It said she was 21 years older than her son, not 21 years older than when she was impregnated, but either way it doesn’t work. 22 + 6 is 28, which is still not 5 times 6.
My mind still rebels - so if a guy looks around and sees ten guys with blue eyes standing around, he thinks to himself, “okay, if somebody around here only sees one other guy with blue eyes, he’ll kill himself tomorrow. I think B thinks C thinks D thinks E thinks F thinks G thinks H thinks I thinks J thinks K is the one and only blue-eyed guy around here.” What, is everyone from D through K blind or something?
Seriously - the situation relies on the lack of deaths on days 1, 2, 3, etc providing information to later days. Doesn’t there come a point where everybody knows that everybody knows that everybody knows that etc etc etc that nobody will die on day 1?
No, it’s just that B and C happen to be wrong. They’re not saying that it’s true that D through K think that. They’re saying it’s possible.
Consider this:
In reality, there is one situation - everybody has blue eyes.
A believes there are two situations - Either he’s brown or blue, but everyone else is blue.
B believes the same thing, but does A report that to the interviewer? No, because he’s unaware of B’s knowledge of A’s eyes.
So, according to A, B holds 1 of 4 possible opinions. Either it’s blue-blue, brown-blue, brown-brown, or blue-brown. Now he knows in reality that only blue-blue and brown-blue are possible, but B doesn’t know that. And A can’t decide between the remaining two because he has no knowledge of his own eyes. The other 98 islanders are, of course, blue because that’s shared info between A and B. A can rule out that B thinks anything other than blue about those 98 others.
So when A thinks about B, he can envision 4 scenarios.
When A thinks about B thinking about C, he can think of 8 scenarios- those same four that could be in B’s head, but one (each) where C is blue and one (each) where C is brown.
Sanity check: We know there is only 1 distribution out there. We know that C really only has 2 possibilities floating around in his head. We know that if you asked B what those 2 were, she could only narrow it down to 4 (out of the 2^100 possible ones). But remember, we’re talking to A here, not B or C. So while B has 4 possible distributions that C could be thinking of, A couldn’t tell the interviewer what they were because he doesn’t know his own eye color (even though B and C both know it).
This doubling continues down the line for each person you’re asked to “think through”. It’s a mix of A not being able to determine what they have in their heads, and the person he’s considering not knowing his own eye color.
So all he’s saying is that it’s possible that B thinks that C thinks that D thinks, etc…we all have brown eyes. It’s essentially a case of A “summing everyone’s ignorance”.
The mother is currently 21 - 3/4 years old and the son will not be born for another 3/4 years (the father is still impregnating the mother). In 6 years, the mother will be 6 + 21 - 3/4 = 27 - 3/4 years old and the son will be 6 - 3/4 years old. 5 * (6 - 3/4) = 27 - 3/4.
It’s not the case that I think B thinks C thinks D thinks E thinks F thinks G thinks H thinks I thinks J thinks K is the one and only blue-eyed guy around here. Rather, what happens is that I think it’s possible, if I myself have brown-eyes (because for all I know I do), that B thinks it’s possible, if B has brown-eyes (because for all B knows, B does), that C thinks it’s possible, if C has brown-eyes (because for all C knows, C does), etc., etc., that J thinks it’s possible that K is the one and only blue-eyed guy around here.
That is, at the start of the puzzle, I don’t know for sure that B knows for sure that C knows for sure that D knows for sure that … that J knows for sure that K is not the one and only blue-eyed guy around here.
Careful. That’s not quite right. Rather, I think it’s possible that B thinks it’s possible that C thinks it’s possible that D thinks it’s possible etc… that the last guy thinks it’s possible we all have brown eyes. At every step in that true statement, the connective is “I think it’s possible that”/“I can’t rule out the possibility that”, rather than “I think”/“I know for sure that”.
I agree. Here’s the though process visually. Pretend that O = Blue and * = Brown. It’s supposed to imitate a filled in circle and an open circle. The first O or * represent’s A’s eye color and the second, B, and so on. Now imagine that A is standing there with a though bubble over his head containing all the possible combinations of eye colors on a 4-person island. How many combos does he have? 2. That’s line one:
A - OOOO *OOO
Now he thinks about what’s in B’s mind. He knows that B actually knows which of the above two is reality, but he doesn’t have that information himself, so he entertains both scenarios. A’s thoughts about B’s thoughts are on line 2:
A - OOOO *OOO
B - OOOO O*OO *OOO **OO
Now, according to this, A thinks there are 4 possible things that B is thinking of. Now here’s the key- we’re all well aware that, just like A, B is only thinking of two possibilities. He’s basically saying “If I’m blue-eyed, then B is thinking the two on the left. I’m I’m brown, then she’s thinking the two on the right. But I don’t know which it is.” Indeed, our interviewer (who sees everyone) is only thinking of one combination- the right one. Thing is, though, A doesn’t know which 2 out of the 4 are the ones that B is actually thinking.
Now here’s what A considers when he thinks of what B thinks of C’s combinations:
OOOO, O*OO, *OOO, **OO,
OO*O, O**O, *O*O, ***O
There are 8 of them. See, A knows that B is only holding two possibilities in her head, but he doesn’t know which two. And B only thinks that C has 2 of 4 possible combinations, but B doesn’t know which 2 those are. And A doesn’t know which 4 those are out of the 8 I just gave.
This occurs for the last guy as well. So on this island of 4, A finds it possible that B finds it possible that C finds it possible that D has 16 different guesses as to the eye color distribution of the island. Does D really? No. He only holds 2, just like everyone else. But C has no way of knowing which 2 those are and can only narrow it down to 4. And B doesn’t know which 4 those are. She can only narrow it down to 8. And of course A can’t know which 8 those are. He can only narrow it down…well, he can’t narrow it down at all. He can only know that it’s one of the 16 possibilities.
Presuming 10 villagers, all blue:
Everyone sees 9 blues.
Everyone knows everyone sees >=8 blues (including themselves).
Everyone knows that everyone knows everyone sees >=7 blues.
So does A think B thinks C thinks D think E thinks F thinks G thinks H sees 2 blues? No. A knows B knows that C knows better - C knows that E knows that F knows that G knows H sees >= 7 blues.
And none of them think anyone thinks anyone will die on day 1, so nobody will learn anything when it doesn’t happen, and so the dominos won’t start falling. Right?
No one thinks that B thinks that C thinks that D thinks that… H sees 2 blues. That’s the point of my last post. Rather, A can’t rule out the possibility that B can’t rule out the possibility that C can’t rule out the possibility that … whoever can’t rule out the possibility that H sees 2 blues.
Let’s consider a very small case: only 4 islanders, A, B, C, and D.
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If D were blue and A, B, and C were brown, then D would not be able to rule out the possibility that everyone was brown.
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If C and D were blue, and A and B were brown, then C would not be able to rule out the possibility that D is blue and A, B, and C are brown. That is to say, C would not be able to rule out the situation analyzed in 1). And remember, one of the things that happens in the situation analyzed in 1) is that D would not be able to rule out the possibility that everyone was brown. Accordingly, C would not be able to rule out the possibility that D would not be able to rule out the possibility that everyone was brown. It’s not that C thinks for sure that D is unable to rule out the possibility that everyone is brown, or that C is unable to rule out the possibility that D thinks for sure that everyone is brown; those statements are both false. It’s just that C is unable to rule out the possibility that D is unable to rule out the possibility that everyone is brown.
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If B, C, and D were blue and A were brown, then B would not be able to rule out the possibility that C and D were blue while A and B are brown. That is, B would be unable to rule out the situation analyzed in 2). And one of the things that happens in the situation analyzed in 2), remember, is that C is unable to rule out the possibility that D is unable to rule out the possibility that everyone is brown. Accordingly, even though it’s not actually true that “C is unable to rule out the possibility that D is unable to rule out the possibility that everyone is brown”, B is unable to rule it out. If B, C, and D are blue and A is brown, then B is unable to rule out the possibility that C is unable to rule out the possibility that D is unable to rule out the possibility that everyone is brown.
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Finally, even everyone is blue, then A is unable to rule out the possibility that A is brown and B, C, and D are blue; that is, A is unable to rule out the case analyzed in 3), and accordingly, A is unable to rule out anything that happens in 3), meaning, in particular, that A is unable to rule out the possibility that B is unable to rule out the possibility that C is unable to rule out the possibility that D is unable to rule out the possibility that everyone is brown.
Whenever I say “A thinks blah”, I mean “A can’t rule out the possibility that blah”. “Think” and “thinks” are just shorter to type.
However, when I type “knows”, I really do mean “knows”, or if you prefer “can rule out the possibility that they’re wrong”. For example, if A sees four blues, then A “knows” there are >=4 blues, and can rule out the possibility that there are <4 blues.
Heh, four blues is the minimum case where the supposed induction starts to break down.
When A sees three blues, he can rule out the possiblity that anyone (including himself) sees less than two blues. And thus he can rule out the possibility that anybody thinks anybody else sees less than one blue. Thus ruling out case 1, ensuring that nobody expects anybody to deduce that they’re the lone blue V sees, and ensuring the village’s continued existence.
Of course A can rule out case 1. Everyone can rule out case 1. And everyone knows that everyone can rule out case 1. But not everyone knows that everyone knows that everyone can rule out case 1. In particular, as demonstrated, A can’t rule out the possibility that B can’t rule out the possibility that C can’t rule out case 1.
Do you agree that in case 4, A can’t rule out case 3? And do you agree that in case 3, B can’t rule out case 2? And do you agree that in case 2, C can’t rule out case 1?
In case 4 everyone can see >=3. Thus, everyone know that everyone can see >=2. Thus, everyone knows that everyone knows know that everyone can see >=1 (because anybody that can see >=2 knows that each of those can see other - and everyone knows that everyone can see >=2). Correct me if I’m wrong, but in case 1 D doesn’t see any blues. Everyone knows that that’s not happening, so they can all to a man rule out case 1, with absolute certainty - and they all know it.
Everyone knows case 1 isn’t happening. That’s true. That’s not being denied.
What’s being denied is that everyone knows that everyone knows that everyone knows that case 1 isn’t happening. This is a very different statement.
Ignore for the moment any larger inference I may want to string these into, and just answer these four specific questions: In case 4, can A rule out the possibility that case 3 is happening? In case 3, can B rule out the possibility that case 2 is happening? In case 2, can C rule out the possibility that case 1 is happening? And in case 1, can D rule out the possibility that everyone has brown eyes?
The nice thing about induction is that you don’t have to think through each individual case.
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If there is exactly one blue-eyed person, he will obviously realize he has blue eyes because he can see nobody else does. So everybody knows if there is 1 blue-eyed person he will die on day 1.
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Assume that everybody knows that if there are n blue-eyed people they will die on day n. Then if there are n+1 blue-eyed people and nobody dies on day n, everyone knows there are n+1 blue-eyed people and they die on day n+1.
Thus n blue-eyed people will die on day n.
The induction can’t suddenly fail when you get to 4, because if people don’t die on day 3, then everyone knows there weren’t just 3 blues, so they know there are 4 and die on day 4.
The proof is very simple, but going through the actual thought process is confusing.
Even though it isn’t quite accurate, I don’t see a problem in thinking of it as though everyone assumes they have brown eyes until proven otherwise.
With 4 people, A assumes he has brown eyes, so that B only sees 2 blues. But he figures that B also assumes he has brown eyes and thus thinks C only sees 1 blue. But he further considers that B will figure that C assumes his eyes are brown, so that B will think C thinks D is the only blue. That is, A thinks B thinks C thinks that D can’t see any blues.
The part that you’re missing is that everyone imagines 2 scenarios, but they don’t know which two scenarios.
Let 1=brown and 0=blue.
The truth is 0000, but…
- A imagines 0000 and 1000.
- B imagines 0000 and 0100.
- C imagines 0000 and 0010.
- D imagines 0000 and 0001.
Now, in line 4, the third slot is 0 for sure. After all, D can see C’s eyes. But C doesn’t know this. C can’t tell what’s in D’s imagination. He knows what D thinks about the A, B, and D slots but not what D thinks about the C slot.
So C thinks “I know what D thinks about A. He’s 0. I know this because we can both see him. Same for B. So we know the pattern starts with 00. But I don’t know what the third slot is, even though D definitely knows. So D is either thinking ‘0000 and 0001’ or ‘0010 and 0011’. I just can’t tell since I don’t know what he sees in my own eyes. I know that the two latter cases are false because I can see 0 in his eyes, but there’s no way he could know that himself. So he is thinking of one of those two pairs- Either the first set or the second set. I just don’t know which.”
But do you see how in all four patterns, the second slot (B) is 0? Well B doesn’t know that. She thinks it could also be a 1. She’s well aware that D has it narrowed down to 2. And she knows that C has D’s combinations narrowed down to 4. But she doesn’t know what those 4 are. They could be the 4 where the second slot (the B slot) is 0 or the 4 where the second slot is 1, but not a combination of the two.
B says to herself, “I know C has D’s combinations narrowed down to 4. I just don’t know which they are. They are either ‘0000, 0001, 0010, and 0011’ if I have 0 eyes, or ‘0100 and 0101’ or ‘0110 and 0111’ if I have 1 eyes. If I were to know my own eye color, I would know which group of 4 that C believes possible for D to hold. But I don’t, so I can at best narrow it down to 8. Of course, I know that C knows that D knows that A has 0 eyes. I know this because I know that C knows that D sees A’s eyes and they’re 0.”
Bear in mind that B can narrow down D’s choices better than 8. She can narrow down the actual possible patterns in C’s mind to 4- namely, 0000, 0001, 0100 and 0101. Problem is, 2 belong to the first group of C’s choices and two belong to the latter. So all it proves is that B knows that C is wrong about half his choices. It does not tell her which choices are wrong. So she can’t determine any info about her own eye color. If she could, for some reason, determine which group of 4 C was actually considering for D’s mind, then B would know her own eye color. But she can’t, so the best she can do is 8 possible thoughts for C to hold about D’s patterns.
Again, B knows that anything where the third slot is 1 is wrong, but she also knows that C is ignorant of that fact and thus can’t rule it out for what C is considering possible in D’s mind.
Notice that all of B’s 8 choices start with 0. That’s because she can see A’s eyes, and she knows that C can too, and C knows that D sees A’s eyes as well. But A doesn’t know that! Sure, B is considering only 8 choices (4 which she knows to be false, but can’t rule out that C thinks them anyway), but A can’t narrow down which set of 8 those are- the one where A is 0 or the one where A is 1.
So A knows:
B is considering one of two set of options regarding C’s analysis of the distributions that D considers possible. Either the set:
- 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111 OR
- 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111
The only difference is his own eye color. Sure, B knows which one she actually is considering. And C knows which 4 he’s actually considering. And D knows which 2 he’s actually considering. And the narrator knows which 1 is actually true.
But A can’t tell if B is considering set 1 or 2, even though we know B is considering set 1 and has discarded 2.
B can’t tell which half of set 1 C is considering to be in D’s mind, the first half or the second half. Of course, we know it’s really the first half (0000, 0001, 0010, 0011).
C knows that it’s the first half because he can see A and B and knows D can too. But he can’t see himself, so he can’t choose which 2 of the 4 patterns D is believing to be possible, 0000 and 0001 or 0010 and 0011.
D knows which pair it is, however. He knows it’s the first pair, 0000 and 0001, that are possible. He just can’t tell which 1 of the two it is.
And we, the outsiders, know which one it is - 0000.
So there you have it. I worked backwards through the logic, then forwards through it again. If anyone still can’t grasp this, then I can’t help you any further, I don’t think. It’s not an issue of how many patterns each thinks is possible, it’s a matter of which patterns they think are possible.
Do not confuse the following two statements:
- A thinks that B thinks there are 4 patterns possible.
- A thinks that B holds 2 out of 4 possible patterns and just doesn’t know which two.
Statement 1 is false and will lead to confusion. Statement 2 is true and explains the situation clearly and easily.
So, what exactly is the new information that the foreigner provides?
Everyone knows there are blue eyed villagers, and everyone knows blue eyed villagers are rarer.
Seems like they should have killed themselves a long time ago.
No they don’t. That’s the thing.
If you are the only blue-eyed villager, you don’t know blue-eyed villagers exist.
If you are one of two blue-eyed villagers, you only know one blue-eyed villager exists.
However, if there was only one blue-eyed villager, he wouldn’t know blue-eyed villagers exist until the tourist tells him. But he didn’t kill himself Day 1. Which means he sees another blue-eyed villager. Since that means there must be two blue-eyed villagers, and you see only one, you’re the other blue-eyed villager.
That’s the new info. That someone outside can confirm there are blue-eyed villagers.
Please remember there are 100 Blue eyed villager and 900 brown eyed. If you make a special case where there are only 1 or 2, then its different. But the conclusions from the 1 or 2 or 3 blue eyed cases are not relevant to the 100/900 case.
There is no new information conveyed by the foreigner.
This has been explained before. Yes, prior to the foreigner’s arrival, everyone knows that there is at least one blue-eyed islander. That’s not new. The new thing is that the foreigner’s statement brings about a situation where everyone knows that everyone knows that everyone knows that … that everyone knows that there is at least one blue-eyed islander. Prior to the foreigner’s arrival, the appropriately long version of this statement was not true.
If the foreigner secretly went up to each islander and privately told them “There is at least one blue-eyed islander on the island”, nothing would change, because everyone already knows that. It’s the fact that he makes this as a public announcement that allows it the required properties; people do not gain new information about whether there are blue-eyed islanders on the island, since everyone knows that already, but there is a change in whether everyone knows that everyone knows that everyone knows… that everyone knows that.
I don’t buy it. 100 blue eyed people on the island. Everyone can see everyone except for themselves.
Every one knows that everyone knows that every knows … that everyone can see at least 98 blue eyed people.
Every single blue eyed person sees 99 blue eyed people. And they know that every single one of those 99 people sees (at least) 98 blue eyed people.
And everyone knows it.