convinced of the n=3 lower bound for making the explorer unnecessary.

1805

Austin, Magnet: I disagree that there is any n for which the pre-announcement position is unstable. Suppose I see three blues on my island. I know that either n=3 or n=4, but I have no way of resolving that uncertainty, and I know that no one else has any way of resolving that uncertainty. When noon comes and goes on any given day without any suicides, that gives me no information. For any number of blues I see, no matter how large, the situation is stable.

The announcement that n>=1 does not seem to add any new information when n>=2, but it does, for any n: it always resolves some layer of uncertainty over who knows what. In the n=2 case, before the announcement, every person on the island knows that there is at least one blue on the island, but every person does not know that every person knows there is at least one blue. If I am one of the blues on isle n=2, I know there is at least one blue, but I do not know that the other blue knows there is at least one blue (because if I am not blue, then the blue person I see themselves see zero blues). The announcement does not tell me that there is at least one blue, I already knew that. The announcement instead tells me something I didn’t know before: that the blue I see knows there is at least one blue. I can now reason that if the blue person I see doesn’t kill themselves one day after the announcement, we must be on island n=2. More generally, after the announcement, but not before, I know that if I see (n-1) blues and I see that no one has killed themselves after (n-1) days, then we must be on island n.

1806

I have been thinking about this all day, and while skedastic’s argument seems to make sense, I can’t shake the feeling that something about it is very wrong and I can’t articulate it.

Maybe it’s that the puzzle refuses to acknowledge that n>0 should already be common knowledge. In other words, why aren’t we allowed to assume that every native knows they have been observed, in public, with at least one blue eyed native?

If the explorer said, “If you think one of your colleagues has never seen a blue eyed native, please meet me privately”, is it reasonable to suggest that someone would actually show up? If not, then everyone already knows that everyone already knows n>0.

And yes, I realize that the reply is that maybe not everyone already knows that everyone already knows that everyone already knows n>0, and so on. But it seems to me that everyone should be able to infer that as well. I just can’t explain how. Man, I hate this puzzle.

1807

X is common knowledge not if I know X and you know X, but rather if I know X and you know X and I know that you know X and you know that I know that you know X and so on and so, ad infinitum.

Again consider island n=2. If I am one of the two blues, I see one blue. I know there is at least one blue on the island, and since we’ve assumed n=2, so too does everyone else. Every person knows n>=1, we are not making some weird assumption that no one’s rational or making observations before the announcement. Every person is fully utilizing the information they have obtained, and rationality is itself common knowledge. The uncertainty is that I know before the announcement that n>=1, but I have no way of knowing whether everyone knows whether n>=1. The announcement does not give anyone the information n>=1 as everyone already knew that, the announcement changes the information set of the two blues: they know after but not before the announcement that the other blue knows that n>=1.

English becomes almost hopelessly convoluted for n>2, but it’s something like: before the announcement I know that n>=1 and I know that everyone else knows that n>=1, but I don’t know that everyone knows that everyone knows that n>=1.

1808

Ok, that n=2 example really helps. I hadn’t considered that even in that case everyone already knew that n>=1, and yet it’s clear that the explorer added something.

Thanks, skedastic.

1809

I still don’t get it. If I know n>=1, and I know everybody else has the same information, and I know (as has been posited by the setup) that everybody is perfectly rational, why can’t I assume that they all make the same conclusion that I did, and further that the made the same conclusion about everybody else knowing that everybody else knows.

It seems like a distinction between all natives being “perfect logicians” vs “pretty good” ones.

1810

Everyone draws exactly the same conclusions. Before the announcement, an islander who sees m blues thinks: “I know that either n=m or n=m+1, but I cannot determine which world we are in, and neither can anyone else.” For the blues, m=(n-1). For the non-blues, m=n. And no one dies.

After the announcement, an islander who sees m blues thinks, “I know that either n=m or n=m+1, and I know that if n=m, then every blue will kill themselves on day m. If that does not happen, I must kill myself on day m+1.” And all the blues shrug this mortal coil on day m if the islander is himself not a blue, and on day m+1 if he is a blue.

1811

That’s why the lower bound for unstable populations would be n=3. The n=2 situation is insufficient for every islander to know that everyone else knows that there are blue-eyed islanders.

In the n=3 case, even the blue-eyed islanders know that every islander knows there’s at least one blue-eyed islander. For n=1, the blue-eyed islander doesn’t even know that A is blue-eyed, for some A. For n=2, everyone knows that A is blue-eyed (for some A), but not everyone knows that A knows that B is blue-eyed. But for n=3, everyone, including the blue-eyed islanders, knows that A knows that B is blue-eyed for all A & some B.

The explorer only provides the additional information that now A knows that B is blue-eyed, for all A & some B, which is redundant information if n>=3. The quantification over all A is important.

Although I could be either mistaken or overlooking something.

1812

The more I read about this blue-eyed thing, the dumber it seems and the more I hate it.

1813

Thank god I’m not the only one.

At this point I just want them to kill themselves so I stop hearing about how they’ll kill themselves.

1814

It isn’t redundent information for any n. Let I(m) denote the information, “I know that n=m if I am not blue-eyed and n=m+1 if I am.” Let K(m) denote the information, “I know there are exactly m blues and I am one of them.” No one kills themselves with an I() information set, but anyone learning a K() information set must kill themselves.

If A sees zero blues, he knows I(0), and without the announcement, he has no way to deduce which world he is in. With the announcement, his information set changes to K(1), and kills himself on day 1.

If A sees one blue, he knows I(1), and he knows that the one blue he sees either know I(0) or they know I(1). Since rationality is common knowledge, he also knows that if the blue he sees knows I(0), then by the preceding argument the blue he sees will kill himself on day 1 with the announcement, but won’t without the announcement. If the blue A sees doesn’t kill himself on day 1, A infers K(2) iff the announcement has been made, and they both kill themselves on day 2.

If A sees two blues, he knows I(2). He also knows that if n=2, then the other blues he sees know I(1), and we determined above that if n=2 and those two blues know I(1), they will both kill themselves on day 2 iff the announcement has been made. So A observes that no one kills themselves on day 1 and that is no surprise at all, but if no one kills themselves on day 2, he infers K(3) and they all die on day 3.

Generally, if A sees m blues, he knows I(m). With or without the announcement, A knows there will be no suicides on days 1 through (m-1). Iff the announcement has been made, A knows that if no one dies on day m, then K(m+1) and he along with the blues die on day m+1.

More simply, if you agree that if n=2 the announcement will cause the two blues to kill themselves on day 2 and that they won’t without the announcement, then I think you are bound to think that there is information provided by the announcement for any n. For n=3, each of the blues knows that if they are themselves are not blue blue, the two blues they see will kill themselves on day 2. If that doesn’t happen, then n=3. But only in the presence of the announcement, the situation is stable in its absence.

Here’s another way to see that there is information in the announcement for any n. Suppose the visitor walks up to an islander who sees m blues and whispers in his ear “there is at least one blue.” This has no effect, even if the visitor secretly whispers that information in every islander’s ear. But if, as we are told in the puzzle, the visitor stands in the middle of the village and loudly announces “there is at least one blue” to the all villagers simultaneously, then all the blues die after either m or m+1 days.

1815

You’re getting hung up on the cases where knowledge of other people’s knowledge is not certain. We’ve already established why there’s insufficient information for n<3. I’m making a claim strictly for n>=3. In those cases, it is common knowledge that everyone sees a blue-eyed islander, which is the only information the explorer contributes.

1816

No, it isn’t common knowledge. When n=3 I know that you know there is at least one blue, but I do not know that you know that I know that you know there is at least one blue. Remember, it’s turtles all the way down with common knowledge. That is, even though we are on island n=3, the blues on the island only know that others know that n is at least 1.

Why exactly is the situation unstable for n>=3 without the announcement? Suppose n=3 and I am one the blues. How exactly do I know that I need to kill myself on day 3? With the announcement, I learn something when the two blues I see don’t kill themselves on the second day. Without the announcement, I don’t learn anything when the second day passes without any deaths.

1817

With love.

1818

/butting in - Wouldn’t it be common knowledge that n{known}=n{visible}-1 though? That is if there are 100 blue eyes you will see either 100 or 99, so you will assume there are either 98 or 99 at least. So everyone knows there are at least 98, right?

1819

It’s true that epistemic logic itself allows for “turtles all the way down,” but the only information the explorer provides is common knowledge that every A knows some B has blue eyes. Since this level of extension is already provided by the n>=3 cases, it is redundant for the explorer to be included.

1820

Everyone knows that n>=98, but not everyone knows that everyone knows that n>=98, so it is not common knowledge that n>98. Similarly, on this island everyone knows that everyone knows that n>=97, but not everyone knows that everyone knows that everyone knows that n>=97. And so on. Before the announcement, no assertion about the number of blues on the island is common knowledge.

Again, asserting “X is common knowledge” is much stronger than asserting “everyone knows X”, and it’s much stronger than “everyone knows X and everyone knows that everyone knows X.” It means everyone knows X, and everyone that knows everyone knows X, and everyone knows that everyone knows that everyone knows X, and so on off to infinity.