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Shadow
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Re: BROWN EYES AND RED EYES  
« Reply #200 on: Oct 24th, 2008, 11:34pm »
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TOWR
 
This is your quote (Being a newbie I am not sure how quoting from others should be formatted)
 
"Yeah, but the point is that it adds to their  
knowledge about the knowledge^N of other monks. That's a point so often repeated in this thread it's hard to believe you missed it.
Even if you had only two REMs; obviously they both know there is at least one REM. But after the public announcement, they now know that the other one knows. It told them nothing about how many REMs there are, it told them something different."
 
 
The point is that  this holds true only if their are 2 REMs(or less). After there is more than 2 monks with red eyes, this premise becomes invalid.  See previous posts.
You're comment makes it seem like I am a complete idiot for not seeing things " your way".  You keep going back to the N-1 theory but completely ignore that this theory does not hold true when there are more than 3 REMs.
« Last Edit: Oct 25th, 2008, 1:10am by Shadow » IP Logged
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Re: BROWN EYES AND RED EYES  
« Reply #201 on: Oct 25th, 2008, 4:03am »
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on Oct 24th, 2008, 9:21pm, Shadow wrote:
This is one monastery.  A knows that both B and C KNOW that at least one monk has red eyes.( these are not separate monasterys)
Yes, but A doesn't that B knows that C knows there is at least one. Because A doesn't know whether he has red eyes or not, and so doesn't know whether B sees two monks with red eyes or one. And if B only sees one monk with red eyes, than B wouldn't know whether C sees any monk with red eyes.
You have to follow the chain of knowledge down to at least N levels.
 
Quote:
Aside: TOWR is very upset that I have questioned his/her opinion.
No, it's just that I, and others, have repeated this discussion so many times with so many people it gets frustrating. Especially since we hope we'd explained it well enough in this thread that other people could also understand it.
At some point, especially late at night, you start to wonder if it's all worth it anymore.
But you're right that I shouldn't vent that frustration at you.
 
Quote:
You have not addressed the opinions expressed by me, but have only attacked my right to express them.
I'm fairly sure I addressed one point.
All monks know that there is at least one REM, that doesn't change. But the knowledge about monks having knowledge about monks having knowledge about ... etc .. does change.
And the simplest way to imagine how that works is to take one REM and have him think "Suppose I have brown eyes, how then would the other monks behave". Which places him in a hypothetical world that after N-1 nights gets disproven, leaving only the other hypothesis, that he does have red eyes.
 
Quote:
You are an Uberpuzzler and a moderator and I respect these credentials as a newbie. I contend that you have a mental model that prevents you from seeing any other option than the one your mental model agrees with.
Yes, I drew it here. It gives the full diagram for 5 red eyed monks about their knowledge and hypotheses of the world (limited to those relevant to the REM problem).
 
Quote:
I respect your opinion, and would also appreciate if you would respect mine.  Look at this from a new beginning and explain where my comments are in error.  I am more than willing to admit I am missing something.
I'll try to be less snarky. But I can't promise I won't at some point give up and throw in the towel.
« Last Edit: Oct 25th, 2008, 4:03am by towr » IP Logged

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Re: BROWN EYES AND RED EYES  
« Reply #202 on: Oct 25th, 2008, 4:41am »
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on Oct 24th, 2008, 11:34pm, Shadow wrote:
This is your quote (Being a newbie I am not sure how quoting from others should be formatted)
You can use the "quote" button at the top right of a persons post.
 
Quote:
The point is that  this holds true only if their are 2 REMs(or less).
Why? You gave a reason why the process wouldn't apply to 3 or more monks; and that exact same reason also applies to two monks. Yet you concede two monks would work out that they have to commit suicide, so obviously the given reason cannot be correct.
 
If you have two or more monks than the statement "there is at least one REM" does not add tell any monk anything with regard to how many monks with red eyes there are. They all already knew there was at least one.  
If two monks nonetheless kill themselves, then this is not a reason you can use for objecting larger amounts of monks.
 
 
Quote:
After there is more than 2 monks with red eyes, this premise becomes invalid.  See previous posts.
What premise? Tell me that first.
 
Quote:
You keep going back to the N-1 theory but completely ignore that this theory does not hold true when there are more than 3 REMs.
I think you simply don't understand what this "theory" is. And when I show you it does work for 3 REMs, you will simply object it doesn't work for 4. And either at some point you'll grasp the concept of induction or I give up.
 
Let's take monks A, B and C. They all have red eyes.
A can see two possibilities:
 world A0 where he has brown eyes, and
 world A1 where he has red eyes.
 
In world A0, monk B (to the best of A's knowledge) may consider himself in
world A0B0, where he and A both have brown eyes, or
world A0B1, where he has red eyes and A doesn't.
 
In world A0B0 (to the best of A's knowledge of B's knowledge), monk C
might hold the hypothesis that he's in world A0B0C0, where none of the monks have brown eyes. This option is eliminated were it to become common knowledge there is at least one monk with red eyes; but at the moment it is possible that A thinks that B think that C thinks this is possible.
There is also the hypothesis A0B0C1, where C is the only monk with red eyes. If it were to become common knowledge that there is at least one REM, then to the best of A's knowledge about B's knowledge C would kill himself after one night. Should this not happen, then the B's hypothesis A0B0 has failed, and only A0B1 is left.
 
So, A considers A0B1. In this case, C might once again consider (according to A's knowledge of B's knowledge) that
either he's in world A0B1C0, where only B has red eyes. Which means that if it is made common knowledge someone has red eyes, C would think (to the best fo A's knowledge of B's knowledge) that B should kill himself that night, because A thinks that B thinks that C thinks that B can't see any other REM. If this doesn't happen this hypothesis is falsified.
The other world A believes it is possible for B to believe that C holds possible is that we're in A0B1C1. So on the second night (A0B1C0 and AB0C1 having been eliminated), because to the best of A's knowledge both B and C would see A has brown yes, they would kill themselves. If this doesn't happen, A can no longer hold this hypothesis, and A0 is eliminated. Which means A can only believe he has red eyes.
 
Now, we have 4 more cases to go; and I'm sick of it already.  (And you can get a further 40 cases by changing the orders of A,B,C.)
 
 
If A sees N-1 REMs, then he believes it possible that B sees N-2, and A believes it is possible that B believes it is possible that C sees N-3, and A believes it is possible that B believes it is possible that C believes it is possible that D sees N-4, etc. Down to the last monk in the chain which is believed to believe there might be none.  
This last monks is eliminated from the graph of hypotheses when it is made common knowledge one monk has red eyes.
« Last Edit: Oct 25th, 2008, 5:09am by towr » IP Logged

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Re: BROWN EYES AND RED EYES  
« Reply #203 on: Oct 25th, 2008, 9:51am »
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The point is that  this holds true only if their are 2 REMs(or less).
Why? You gave a reason why the process wouldn't apply to 3 or more monks; and that exact same reason also applies to two monks. Yet you concede two monks would work out that they have to commit suicide, so obviously the given reason cannot be correct.
 
If you have two or more monks than the statement "there is at least one REM" does not add tell any monk anything with regard to how many monks with red eyes there are. They all already knew there was at least one.  
If two monks nonetheless kill themselves, then this is not a reason you can use for objecting larger amounts of monks. "
 
Still can't get the grasp of the quote in top right corner.  I highlight the text I want to quote and then hit this Quote link and it takes your entire response into a reply box(not just the the text I had highlighted, but the entire response)
 
The reason it works for 2 monks is that A might (and hopes he has)  brown eyes, so he doesn't know that all other monks know that at least 1 monk has red eyes(he hopes monk B sees him with brown eyes).  As soon as there are 3 monks with red eyes, all 3  monks with red eyes know that the other 2 with red eyes already know that at least one monk has red eyes.  A tourist telling this already known fact should not start the sequence described since this "fact" was already known by all monks.  
 
I sense the frustration in your responses. I just do not understand the cascading principal you keep describing.
 
My intent is not to keep you responding as above.  Your responses are very intricate and detailed, but rely on the fact that all monks do not realize the fact all other monks do not know that all other monks already know that at least 1 monk has red eyes.  My observation is that this is already known by all monks when there are 3 or more. Perhaps we should just agree to disagree?
« Last Edit: Oct 25th, 2008, 10:10am by Shadow » IP Logged
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Re: BROWN EYES AND RED EYES  
« Reply #204 on: Oct 25th, 2008, 10:44am »
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on Oct 25th, 2008, 9:51am, Shadow wrote:
Still can't get the grasp of the quote in top right corner.  I highlight the text I want to quote and then hit this Quote link and it takes your entire response into a reply box(not just the the text I had highlighted, but the entire response)

Once you have the quoted text included, you can delete the parts you don't want just as you would any other text in your post.
Quote:
The reason it works for 2 monks is that A might (and hopes he has)  brown eyes, so he doesn't know that all other monks know that at least 1 monk has red eyes(he hopes monk B sees him with brown eyes).  As soon as there are 3 monks with red eyes, all 3  monks with red eyes know that the other 2 with red eyes already know that at least one monk has red eyes.  A tourist telling this already known fact should not start the sequence described since this "fact" was already known by all monks.

 
With two REMs, B knows that A has red eyes, and A knows that B has red eyes, but B doesn't know that A knows B has red eyes.
 
With three REMs, C knows that B and A have red eyes, B knows that C and A have red eyes, and A knows that C and B have red eyes, so C knows that B knows that A has red eyes, and C knows that A knows that B has red eyes, but C also knows that B doesn't know that A knows that B has red eyes, and doesn't know that B knows that A knows that C has red eyes, so, while B knows that A knows there's at least one REM (C), and C knows that A knows that there's at least one REM (B), C doesn't know that B knows that A knows that there's at least one REM
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Re: BROWN EYES AND RED EYES  
« Reply #205 on: Oct 25th, 2008, 11:18am »
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on Oct 25th, 2008, 9:51am, Shadow wrote:
 Your responses are very intricate and detailed, but rely on the fact that all monks do not realize the fact all other monks do not know that all other monks already know that at least 1 monk has red eyes.
But you're leaving out a lot of the links here. We're concerned with N levels of knowledge, and you're dealing with just two here.
 
Quote:
My observation is that this is already known by all monks when there are 3 or more. Perhaps we should just agree to disagree?
For three monks, all monks do indeed know that all other monks know that there is at least one REM. That much I completely agree with. But they don't know that all other monks know that all other monks know there is at least one REM. For three monks, it's this third level knowledge that gives the breakthrough.
All three know there is at least one; all three know that all three know there is at least one; but it's not the case that all three know that all three know that all three know.
 
And for every extra red eyed monk, we'd get another level.
 
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Re: BROWN EYES AND RED EYES  
« Reply #206 on: Jan 28th, 2009, 2:16am »
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soo..
 
0 reds = evil tourist. mass suicide on 1st night (all think they are the red eye monk)
 
1 red = red eye monk thinks "i don't see any monks w/ red eyes.." suicide on 1st night
 
2 reds = all red eye monks think "i see one but he didn't kill himself last night. uh oh.." deaths on 2nd night
 
3 reds = all red eye monks think "i see two, but they didn't kill themselves last night. uh oh.." deaths on 3rd night.
 
etc...
 
effects of tourist's statement:
 
-brings the red eye issue to the front of all monks' minds.
 
-sets a reference in time for all monks to count and figure out how man red eyes there are.
 
-if there is only one red eye monk, he is introducing new info to that one monk.
 
brown eye monks always count one more red eye monk than red eye monks count(except in the case of zero red eye monks), so they wait one day longer before thinking about killing themselves.
when the suicides occur the day before.. all the brown eyes can relax.
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Re: BROWN EYES AND RED EYES  
« Reply #207 on: Jan 28th, 2009, 5:43am »
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That just about covers it, yep. Smiley
 
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Re: BROWN EYES AND RED EYES  
« Reply #208 on: Apr 13th, 2009, 2:41pm »
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I think you got through to my thick skull.  Still a bit confusing, but I finally do see the light. ( I think)
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Re: BROWN EYES AND RED EYES  
« Reply #209 on: May 6th, 2009, 10:11pm »
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OK the logic is clear. The riddle states that something drastic happens. The most drastic result is the mass suicide. Is it safe to draw a conclusion that the best answer for this riddle is a lying tourist starting a tragic mass suicide?
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Re: BROWN EYES AND RED EYES  
« Reply #210 on: Oct 28th, 2009, 1:37am »
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where's the problem statement .. i don't see it in the first page ?
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Re: BROWN EYES AND RED EYES  
« Reply #211 on: Oct 28th, 2009, 1:56am »
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on Oct 28th, 2009, 1:37am, Ankitt_sk wrote:
where's the problem statement .. i don't see it in the first page ?
It's at http://www.ocf.berkeley.edu/~wwu/riddles/medium.shtml#brownEyesRedEyes
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Re: BROWN EYES AND RED EYES  
« Reply #212 on: May 13th, 2013, 2:38pm »
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Cool riddle
 
Got me puzzled for some time, but now im pretty sure that
COMMON SOLUTION IS WRONG
(meaning n monks die on n-th day)
 
Proper solution in my opinion is
1, 2, 3 red eyed Monks - they die
4 or more red eyed monks - noone dies
 
Hint: Why 4 red eyed monks expect 3 other to die on 3rd night?
« Last Edit: May 13th, 2013, 2:40pm by riddler358 » IP Logged
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Re: BROWN EYES AND RED EYES  
« Reply #213 on: May 13th, 2013, 10:14pm »
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Hint: For the reason why you think three monks would die on the third night if there were three.
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Re: BROWN EYES AND RED EYES  
« Reply #214 on: May 14th, 2013, 8:46am »
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This reason doesn't hold for 4 monks
 
I guess noone will even try to think about this riddle again so i'll just put an explanation here
 
 
--- EXPLANATION:
 
Something will happen only if any monk might think that some monk sees just one red eyed monk and will expect him to react on new knowledge
 
Otherwise no monk would ever think that any monk might think that someone will suicide on the first night.
 
 
--- 3 and 4 CASES from point of view of red eyed:
 
3 case - I see 2 red eyed monks and if he have brown eyes, they would die on 2nd night. WE DON'T STOP OUR REASONING HERE. Why I think these 2 would die? Because each of them would see just 1 in that case. Therefore each of them would expect this other guy to die. We can deduce from this point.
 
4 case - i see 3 red eyed monks, if i have brown eyes they would die on third day. AGAIN WE DON'T STOP HERE. Why I think these 3 would die? Because each of them would see just 2 in that case and expects them to die on 2nd day. AGAIN WHY? Because each of these 2 would see just 1 in that case. WE STOP HERE. That's impossible. I know they all see at least 2. Noone will ever die.
 
 
--- IN OTHER WORDS AGAIN
 
With 4 red eyed monks it's not possible to make assumption that anyone might think that anyone might think that there is just one red eyed monk. We can paste this phrase more times it doesn't change the fact. With 4 red eyed monks it's not possible to make assumption that anyone might think that anyone might think that anyone might think that anyone might think that anyone might think that anyone might think that there is just one red eyed monk. Every monk knows that every monk knows that every monk sees at least 2 red eyed monks. Every reasoning based on the assumption that anyone might think that anyone might think that someone will die on first night is not valid.
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Re: BROWN EYES AND RED EYES  
« Reply #215 on: May 14th, 2013, 9:04am »
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The thing you fail to understand is that for the three monks that the four monks hypothesize, two monks can be hypothesized that hypothesize there is only one. And each day one of these hypotheses fails.
If I'm the fourth monk, than I imagine that if there were only three, they would kill themselves on the third day; because if there were only three monks, and I was the third, I'd imagine that if there were only two monks, they'd kill themselves on the second day and so if they didn't I'd kill myself along with them on the third. So when no one kills themselves on the third day, I know I must be a fourth red-eyed-monk, because I can only see the three that would have killed themselves if there wasn't a fourth.
 
My master's thesis was on modal logic, trust me, the canon solution is correct. Just be sure to hypothesize a separate world with n-1 REMs to figure out what the nth REM would conclude.
I think that somewhere in this thread there's an illustration for up to 5 REMs. (Or in a similar thread; since this puzzle comes in many guises.)
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Re: BROWN EYES AND RED EYES  
« Reply #216 on: May 14th, 2013, 9:12am »
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Another consideration:
 
If there are 4 REMs
every non-REM knows everyone can see at least 3 REMs
every REM knows only everyone can see at least 2 REM
 
every non-REM knows that everyone knows that everyone can see at least 2 REMs
every REM knows only that everyone knows that everyone can see at least 1 REM
 
every non-REM knows that everyone knows that everyone knows that everyone can see at least 1 REM
But no REM knows that, so it is NOT common knowledge.
 
No non-REM knows that everyone knows that everyone knows that everyone knows that everyone can see at least 1 REM
 
 
That's why it matters when it is announced, because that makes it common knowledge, in the sense that only after that moment everyone knows that everyone knows that everyone knows that everyone knows that everyone knows ... that there is at least 1 REM, which starts the inevitable fatal cascade.
« Last Edit: May 14th, 2013, 9:26am by towr » IP Logged

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Re: BROWN EYES AND RED EYES  
« Reply #217 on: May 14th, 2013, 10:30am »
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You know, the thing I find really weird, is that at the same time you say that if there are three red eyed monks, they'll kill themselves on the third night. But if I see three red-eyed monks and they're alive after the third night, you don't think it's a valid conclusion that I must be a fourth.
That's not consistent.
 
 
It really is this simple:
* If I don't see any REM, even though I know there must be one, then it must be me, and I kill myself.
>> Therefore if there is one REM, he kills himself on the first night.
 
* If I see one REM, and he doesn't kill himself on the first night, even though I know that if there is only one REM he will kill himself on the first night, then I must be a second REM, and thus I kill myself on the second night (along with the other).
>> Therefore if there are two REMs they all kill themselves on the second night.
 
* If I see two REMs, and they don't kill themselves on the second night, even though I know that if there are only two REMs they will kill themselves on the second night, then I must be a third REM, and thus I kill myself on the third night (along with the others).
>> Therefore if there are three REMs they all kill themselves on the third night.
 
* If I see three REMs, and they don't kill themselves on the third night, even though I know that if there are only three REMs they will kill themselves on the third night, then I must be a fourth REM, and thus I kill myself on the fourth night (along with the others).
>> Therefore if there are four REMs they all kill themselves on the fourth night.
 
* If I see N-1 REMs, and they don't kill themselves on the N-1st night, even though I know that if there are only N-1 REMs they will kill themselves on the N-1th night, then I must be an Nth REM, and thus I kill myself on the Nth night (along with the others).
>> Therefore if there are N REMs they all kill themselves on the Nth night.
 
Simple induction is the easiest proof.
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Re: BROWN EYES AND RED EYES  
« Reply #218 on: May 15th, 2013, 6:58am »
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on May 14th, 2013, 8:46am, riddler358 wrote:
--- 3 and 4 CASES from point of view of red eyed:
 
3 case - I see 2 red eyed monks and if he have brown eyes, they would die on 2nd night. WE DON'T STOP OUR REASONING HERE. Why I think these 2 would die? Because each of them would see just 1 in that case. Therefore each of them would expect this other guy to die. We can deduce from this point.
 
4 case - i see 3 red eyed monks, if i have brown eyes they would die on third day. AGAIN WE DON'T STOP HERE. Why I think these 3 would die? Because each of them would see just 2 in that case and expects them to die on 2nd day. AGAIN WHY? Because each of these 2 would see just 1 in that case. WE STOP HERE. That's impossible. I know they all see at least 2. Noone will ever die.

 
But what about the 3 case from the brown-eyed viewpoint?
 
Red eyes - I see 2 red eyed monks and if I have brown eyes, they would die on 2nd night. WE DON'T STOP OUR REASONING HERE. Why I think these 2 would die? Because each of them would see just 1 in that case. Therefore each of them would expect this other guy to die. We can deduce from this point.
 
Brown eyes - i see 3 red eyed monks, if i have brown eyes they would die on third day. AGAIN WE DON'T STOP HERE. Why I think these 3 would die? Because each of them would see just 2 in that case and expects them to die on 2nd day. AGAIN WHY? Because each of these 2 would see just 1 in that case. WE STOP HERE. That's impossible. I know they all see at least 2. Noone will ever die.
 
So in the 3 REM case, the brown-eyed monks have "proved" that no-one will die. That's a problem because we've just got done proving that the three REMs will die, so something's wrong - either the three REMs won't die because the BEMs are right, or the REMs are right and will die. Either way, the logic of the other group must be wrong, but we've assumed that the monks are all (and know each other to be) expert logicians, so any conclusions reached by any monk must be correct - the monks have different information to start with, but that just means that their correct conclusions don't have to be identical - they each have to include the true outcome, so they still have to overlap.
 
In the 3 case, each REM concludes: Either those two die on the second night, or those two and myself die on the third night. In the usual solution, each BEM concludes: Either those three die on the third night, or those three and myself die on the fourth night. In your solution, each BEM concludes: No-one dies.
 
The usual solution provides an overlap between the two conclusions - that all three REMs die; your solution does not. That doesn't prove that the usual solution is correct, but it does prove that your solution is wrong.
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Re: BROWN EYES AND RED EYES  
« Reply #219 on: May 16th, 2013, 9:55am »
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on May 14th, 2013, 10:30am, towr wrote:
You know, the thing I find really weird, is that at the same time you say that if there are three red eyed monks, they'll kill themselves on the third night. But if I see three red-eyed monks and they're alive after the third night, you don't think it's a valid conclusion that I must be a fourth.
That's not consistent.

 
It makes it obvious after reading this. Thanks.
 
It's still a bit confusing for me because in cases 4 and more evey monk knows that every other monk knows that noone will die on first day. And still they use this possibility to contradict their nested hypothesis.
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Re: BROWN EYES AND RED EYES  
« Reply #220 on: May 16th, 2013, 1:21pm »
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Well, look at it this way; if you see three red-eyed monks, you're in one of two possible worlds:
either there are 4 red-eyed monks, and you're one of them,
or there are three red-eyed monks and you're not one of them.
 
For every red-eyed monk in the first possible world, the options they can consider are exactly the same; they would see what you see think what you think.
But in the second case, the red-eyed monks are in a different situation; so not everything you know and see and think necessarily applies to them. They would consider either a world where they are one of three red-eyed monks, or one where they're not one of two red-eyed monks.
 
Basically, the same knowledge does not automatically apply to the world you're in and the worlds you can imagine someone else might think to be in.
The three REMs consider there might be only two REM, who imagine there might be only one REM, who might have imagined there wasn't one until some bloody tourist ruined his ignorance. Each shift in perspective and into a deeper nested possible-world can change what knowledge is available.
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Re: BROWN EYES AND RED EYES  
« Reply #221 on: May 18th, 2013, 4:53am »
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I know it works like you said,
 
still
on May 16th, 2013, 1:21pm, towr wrote:
The three REMs consider there might be only two REM, who imagine there might be only one REM, who might have imagined there wasn't one until some bloody tourist ruined his ignorance.

for me that's the confusing part, because on the other hand everyone knows that everyone knows that this is only imaginary situation (for cases 4 and more)
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Re: BROWN EYES AND RED EYES  
« Reply #222 on: May 18th, 2013, 1:01pm »
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Sorry, I come in a bit late, but maybe here is a less confusing way to explain the solution.
 
Imagine there are N islands, numbered from 1 to N and on island n there are n REMs.  But nobody knows on which island they are, except that if they see k REMs, they can tell they are on island k or k+1.  As before, the tourist goes to each island and tells there is at least one REM.  The bastard.
 
On the first night, the REM on island 1 commits suicide.
 
Now everybody on island 2 knows they are not on island 1.
 
On the second night, the 2 REMs on island 2, seeing only 1 REM and knowing they are not on island 1, conclude they are a REM and commit suicide.
 
Now, everybody on island 3 knows they are not on island 2.
 
On the third night, the REMs on island 3, seeing only 2 REMs and knowing they are not on island 2, conclude they are a REM and commit suicide.
 
Now, everybody on island 4 knows they are not on island 3.
 
Etc.
 
This makes clear that the hypothetical case you reason is about another world that can have a number of monks that contradicts what you observe.
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Re: BROWN EYES AND RED EYES  
« Reply #223 on: May 19th, 2013, 4:04am »
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Yeah but monks on island 4 all knew at the very beginning that everyone on island 4 knows that on island 1 and 2 noone will die, that's the confusing part
 
my initial mistake was by this reasoning - why would 4 monks bother at all what would happen to 1 red eyed monk if they all know that everyone knows it's not possible there is only one
 
and now the best explanation i got for ppl who think the same is - ok, 4 monks might bother only about 3 monks case, but these 3 monks (imagined by 4 monks) bother about 2 monks case, and these 2 monks (imagined by 3 monks that are imagined by 4 monks) they bother about 1 monk case, therefore 4 monks are indirectly bothered by 1 monk case.
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Re: BROWN EYES AND RED EYES  
« Reply #224 on: May 20th, 2013, 10:44am »
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The "simplest" explanation is probably the superficial: "A monk who sees 3 REMs knows that either those 3 will die on the third night, or he's the fourth REM".
 
Provided you can prove that, when there are only 3 REMs, they will die on the third night, the above explanation is mathematically rigorous (okay, you also have to assume that all the monks are at least as good at logic as you are, and that it's common knowledge among the monks that that is the case - but you need that assumption to get anywhere anyway - if the monks don't trust each other's ability to follow the logic, then with more than a couple of REMs, everyone lives anyway)
 
That simple proof also generalises to an inductive proof that, when there are n REMs, they will all die on the nth night.
 
You can also argue directly that 4 REMs each hope that the other 3 only see 2 REMs, so hope those 2 each only see 1 REM (despite the original REM knowing that that couldn't be true) and the hypothetical 2 monks would each hope that the other saw no REMs and would face suicide that night, only to be disappointed, and suicide the second night, but the 3 monks would be disappointed and suicide the third night, but the original REM is, himself, doomed to further disappointment when the other 3 don't suicide, so kills himself on the fourth night.
 
On the other hand, any explanation like that is hard to follow, so people tend to find it unconvincing - which is fair enough - a long and confusing argument is more likely to slip in a subtle flaw without it being noticed, so, all else being equal, should be less convincing than one you can grasp in its entirety...
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