1

For $k = 2$, it is merely "first-order" knowledge. Each blue-eyed person knows that there is someone with blue eyes, but each blue eyed person does ''not'' know that the other blue-eyed person has this same knowledge. (from http://en.wikipedia.org/wiki/Common_knowledge_(logic))

I am not getting this. If there are more than one people that have blue eyes, each person can see that there is a person with blue eyes and people that have green eyes can clearly see that there are people with blue eyes. So even before the common knowledge annoucement, isn't it natural to say that everyone knows that everyone knows there is at least one person with blue eyes - common knowledge?

How am I mistaken? I am not getting how announcement sets common knowledge - as it seems for me that there already is common knowledge.

Edit: OK, I get it for $k = 2$. But what about $k>2$? Then, everyone would be sure to know that everyone knows that there at least exists one person with blue eyes, right? Doesn't this already constitute as common knowledge?

Common Knowledge
  • 503
  • Something I'm quite sure could be helpful. http://isites.harvard.edu/fs/docs/icb.topic138342.files/lecture7.pdf http://www.ma.huji.ac.il/raumann/pdf/Interactive%20epistemology1.pdf – Metta World Peace Feb 07 '13 at 17:56
  • 1
    I'm not an expert, but according to wikipedia if everyone knows X and everyone knows that everyone knows X, it doesn't mean that X is common knowledge. For common knowledge, everyone should also know that everyone knows that everyone knows X, and so on for any finite "depth of nesting". – Dan Shved Feb 07 '13 at 18:03
  • I would recommend you to have a look at the 'story' of 40 married couples in a village. The only link I found is the following, it's the first example in the first page. http://ocw.mit.edu/courses/economics/14-126-game-theory-spring-2010/lecture-notes/MIT14_126S10_lec14.pdf – memo Feb 08 '13 at 00:21

1 Answers1

5

Consider the two persons $A, B$ with blue eyes.

$A$ sees $B$, so she knows there is somebody with blue eyes. (And so does $B$.)

But $A$ has to consider the possibility that $B$ is the only person with blue eyes. (After all, $A$ does not know the colour of her own eyes.)

In this case (that is, if $B$ is the only person with blue eyes, a case $A$ cannot rule out), before the announcement, $B$ wouldn't know that there are people with blue eyes. So $A$ cannot be sure whether $B$ knows.

The same holds for $B$ concerning $A$.

So before the announcement $A$ and $B$ do not know whether everyone knows that there are blue-eyed people - they are precisely in doubt about what the other knows.

Andreas Caranti
  • 68,873
  • OK, then. What about $k >2$? This case seems different from $k=2$... – Common Knowledge Feb 07 '13 at 18:10
  • 1
    It's the same, just one layer more. $A, B, C$ do not know whether everybody knows whether everybody knows that there are blue-eyed people. Use precisely the same argument I just gave. $A$ knows that everybody, including $B$ and $C$, know that everybody knows that there are blue-eyed people. But if $B$ and $C$ are the only ones with blue eyes (a case $A$ cannot rule out), then by the previous case, before the announcement $B$ and $C$ wouldn't know whether everyone knows... – Andreas Caranti Feb 07 '13 at 18:14
  • @CommonKnowledge take $k=3$. Each blue-eyed person sees 2 people with blue eyes. They don't know if there are actually 2 people with blue eyes, or 3 (including them). So before they do anything they'd have to rule out the two person case. If there were two of them, then each of those would only see one. If that was the only one, the person who only saw green eyes would die the next day. If they don't, then both people with blue eyes know that there are at least two people with blue eyes. If they only see one, they know that they must be the other and they'll die the next day. – Robert Mastragostino Feb 07 '13 at 19:02
  • @CommonKnowledge If none of that happened, then the k=1 and k=2 cases can't be true and are ruled out. If I only see two people with blue eyes, that must mean the k=3 case is the only one possible, I have blue eyes, and I die. – Robert Mastragostino Feb 07 '13 at 19:02
  • @Robert: I don't understand what your comments are getting at. Andreas had already explained why the announcement is required in the $k=3$ case. It seems that your comments explain not why the announcement is required but how the deduction works given the announcement -- but if I understood the question correctly that wasn't what it was about. – joriki Feb 07 '13 at 23:23
  • 1
    @joriki my point was to clarify that each level depends on the previous ones, and so can't possibly be a form of common knowledge on its own. That each step lasts because of the ambiguity (do $n$ or $n-1$ people have blue eyes?) and the logical chain isn't triggered if there's no announcement to rule out the 0-person case. I can delete if it seems too out of place. – Robert Mastragostino Feb 07 '13 at 23:45
  • @Robert: I still can't see that in rereading the comments, but maybe it's just me; no need to delete, I think. – joriki Feb 07 '13 at 23:50