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S. Owen
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Re: BROWN EYES AND RED EYES  
« Reply #25 on: Jul 29th, 2002, 5:58am »
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Not true - say there are three red-eyed monks. They each see two others. It's quite possible, to each, that the other two are the only one red-eyed ones. The other two would not be able to figure out whether they both had red eyes or not, as you argue. None can deduce anything from this - this is stable.
 
0 red-eyed monks is stable, but not part of this problem.
1 is stable, until the visitor arrives!
n is stable as long as n-1 is, along the same lines as above.
 
So indeed the whole thing falls down when the visitor arrives, and monks can start making conclusions about the state of things.
 
And yes, I think it is assumed that they have all seen each other. They have to eat and do monk-chores every day! I don't think they have to see each other daily, but it doesn't matter... this is not important to the problem.
 
on Jul 29th, 2002, 1:16am, Frost wrote:
If, at any time, the monestary has >2 red-eyed monks, each monk can see at least two red-eyed monks. From this information they know each other monk knows at least one of them has red eyes. This results a Nth-day suicide.
...

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Re: BROWN EYES AND RED EYES  
« Reply #26 on: Jul 29th, 2002, 10:06am »
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I agree. Any system before the tourist's magic phrase is stable.
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pa0pa0
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Re: BROWN EYES AND RED EYES  
« Reply #27 on: Jul 30th, 2002, 8:53am »
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on Jul 29th, 2002, 5:58am, srowen wrote:
And yes, I think it is assumed that they have all seen each other. They have to eat and do monk-chores every day! I don't think they have to see each other daily, but it doesn't matter... this is not important to the problem.
 

 
Hmm, that's a fascinating question.  I *think* that the following is the case:
 
  (1)   IF death-yells are guaranteed to resound throughout the monastery, AND they are willing to accept the risk of the logic going haywire should all the red-eyed monks die in the interim of natural causes, THEN no one needs to see each other, or have any communication whatsoever, other than listening out for and if necessary making death-yells, once the tourist has departed.  (They do all have to be together when the tourist makes his statement, if they are to start with common knowledge of his statement.)
 
Admittedly,  the situation is quite bizarre.  Suppose there are 100 monks,  of which 42 have red eyes.  When the tourist departs, everyone can go to their separate cells, and stay there.  Absolutely nothing appears to happen, until midnight on the 42nd night, when unearthly yells start resounding through the monastery.  (The yells don't need to be synchronized, nor do the remaining monks need to try counting the yells;  everyone already knew that  if any yell occurs on the 42nd night, then there will be 42 yells.)
 
The monks' algorithm is even more simple than the above might suggest:  actually, each monk only has to listen out for death-yells on one particular night.  His algorithm is:  if at the tourist's gathering (on day 1) I saw R red-eyed monks, and there are no death-yells on night R, then I die on night R+1.
 
One might ask, how can that possibly be?  How can it take 42 days for something to happen, when nothing appears to happen in the interim to provide any relevant lead-up?  It's not as though *absence* of yells on the 17th night (for example) is needed in order to increase anyone's knowledge on day 18.  Everyone already knew that there wouldn't be any yells on the 17th night.
 
Bizarre, yes.  But anyway, when one thinks about it a bit more, the situation isn't really any more bizarre than (and is in fact identical to) the normal scenario where the monks see each other (but don't say anything) every day.
 
 (2)  If death-yells don't work, then I think the situation is much more complicated.  Let us suppose, for definiteness, that the only time when the monks can be sure of seeing each other (if they are still alive) is at breakfast.  And again, let us have 42 red-eyed monks (Rs) and 58 brown-eyed monks (Bs).
 
Now, before the tourist arrives, each R knows that there are either 41 or 42 Rs, and each B knows that there are either 42 or 43 Rs.  That knowledge stays the same after the tourist's statement on day 1, but they now additionally know the following:
 
Each R knows that:
 
   Either at the end of day 41, 41 monks (not including  
   himself) will die;  or at the end of day 42, 42 monks  
   (including himself) will die.
 
And each B knows that:
 
   Either at the end of day 42, 42 monks (not including  
   himself) will die;  or at the end of day 43, 43 monks  
   (including himself) will die.
 
Thus everyone knows that no one will die (unless of natural causes) until the end of day 41 at the earliest.  So does anyone actually need to see each other before that time?  The natural inclination of monks being to fast, couldn't they all just skip breakfast until day 41?  Or even day 42?  After all, each individual knows that although breakfast on day 41 might be interesting to someone else, it won't be interesting to himself.
 
Not only that, but perhaps the Bs don't even need to turn up until day 43?  (The fact that they don't know yet that they are Bs is not relevant, at least not in any straightforward way.   And if they don't turn up on day 42, that will certainly tell the Rs something, but it is something that those Rs are going to discover anyway.)
 
I'm still thinking about that.  Does anyone have the answer?  I have a strong feeling that fasting is illegal - not because it disturbs the logic of the riddle, but because it violates the rule that the monks mustn't reveal anything about each others' eye colour.  Except that if it does violate that rule, it could cause an R to die too early, which *would* disturb the logic of the riddle.  Hmm...
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pa0pa0
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Re: BROWN EYES AND RED EYES  
« Reply #28 on: Jul 30th, 2002, 9:17am »
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P.S. to my note a moment ago:  upon reading it through now that it's all nicely formatted by the forum software, I've suddenly realized how the absence of any yells on night 17 *does* increase everyone's knowledge on day 18, even though everyone already *knew* there wouldn't be any yells on night 17.
 
Consider propositions of the form "everyone knows that everyone knows that ... (J times) ... that there won't be any yells on night 17".
 
Before night 17, those propositions were true for values of J only up to some critical value K (round about 16, but if I try to work it out exactly, I'll get it wrong).
 
On day 18, those propositions are true for *all* values of J, since the absence of yells on night 17 has now become common knowledge.
 
(In fact, I think the logic of the riddle only requires that the proposition become true for the single extra value of K+1, so the absence of those yells is telling the monks more than they need to know!)
 
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Re: BROWN EYES AND RED EYES  
« Reply #29 on: Jul 30th, 2002, 4:19pm »
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Assuming that suicide is a sin if you do not have red eyes, then the only way a monk would kill himself after that statement is if he was the only one with red eyes.  Otherwise he would see someone else with red eyes, assume it was not him, and not suicide.
 
If the tourist was looking at a group photo, and saw red eye, not red eyes, then they would all suicide because they would see no one else with red eyes and have to assume it was themselves.
 
If more than one had red eyes, both would assume the other was alone and neither would suicide.
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S. Owen
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Re: BROWN EYES AND RED EYES  
« Reply #30 on: Jul 30th, 2002, 4:56pm »
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See the preceding discussion - we've been over this.
 
Imagine you see one other red-eyed monk. What do you think to yourself when, the following morning, you see him again in the courtyard? What must he see, and what does it mean about you? Extend from there.
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Re: BROWN EYES AND RED EYES  
« Reply #31 on: Aug 2nd, 2002, 9:30am »
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While I don't dispute the fact assuming that all of the monks have seen each others eyes and all of the monks heard the tourist, what we must also assume is that these monks must never gesture or write someone a note, or hand them a knife or a rope, letting them know that they have red eyes.  I'm willing to assume that this red-eye curse thing is a tenent of their religion, that as monks(I'm assuming they would be devout followers of said religion) they would be morally obligated to somehow signal the red-eyed monks what their true eye-color is.  With that in mind, I see only two possiblities:
 
1) the monks are devout followers of their religion, the tourist lied and they commit mass suicide.
 
2) no one commits suicide. the monks with red eyes have been living there for this long with all of the other monks knowing that their eyes were red not making a big deal about it, the tourist didn't know anything about the eye curse, and with the monk's all having taken a vow of silence, who's going to say anything?
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Tony Mountifield
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Re: BROWN EYES AND RED EYES  
« Reply #32 on: Aug 13th, 2002, 8:22am »
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But surely:
If there were 2 or more monks with red eyes, and they could all see each other, they ALL already knew that "at least one of them has red eyes". So the visitor did not add any new information. He could not then have kickstarted a countdown of days.
 
The only way the visitor increased the information known by any of the monks was if there was only one with red eyes. The red-eyed monk would then know that it was him, since he could only see brown eyes, and commit suicide on the first night.
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S. Owen
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Re: BROWN EYES AND RED EYES  
« Reply #33 on: Aug 13th, 2002, 8:51am »
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Good point: the key is that the visitor's comment tells them more than that: all the monks now know for sure that all the monks must know that at least one has red eyes.
 
In the case of two monks with red eyes, agreed - both already know that at least one has red eyes. We, the riddlers that posit this situation, "know that."
 
But crucially, each of those two monks doesn't know that the other knows that, because each is not sure of his own eye color. Monk A might see that B has red eyes, but doesn't know if he himself has red eyes. Therefore he isn't sure if B knows that "at least one monk has red eyes."
 
Once the visitor makes his comment, things change, not because they now know "at least one monk has red eyes" - they each knew that - but because they are sure the other knows it. Then the chain of reasoning starts to get somewhere.
 
This holds for any number of red-eyed monks.
« Last Edit: Aug 13th, 2002, 8:52am by S. Owen » IP Logged
tot
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Re: BROWN EYES AND RED EYES  
« Reply #34 on: Aug 13th, 2002, 1:35pm »
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If the monks see each other every day, they already knew for sure that there is at least one monk with red eyes.  The visitor did not give any new information to the monks.
 
What he really did was a reference point in time that is known to everybody.  They needed day 0 to make the Nth day mass suicide.
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S. Owen
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Re: BROWN EYES AND RED EYES  
« Reply #35 on: Aug 13th, 2002, 1:44pm »
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on Aug 13th, 2002, 1:35pm, tot wrote:
If the monks see each other every day, they already knew for sure that there is at least one monk with red eyes.  The visitor did not give any new information to the monks.

 
Yes, the visitor did give them new information. See above - after that they all know that all monks know that at least one has red eyes. This is what allows them to start deducing things about themselves.
 
This "marking a reference point in time" changes nothing about their deductions or information, so that alone cannot be the new thing that makes some of them deduce that they have red eyes.
« Last Edit: Aug 13th, 2002, 1:46pm by S. Owen » IP Logged
Tony Mountifield
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Re: BROWN EYES AND RED EYES  
« Reply #36 on: Aug 13th, 2002, 1:48pm »
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OK, I don't get the bit about a time reference, but I can now see S Owen's reasoning for 2 red-eyed monks. Each of them could only see one red-eyed monk. But they didn't know whether they were also red-eyed. So A would reason: "if I have brown eyes, then B can't see a red-eyed monk - now he knows there is one, he has to conclude it is himself and commit suicide tonight. If he doesn't, then I must have red eyes too." B reasons similarly, and the following day both A and B see the other alive and realise they each have red eyes.
 
I'm still trying to fathom how it works with 3 red-eyes, because the ALL the monks already could see more red-eyes than the visitor mentioned, AND knew that all the others could already see at least one.  Undecided
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S. Owen
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Re: BROWN EYES AND RED EYES  
« Reply #37 on: Aug 13th, 2002, 1:53pm »
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Well imagine there are three red-eyed monks. Each sees two others. Each is not sure whether the others are the only two or not.
 
You see how it works for one and two red-eyes monks, yes? One kills himself the first night; or both kill themselves on the second night, respectively.
 
So if on the third morning, you still see those same two red-eyed monks, then you know they must be seeing another set of red eyes - yours!
 
The reasoning in this situation still relies indirectly on knowing that everyone knows there is at least one red-eyed monk. Until you're sure everyone knows that, it doesn't mean anything that those two other red-eyed monks haven't killed themselves. Once everyone knows that, you can infer things from who is and isn't killing themselves.
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Re: BROWN EYES AND RED EYES  
« Reply #38 on: Aug 13th, 2002, 2:02pm »
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on Aug 13th, 2002, 1:48pm, Tony Mountifield wrote:
I'm still trying to fathom how it works with 3 red-eyes, because the ALL the monks already could see more red-eyes than the visitor mentioned, AND knew that all the others could already see at least one. Undecided

 
OK, I get it now. For 3 red-eyes: A reasons "if B and C are the only two, they will know that tomorrow morning, and will die tomorrow night." When B and C survive another night, A knows he has red eyes too. Same for B and C, by symmetry.
 
For 4 red-eyes, he can see the other three. When they survive a day longer than expected, he and they all know.  
 
Yes?
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Re: BROWN EYES AND RED EYES  
« Reply #39 on: Aug 13th, 2002, 2:04pm »
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(still waiting for my password)
 
Yes, I'm glad to say I worked it out and didn't see S Owen's last post (thanks anyway!) until I'd posted mine! Right onto the next.....
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Re: BROWN EYES AND RED EYES  
« Reply #40 on: Aug 14th, 2002, 4:46pm »
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on Aug 13th, 2002, 1:48pm, Tony Mountifield wrote:
OK, I don't get the bit about a time reference, but I can now see S Owen's reasoning for 2 red-eyed monks.   Undecided

 
What I mean by time reference, is that every monk needs to have a day 0 to start the logic that S. Owen has already defined.
 
It really does not matter exactly what the visitor says, as long as it is a clear reference to all the monks start to wonder why the red-eyed monks have not yet committed suicide, even they should have done it already.  
 
The riddle just says that the current situation is what it is and that there are red-eyed monks in the island.  All that is needed is some way to start the counting of nights to find out how many red-eyed monks everyone else sees.
 
Of course, if the red-eyed monks do not follow this logic but the brown-eyed ones do, it could be that brown-eyed ones commit suicide instead on Nth + 1 night, or just part of the red-eyed do,  or that everybody commits suicide next night thinking they are the missing red-eyed one.  But I believe that the intetion of the riddle is the Nnth night sucide.
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S. Owen
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Re: BROWN EYES AND RED EYES  
« Reply #41 on: Aug 15th, 2002, 7:35am »
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on Aug 14th, 2002, 4:46pm, tot wrote:

The riddle just says that the current situation is what it is and that there are red-eyed monks in the island.  All that is needed is some way to start the counting of nights to find out how many red-eyed monks everyone else sees.

 
No, a point from which to count nights is decidedly not all that they need.
 
If the visitor had simply said, "Hey monks! Tonight is night zero, if you know what I mean!" then nothing would have happened. No monk would then be able to deduce anything more about his own eye color.
 
It does matter what the visitor says - the crucial fact that he communicates to all the monks is that "all monks now know that at least one monk has red eyes." This new information, not the simple fact that anything at all was said, allows a chain of reasoning to begin.
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pa0pa0
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Re: BROWN EYES AND RED EYES  
« Reply #42 on: Aug 15th, 2002, 1:06pm »
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S.Owen, I don't think you've quite got the nub of it yet.
 
on Aug 15th, 2002, 7:35am, S. Owen wrote:

It does matter what the visitor says - the crucial fact that he communicates to all the monks is that "all monks now know that at least one monk has red eyes." This new information, not the simple fact that anything at all was said, allows a chain of reasoning to begin.

Well, that was not new information if there are at least 2 red-eyed monks.
 
You, 13th Aug
 
>Yes, the visitor did give them new information. See above -
> after that they all know that all monks know that at least  
>one has red eyes. This is what allows them to start  
>deducing things about themselves.  
 
Better, but no cigar.  That was not new information if there were at least 3 red-eyed monks.
 
If you had said "The visitor did give them new information:  after his statement they all know that all monks know that all monks know that at least one has red eyes" - still no cigar.  That was not new information if there were at least 4 red-eyed monks.
 
It is precisely this difficulty which is causing people in this thread to keep going back to square 1 and ask what new information did the tourist really give.  Please think carefully about the answer to this.
 
Does anyone have any views on whether the monks are allowed to fast? (ref my earlier note, July 30th)
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Jonathan_the_Red
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Re: BROWN EYES AND RED EYES  
« Reply #43 on: Aug 15th, 2002, 1:32pm »
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Yes, thinking about the problem this way really highlights just how delicate the induction is.
 
Suppose there are N monks with red eyes. Pick a random red-eyed monk (REM). He knows there are either N or N-1 REMs. He also knows:
 
If there are N-1 REMs, they know there are either N-1 or N-2 REMs, and they also know:
If there are N-2 REMs, they know there are either N-2 or N-3 REMs, and they also know:
If there are N-3 REMs, they know there are either N-3 or N-4 REMs, and they also know:
...
If there are 2 REMs, they know there are either 2 or 1 REMs, and they know also know:
If there is 1 REM, he knows there are either 1 or 0 REMs.
 
The tourist's statement settles the indecision of this hypothetical lone REM at the very bottom of this long chain of "if"s. One by one, day by day, the premises are shattered, until the chain reaches the top and all the REMs realize that there are in fact N REMs, not N-1, and they all kill themselves.
 
Or... the monks, being smart, realize what's going to happen, all draw straws, with the one who picks the short straw selecting a REM at random and telling him: "You have red eyes." That one guy kills himself, the induction is broken, and the rest of the monks live happily ever after, mourning their lost brother but celebrating his sacrifice for the good of all. A month later, they get sensitivity training, get over their REMophobia, and learn that it's okay to have red eyes and nothing to kill oneself over.
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S. Owen
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Re: BROWN EYES AND RED EYES  
« Reply #44 on: Aug 15th, 2002, 2:23pm »
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on Aug 15th, 2002, 1:06pm, pa0pa0 wrote:
Better, but no cigar.  That was not new information if there were at least 3 red-eyed monks.
 
If you had said "The visitor did give them new information:  after his statement they all know that all monks know that all monks know that at least one has red eyes" - still no cigar.  That was not new information if there were at least 4 red-eyed monks.
 
It is precisely this difficulty which is causing people in this thread to keep going back to square 1 and ask what new information did the tourist really give.  Please think carefully about the answer to this.

 
OK, and I ask that you think carefully about this! I agree that my argument needs a more complete statement, but I maintain that it is correct.
 
I understand and agree with your point about new information in the case of more than 2 red-eyed monks.
 
For more than 2 red-eyed monks, I agree that the "new information" does not directly affect the reasoning of each monk - it does however affect it indirectly, if you will.
 
Take 4 red-eyed monks.
Each of them doesn't know whether there are a total of 3 or 4 red-eyed monks. Each is wondering whether the other 3 are the only red-eyed monks.
And thus, they wonder whether those 3 are themselves wondering whether there are 2 or 3 red-eyed monks.
And thus, whether those 3 monks are wondering whether the 2 others are wondering whether there are 2 or 1. And thus... whether each of those 2 is wondering whether the other is the only one or not.
 
But that's where it all stops, at this "base case" of 2 monks wondering about each other. They can't figure anything out, and so won't do anything. Knowing this, the supposed 3 monks wondering about the other 2 won't do anything. Knowing this, the 4 monks wondering about the other 3 won't do anything.  
 
Therefore, the fact that nobody is committing suicide does not provide any information.
 
But when they learn from the visitor that "all monks know that at least one monk has red eyes" (and I stress that this is all they need), then the base case is no longer indeterminate. Now two monks can figure it out. Now it does mean something when they do or do not kill themselves.
 
Thus it is precisely the new fact that "all monks know that at least one monk has red eyes" that drives the chain of reasoning.
 
Or to put it another way, what precisely are you arguing? That there is not a mass suicide on the Nth night? Or that there is, but that it's somehow the "marking a point in time" that makes it possible?
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Re: BROWN EYES AND RED EYES  
« Reply #45 on: Aug 16th, 2002, 7:23am »
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on Aug 15th, 2002, 2:23pm, S. Owen wrote:

 
...But when they learn from the visitor that "all monks know that at least one monk has red eyes" (and I stress that this is all they need), then the base case is no longer indeterminate.  
...  
Thus it is precisely the new fact that "all monks know that at least one monk has red eyes" that drives the chain of reasoning.

 
Hmm.  I invite you to reconsider your "I stress that this is all they need".  In particular, I argue that to make the riddle work, it is not sufficient that they "learn from the visitor that all monks know that at least one monk has red eyes", as you put it.
 
To try to make this clear, let me introduce a sequence of scenarios.  To make these scenarios work, it has to be the case that there is no question about the trustworthiness of the tourist (that also has to be the case in the original riddle).
 
Scenario 1:  the tourist goes round to each monk, whispering to each "at least one monk has red eyes" and then leaves.
 
Scenaro 2:  as scenario 1, but before leaving the tourist goes around again, whispering to each "in my first round I whispered the same thing to each monk, namely that at least one monk has red eyes".
 
Scenario 3: as scenario 2, but before leaving the tourist goes around again, whispering to each "in my second round I whispered the same thing to each monk, namely that in my first round ..."
 
I hope you will agree with the following:
 
- that in scenario 1, each monk learns from the visitor that at least one monk has red eyes;
- that in scenario 2, each monk learns from the visitor that all monks know that at least one monk has red eyes - which according to you is sufficient to make the riddle work.
 
I hope you will also agree
 
- with scenario 1, if there is more than 1 red-eyed monk, then nothing will happen;
- with scenario 2, if there are more than 2 red-eyed monks, then nothing will happen;
- with scenario 3, if there are more than 3 red-eyed monks, nothing will happen.
 
(In thinking this through, you have to keep bearing in mind that each monk learns the contents of all whispers in non-final rounds, but knows the content of only one whisper in the final round.  Suppose that there are 3 red-eyed monks A, B and C.  The problem is that in scenario 2, C never knows whether B knows that A knows that there is at least one red-eyed monk, because he is not privy to the tourist's final whisper to B.)
 
Implication: the monks must be learning something more from the tourist than you have nailed down.
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Re: BROWN EYES AND RED EYES  
« Reply #46 on: Aug 16th, 2002, 8:19am »
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I agree with your analysis of your scenarios, but aren't they slightly different than the riddle's scenario?
 
In the original riddle, the visitor tells them all at once that "at least one monk has red eyes", presumably in a group. Your scenarios make a different assumption.
 
That is, in the riddle they all have all the first-order, second-order, etc. knowledge at the outset, instead of in progressive rounds.
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Re: BROWN EYES AND RED EYES  
« Reply #47 on: Aug 16th, 2002, 11:45am »
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Jonathan has the right logic.  
 
The new information is that all monks know that "If there were 1 REM, he would know it now". If there are >2 REMs then they will all know that this information doesn't immediately apply to them, but it is still new information. Each morning they learn new information about a hypothetical which they know isn't realized. Note that they will all be able to predict what they will learn up until the next to last morning since they will know the hypothetical for that morning is false, but thats not the same as actually learning it because the statement wouldn't be true until that morning. The morning prior to the suicides, the hypothetical they learn about will be one which the REMs will know is realized.
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Re: BROWN EYES AND RED EYES  
« Reply #48 on: Aug 16th, 2002, 11:49am »
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on Aug 16th, 2002, 8:19am, S. Owen wrote:
Your scenarios make a different assumption...
 ... in the riddle they all have all the first-order, second-order, etc. knowledge at the outset, instead of in progressive rounds.

Yes, that's exactly it, and the point of my scenarios was to illustrate how the essential effect of the tourist's visit in the original riddle is to give them this nth-order etc. knowledge without which the riddle doesn't work.
 
(Incidentally, for any particular number of red-eyed monks, there is a number of rounds in my scenario which would make the riddle work.)
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Re: BROWN EYES AND RED EYES  
« Reply #49 on: Aug 16th, 2002, 12:01pm »
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Thats true. The "if there were 1 he would know it now" has to be known to the nth order otherwise we stop learning about hypotheticals before we reach the one that applies.
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