Date: October 7th, 2014 5:37 PM Author: Flirting plum public bath
stops the recursion, right? A dragon doesn't sparrow when he is no longer certain that he doesn't have green eyes, only when he knows that he has green eyes. As such, and dragon that is uncertain whether he has green eyes wouldn't sparrow.
From the Harvard explanation:
If N = 3, let the dragons be called A, B, and C. After your announcement, C will think to himself, “If I do not have green eyes, then A and B can see that I don’t, so as far as they are concerned, they can use the reasoning for the N = 2 situation, in which case they will both turn into sparrows on the second midnight.” Therefore, if A and B do not turn into sparrows on the second midnight, then on the third day C will conclude that he himself must have green eyes, and so he will turn into a sparrow on the third midnight. The same thought process will occur for A and B, so they will all turn into sparrows on the third midnight. The pattern now seems clear.
Why would perfectly logical dragon C think “If I do not have green eyes, then A and B can see that I don’t." In this situation, dragon C would think "I may or may not have green eyes, and therefore I cannot remove myself from the pool of dragons that may have green eyes." Based on this, dragon C cannot infer anything about the other dragons' knowledge when they are still around on the third night since from dragon C's perspective, it is possible, but not certain, that dragon A is not sparrowing because he believes that one other dragon has green eyes or that two other dragons have green eyes. Given this indeterminacy that is allowed by the logic in the problem , I don't understand how the N=2 deduction can be applied to N=3, which in turn prevents the recursion.
Date: October 7th, 2014 5:49 PM Author: Flirting plum public bath
I have been stuck on N=3 from the beginning. The Harvard answer relies on dragon C affirmatively believing it has non-green eyes. I don't see why dragon C would have this belief since its continued existence is not premised on this. All it needs to believe is that its eyes may or may not be green. Given this uncertainty, dragon C cannot divide dragon A and B into a separate population.
Date: October 7th, 2014 5:59 PM Author: beta university
Don't get caught up in the "remove yourself from pool" explanation, it's not helpful and glosses over the conceptual difficulty, it's just technically correct from a formalistic proof perspective.
The baseline belief for each of A, B, and C, pre-announcement, is this:
(1) I may have Brown (non-Green) eyes, but the other two have Green eyes.
(2) Each of the other two sees 1 definite Green eye, 1 maybe Brown eye (mine), and thinks that he himself has 1 maybe Brown eye.
(3) this is where it gets tricky: Each of the other two thinks that the other one sees 2 Brown eyes, and thinks that he himself has 1 maybe Brown eye.
Post announcement, each knows that (3) is no longer true.
(3) is very tricky because clearly, A knows that C sees at least one green eye, but A also thinks that B is an idiot who thinks that C sees 2 brown eyes. This is implausible on an intuitive level, but is logically allowed based on the premise of the story.
Date: October 7th, 2014 6:53 PM Author: indecent juggernaut
also there are 100 dragons on the island. they'd have to monitor each other very closely and think very carefully about the issue to turn into sparrows
if one of the dragons was sealclubber, he'd never turn into a dragon
Date: October 7th, 2014 5:51 PM Author: bright psychic coffee pot
N=3 and N=4 are easy for me. My problem is with N=100. I just can't fathom how the chain allows for any conceivable belief that the dragons all have brown eyes except for him.
Date: October 7th, 2014 6:03 PM Author: bright psychic coffee pot
So I gather, but even though the induction seems obvious it seems to lead to a very bizarre conclusion.
Actually, I think it can be expressed in this way: With 100 Dragons, Dragon 1 supposedly can envision Dragon 2 thinking something about Dragon 3 who can imagine somethign about Dragon 4, etc. The very end of this chain is supposedly some dragon who could conceivably not have realized there was a green-eyed dragon until the bitch said so. But this doesn't follow because ANY of the Dragons could have been Dragon 2. If we know that any Dragon could be Dragon 2, and that Dragon 2 sees many Green-eyed dragons, how can we end up reducing to the base case?
Date: October 7th, 2014 6:05 PM Author: Supple base corn cake
Try Judas Jones' way of thinking about it below. Imagine the dragons' reasoning takes the following form: "If I didn't have green eyes, X would have happened. But X didn't happen. Thus I have green eyes." Just solve for X.
Date: October 7th, 2014 6:09 PM Author: bright psychic coffee pot
It's all very easy as long as you can reduce it to the 3-dragon case, for me at least. If you can imagine at least one of the dragons believing he's in a three-dragon scenario looking at two other green-eyed dragons, it all checks out.
I see no such avenue to that three-dragon scenario once we're at a high number of dragons.
Date: October 7th, 2014 6:27 PM Author: Supple base corn cake
Can the four imagine the three? Can the five imagine the four? Can the ninety-nine imagine the ninety-eight? You are basically doubting that mathematical induction works. Don't doubt that.
Date: October 7th, 2014 6:39 PM Author: bright psychic coffee pot
I can't help it. No matter how many times I read the basic case, it seems utterly ludicrous for me to try conceiving of a circumstance where 20 dragons, all green-eyed, can have a cascade set off by somebody pointing out that one of them has green eyes, something they ALL know and ALL KNOW the other dragons know.
Date: October 7th, 2014 5:59 PM Author: maroon persian
At least 1 dragon has green eyes.
1 dragon = Dragon 1 knows its him immediately on day 1
2 dragons = a dragon (Dragon 2) watching Dragon 1 who would know immediately he has green eyes if Dragon 2 had non-green eyes. When Dragon 1 doesn't change at end of day 1, Dragon 2 knows he too has green eyes.
3 dragons = a dragon (Dragon 3) watching the 2 dragon scenario, and when neither change at the end of Day 2, he knows he also has green eyes
4 dragons = a dragon (Dragon 4) watching the 3-dragon scenario...at end of day 3 knows he's a dragon
Date: October 7th, 2014 6:10 PM Author: maroon persian
Dragon 3 knows both Dragon 1 and Dragon 2 have green eyes. If Dragon 3 had non-green eyes, the 2 green eyed drago scenario would have played out because Dragon 3 is a nonfactor.
Date: October 7th, 2014 6:11 PM Author: beta university
That's just dragon 3's perspective. It holds for dragon 1 and dragon 2's perspective as well. Each of them thinks that the other two are having the stare down.
Date: October 7th, 2014 6:19 PM Author: motley lodge
because they have to "know" their eyes are green to turn so they exhaust the possibilities of each other dragon. if he assumes he's in the stare down then he doesn't know it has to be him yet. after the 99th midnight every dragon has blamed the green on every other dragon and there's no one left but himself
Date: October 7th, 2014 6:09 PM Author: maroon persian
refer to the 2-dragon scenario. If Dragon 3 had non-green eyes, the 2-dragon (both green-eyed) scenario would have played out. Dragon 3 knows this because he is logical as fuck.
ok that makes sense. I wonder why it's so mentally cumbersome to perform the same operation at higher # of dragons in your head. it makes me want to smash my head against a wall
Date: October 7th, 2014 6:18 PM Author: bright psychic coffee pot
It's a little easier to think of it in terms of "Dragon A thinks he is a bystander to the 99-dragon scenario, who realizes Dragon B would then consider himself a bystander to the 98-dragon scenario, who realizes Dragon C would then consider himself a bystander to the 97-dragon scenario," all the way down. I get it. I get the inductive logic. But it still seems fundamentally paradoxical that "one of you has green eyes" is adding any kind of new information, when everybody knows that everybody knows that.
Date: October 8th, 2014 1:47 PM Author: frozen ticket booth
assuming they are in constant contact which isn't in the original question,
a dragon knows that every other dragon can see exactly what he sees with one exception, you can see his eyes and he can see yours
the other dragons just become a number that every dragon already knows.
you can still hold out hope that you are the lone exception, until every other dragon sticks around past the point that you could have been the exception.
Date: October 7th, 2014 6:14 PM Author: beta university
This problem illustrates the difference between (1) everyone in the group being rational and (2) everyone in the group being rational, and knowing that everyone else is rational as well.
The human's announcement puts the dragon population from (1) to (2). Before, they were just in (1).
Date: October 7th, 2014 6:19 PM Author: bright psychic coffee pot
No, this is not the case. It's already understood by every dragon that all dragons are rational. Nothing the bitch says changes that, because if the dragons think that other dragons may be irrational it's totally possible that Dragon B could simply be thinking "coat hangers."
Date: October 7th, 2014 6:27 PM Author: beta university
Ok, this is probably the way to address the implausible base scenario, i.e., each of the dragons thinking that in the chain of 100 dragons, there's one dragon who sees all brown eyes:
It's because at each level of frame jumping (to another dragon's perspective), one more mistake gets added, i.e. the mistake of that one dragon's lack of knowledge about his own eye color. So at the 100th level, there's 100 mistaken beliefs accumulated from each dragon.
Date: October 7th, 2014 6:24 PM Author: mildly autistic exhilarant black woman
There is nothing in the terms of the problem as stated that requires each dragon to assume as a fact that every other dragon sees 99 green eyed dragons.
Indeed wouldn't it be the case that if this was a fact then every dragon should have sparrowed long ago because if every dragon knew as a fact that every other dragon was also seeing 99 green eyed dragons then one of those dragons must be himself?
Because the problem allows the dragons to assume other dragons hold mistaken beliefs then the dragons can construct the chain of mistaken beliefs and rely on that not to sparrow. The new fact destroys the chain.
Date: October 7th, 2014 6:29 PM Author: beta university
no she's right. And it's not just the fact that the problem "allows" dragon to believe the others hold mistaken beliefs, it's that the problem inevitably forces the other dragons to believe that the others hold mistaken beliefs because the available evidence to the dragons at the time does not allow them to hold any other beliefs, logically speaking. It's quite neat.
Date: October 7th, 2014 6:34 PM Author: Supple base corn cake
The dragons are infinitely logical and hold no mistaken beliefs. The solution is basic mathematical induction. The idea of a "belief" in the first place is foreign to it. The dragons are completely accurate and thorough in their estimation both of the evidence and of their counterparts' reasoning.
Date: October 7th, 2014 6:37 PM Author: mildly autistic exhilarant black woman
Yes that is much more precise. That other dragons must hold mistaken beliefs is a necessary consequence of the beliefs each dragon must hold in order not to have sparrowed.
Isn't there some old thread where she tries to discuss some fairly complicated proofs or something?
I realize this goes nowhere in terms of proving she's a real person or not the shrewmo/SB/mydoghasmoredates troll. If I recall, the 'medea' poster claimed to be some sort of engineer and could capably discuss math too.
Also would like for her to hold forth on some geeky international tax law bullshit.
Date: October 7th, 2014 7:29 PM Author: Supple base corn cake
I don't know about the old proofs thread. I was not a math major but could probably pass for one here among non-STEM types. A complete failure to do mathematical induction without questioning the hypo doesn't speak well to her qualifications.
Exactly. A math major should be the person who looks at this and sees the problem for what it is, not bucks against it with some boneheaded rhetorical waxing about the dragons' subjective state of mind.
Date: October 7th, 2014 7:41 PM Author: mildly autistic exhilarant black woman
Why be so rude?
I agree "belief" is imprecise because the dragons being totally logical creatures must simply adopt the necessary state of mind that logic requires, they don't get to choose what to believe, but "belief" is convenient shorthand for "state of mind that logic requires the dragon to hold".
Date: October 7th, 2014 7:56 PM Author: mildly autistic exhilarant black woman
Well if you think its reasonably straightforward then you're obviously far more intelligent than me. I thought it was pretty difficult but I gave it my best shot yesterday.
Maybe my solution is wrong, though at least one poster who's deep into the issue thinks its right, or maybe its just the same reasoning as the published proof but approaching the problem the reverse way by reasoning down from 100 instead of up from 1. I've glanced at the published proof but I haven't directly compared them.
If you think its all straightforward then you can tell me where I went wrong yesterday.
You took the time to reason through and write that explanation, but only 'glanced' at the page and a half answer?
Good lord. Twist has explained why you're deviating from the rules of the problem a dozen times in a dozen ways. You can make the problem much more difficult and ambiguous by injecting assumptions are purposefully playing rhetorical games with the rules, but you're not supposed to do that. This isn't an original or debated problem. It's asking you to do specific things and the result is you arrive at the inductive chain outlined in the answer.
What kind of maths did you major in, just out of curiosity?
Twist has a view and lolcats4eva has a view, and I happen to agree with lolcats4eva. The competing arguments are all set out in the posts following and I don't feel like summarising them again.
If you think Twist is right go and argue with lolcats4eva and tell him he is wrong because I am not interested in using discussion over this problem as an excuse for you to insult me.
You seriously take my post as an excuse to insult you? The question about your major was entirely sincere.
I have read the debates and you can see I'm not entering them because I really have nothing to add. I am simply pointing out that as a math major and a fucking tax lawyer, your rejection of the rules clearly set out by the problem is pretty unusual. Read the actual answer and then state in three sentences or less why your reasoning is at all valid or makes more sense.
Like I said before, this isn't some kind of rare or controversial problem, nor is it a trick question. The rules set you up for the induction chain.
Date: October 7th, 2014 8:36 PM Author: mildly autistic exhilarant black woman
You want to be a dick to me because I had a serious go at trying to solve a problem 95% of xo got nowhere with then fine, go ahead, but in future please just pretend I don't exist and I'll extend you the same courtesy.
That way you won't need to deal with my stupidity and I won't need to deal with your rudeness, which I may say is completely uncalled for as I have never been anything other than unfailingly polite to you.
I'm not a mathematician and I haven't studied mathS for almost 15 years so I am not ashamed that my attempt isn't up to the standard of Harvard mathematicians.
WTF is this rage over this? Explain how I am being a dick. I'm not calling you stupid or holding you to ridicule. I said I'm surprised by your response and reasoning, especially since you should have some familiarity with this type of problem.
The problem is from a Harvard class, but you don't need to be a Harvard mathematician to grasp it or solve it. It's just a rephrasing of a common setup that you can probably find in hundreds of syllabi every year.
This outsized reaction isn't at all in keeping with the 'twentynine' posting presence I'm familiar with. I have seen dozens of poster rail at you for being the most autistic and bland poster in the history of the board and your reaction is essentially "eh, ok." Very odd case.
I swear, I don't think alia ever once got me. But for years I have been so assured that twentynine is who she claims. This gay earth, let it be fucked.
Date: October 8th, 2014 10:24 AM Author: Flirting plum public bath
"There is nothing in the terms of the problem as stated that requires each dragon to assume as a fact that every other dragon sees 99 green eyed dragons."
This isn't correct. The problem states "Then you decide to tell them something that they all already know (for each can see the colors of the eyes of the other dragons). You tell them all that at least one of them has green eyes."
Date: October 7th, 2014 8:08 PM Author: Confused cruise ship
1) Bitch should have given a "trigger warning" before opening her mouth.
2) Is this what the difference of universal knowledge and common knowledge is? Everyone knows there are 99 green dragons is universal knowledge, everyone knows that everyone knows that everyone knows... that there is at least one green dragon is the common knowledge that is new.
3) You guys are being dicks to 29. She's a terrible poster usually but she seems get it much more than almost any other woman on here would.
Date: October 7th, 2014 9:16 PM Author: spectacular plaza weed whacker
ok.. thanks. now im mad.. by deduction he sees 99 dragons with green eyes in denial, that are not turning into sparrows.. he should conclude he is in denial and poof a sparrow.
Date: October 7th, 2014 9:09 PM Author: Supple base corn cake
Questions in analytic philosophy are usually of the form "What is X?" for certain silly Xs (knowledge, existence, reference, consciousness, the good, etc.). The dragon problem is solvable using purely mathematical methods and broaches no particularly philosophical topics.
I have resigned to just bobbing around twists COCKcarousel ITT. Not sure why everyone in this elite cyber clubhouse wants to inject some kind of philosophical element into a goddamn math problem. But it is a very kooky, very odd case.
Date: October 7th, 2014 9:09 PM Author: cerise trailer park masturbator
how the fuck are you guys still talking about this?
all of this talk about who knows what, look--there is some THING that causes the induction that we all agree happens. maybe it's that they didn't have a concept of 'green eyes', and now they do. maybe it's the introduction of other-knowledge.
the first domino is the first domino; they point is the exposition of the induction, which just about everyone gets. don't sweat the details. *eats feces*
Date: October 7th, 2014 9:22 PM Author: bright psychic coffee pot
A strange variant I've thought of and don't really know the answer to:
Suppose there's no visitor, and instead each dragon publicly announces how many pairs of green eyes he can see, but only after randomly subtracting between 0 and 95 from his total. By happenstance, no dragon actually has zero subtracted, so we get a range of values from 4 to 98. This also appears to just confirm what the dragons already know, albeit in a more exotic way.
Date: October 8th, 2014 10:37 AM Author: beta university
The critical element of the announcement is that it's taken as truth, and the dragons know that all of the other dragons would take it as truth.
Your announcement method would have conflicting numbers, so they're obviously not true. Thus, each dragon could still hold a recursively imagined other dragon who believes that it sees 99 non-green eyes.
Date: October 8th, 2014 10:52 AM Author: beta university
I feel like 29 and I have explained how new information is added to the base scenario to cause the cascade sufficiently well.
After thinking about it some more, I'm pretty confident it's correct.
I don't poast very often and don't know you as a poaster, but I recall that you got that lion-sheep game spectacularly wrong, so maybe give me the benefit of the doubt for this one and go back to what I've said.
Date: October 8th, 2014 11:05 AM Author: Supple base corn cake
The point of the new hypo is that you get the same information from it (at least one dragon has green eyes), and publicly.
I didn't get the lion-sheep riddle wrong, I just missed the list step. You still haven't gotten this one right. Everything 29 has posted has been 10 times as wordy as necessary and 1/10th as correct. She's also a known sociopath.
For example, you said above that before the human said something, the dragons didn't know the other dragons were rational. But they did. "At least one of you has green eyes" couldn't possibly communicate "You're all rational". That's nuts.
Date: October 8th, 2014 11:23 AM Author: beta university
I already said I was wrong on that, which is clearly documented. It's not about rationality at all.
You're not getting the same information from it because 100 dragons saying (a - zz) is very different from one person saying (at least 1). It's literally impossible to believe (a - zz) simultaneously. The problem implicitly states that all of the dragons believe that the other dragons believe the utterance being made.
Edit: mathematical induction is of course the correct formal proof of the solution, but it does not address the question of what new information is added that would cause the cascade.
Date: October 8th, 2014 11:27 AM Author: beta university
Everyone knows what's going on. Everyone does not know that everyone else knows what's going on. In fact everyone knows that everyone else does NOT know what's going on because by definition (or premise), each dragon "believes" that he has non-Green eyes (or whatever formulation you want to use about belief).
Date: October 8th, 2014 11:40 AM Author: beta university
How does speaking a known fact publicly cause the cascade? I've described a mechanism of nested beliefs, which I think works. Do you not think it works? Do you have a superior alternative explanation?
Date: October 8th, 2014 11:43 AM Author: Supple base corn cake
I've said before it provides the base case for the induction. Since the dragons are doing the same proof by induction we are, that's crucial. But that's not what the Harvard site said, so you should read that instead.
Date: October 8th, 2014 11:54 AM Author: beta university
sorry can you clarify what you mean by "it" in "it provides the base case for the induction?" Also I've read the Harvard site, it's a very basic proof by induction but it does not speak as to what new information is being added.
If by "it" you mean the publicly announced known fact that there is at least one green eyed dragon, then yes I agree, but not just the fact itself, but the fact that it is accepted by other dragons, which was previously unknown by the dragons.
Take the two dragon scenario. It is publicly known that there is at least one green-eyed dragon, but it is NOT the case that Dragon A thinks that Dragon B knows this. Why? Suppose the opposite - if Dragon A thinks, before the announcement is made, that Dragon B knows that there is at least one green-eyed dragon, then Dragon A would expect Dragon B to sparrow (because Dragon A believes that he himself is not green-eyed). Since pre-announcement Dragon A had no such expectation, then it must not be the case that Dragon A thinks that Dragon B knows the public information.
Date: October 8th, 2014 11:56 AM Author: Supple base corn cake
We aren't in the two-dragon scenario, but you seem to be saying what I'm saying: there is new information in counterfactual cases of low N, which means we can get the inductive proof going.
Date: October 8th, 2014 12:00 PM Author: beta university
It seems like we're not REALLY disagreeing. You seem to accept the nested belief explanation for the low-N scenarios, but resist for the high N scenarios. I agree the inductive proof is the most elegant proof of the solution, but it doesn't really explain what's going on at the mechanical level. Can you explain why the nested belief explanation doesn't hold for the high-N scenarios? At what point does it cease to not be valid and why?
Date: October 8th, 2014 12:09 PM Author: beta university
You're absolutely correct, but you're also missing something, just like the lion-sheep problem.
At N = 3, for example, you're right, Dragon A (holds for each dragon), KNOWS that both Dragon B and Dragon C can see at least 1 green eye.
BUT, Dragon A thinks that Dragon B thinks that Dragon C does not see ANY green eyes.
THAT'S what causes Dragon A to think that, post announcement, Dragon C (and Dragon B) will sparrow, it's because after day 1, Dragon A will expect dragon B to expect Dragon C to sparrow, and when Dragon C doesn't sparrow, Dragon A will expect Dragon B to sparrow.
Do you see how the expectation chain (which the Harvard explanation also describes and which you must agree), must be collapsed through the original build up of the nested beliefs?
Date: October 8th, 2014 12:35 PM Author: Supple base corn cake
You're still describing it slightly wrong, which is what's giving some of us problems.
It's not that A thinks anything about what B thinks or whatever.
A thinks: "If I did not have green eyes, B would think: 'If I did not have green eyes, C would think: "If..."'", etc. down to "Z would think: 'I must turn into a sparrow tonight.'" It is not a set of nested mistaken beliefs, merely nested conditional reasoning. It's the nesting of conditions that allows for so many simultaneous "mistakes" to be made.
Date: October 8th, 2014 12:59 PM Author: Flirting plum public bath
From the beginning, I have thought of the problem as set of steps consisting of conditional statements both with respect to what each dragon believes about the observable world and what each dragon believes other dragons believe about the observable world.
In the case of N=3, I understand that after the first night, no dragon sparrows because they each see at least one dragon with green eyes and therefore cannot deduce from the at least 1 dragon has green eyes rule that they are that dragon. I also understand that this imparts information to each dragon and that all of dragons would realize this about the other dragons.
After the first night:
1.A knows that B & C have green eyes.
2.A knows that B knows that C has green eyes.
3.A knows that C knows that B has green eyes.
4.A does not know what B knows about A's eyes.
5.A does not know what C knows about A's eyes.
6.From 4 and 5, A does not know if B & C see 1 or 2 sets of green eyes.
What I am failing to understand is how the second night adds any certainty from the perspective to any individual dragon.
Date: October 8th, 2014 1:03 PM Author: Supple base corn cake
A says to himself:
"Assume I have non-green eyes. Then B will say to himself,
'Assume I have non-green eyes. Then C will say to himself,
"Assume I have non-green eyes. Then there are no dragons with green eyes. But I know there are. Therefore I must have green eyes and must sparrow tonight."
But C didn't sparrow. Therefore I must have green eyes and must sparrow tonight.'
But B didn't sparrow. Therefore I must have green eyes and must sparrow tonight."
Date: October 8th, 2014 1:05 PM Author: Chestnut Heady Center Telephone
A knows that if B or C sees A's brown eyes, then B or C will only see one pair of green eyes. If B or C also had brown eyes, then the green eyed dragon (which they see) will know it is the one with green eyes and will sparrow.
Because B or C doesn't sparrow, they can tell that both of them saw green eyes in each other, and then on night 2 they will sparrow.
Date: October 8th, 2014 1:45 PM Author: Flirting plum public bath
OK, let's assume the counterfactual. A has brown eyes, and B and C have green eyes.
Observations during the first day (before the first midnight):
A: two sets of green eyes and 3 dragons present;
B: 1 set of green eyes, 1 set of brown eyes, and 3 dragons present;
C: same as B.
Result of application of conditional statement on first night: 3 dragons.
Observations on the second day:
Same as day 1, but from the presence of 3 dragons after the application of the conditional statements, the dragons have acquired more knowledge:
A:Same as on day 1;
B:would know that C saw a set of green eyes staring back when C looked at B;
C: would know that B saw a set of green eyes staring back when B looked at C.
Result of application of conditional statement on second night: 2 dragons sparrow.
Observations on the third day:
A: observes 1 dragon (himself) and deduces that B & C saw only 1 set of green eyes staring back and therefore that he does not have green eyes.
Now, applying this to our situation where all 3 dragons have green eyes, none would have sparrowed after the second night. After observing no sparrowing after the second application of the conditional statement, the dragons would know throughout the third day that they all had green eyes and all would sparrow the following midnight.
I was getting thrown off by people saying in the case of N=3, dragons would sparrow on the second night.
Date: October 8th, 2014 1:03 PM Author: beta university
yes it's some type of conditional reasoning, but I find it useful to use the "belief" operator
The reason is that: suppose Dragon A thinks that Dragon B thinks that Dragon C sees at least 1 green eye, it must be the case that Dragon A thinks that Dragon B thinks that Dragon C sees Dragon B having green eyes (because dragon A himself would not think of himself as having green eyes without sparrowing). But if Dragon A thinks that Dragon B thinks Dragon C sees B having green eyes, then Dragon A would expect Dragon B to sparrow the next time. But of course this is all pre announcement, so Dragon A does not have any such expectation. Therefore it must be the case that Dragon A thinks that Dragon B thinks that Dragon C does not see ANY green eyes.
Date: October 8th, 2014 11:48 AM Author: turquoise irradiated space place of business
Maybe I'm outting myself as an idiot here, but how does the 100 day progression work? Aren't all the dragons seeing all 100 dragons from day 1? So how do they isolate cases of n=2. n=3, n=4, etc over a series of days? Surely there's overlap because in any set "n", the number of eyes a dragon will observe is fixed at 99.
Date: October 8th, 2014 1:06 PM Author: motley lodge
on day 99 you will have exhausted the 99 other dragons who could satisfy the at least one green rule and the other 99 in turn will have done the same so on day 100 everyone says i guess it's on me and changes into a sparrow
Date: October 8th, 2014 1:13 PM Author: turquoise irradiated space place of business
i don't get how you can exhaust dragons. given that you start off with 100 dragons, you being dragon #100, how can you sequentially exhaust dragons given that there's no way to isolate the n=2 case (only dragons 1 and 2), the n=3 case (dragons 1, 2, and 3), and so on? it's not like you're geographically separated from these dragons and get introduced to one new one every day. you see 99 other dragons from day 1, so when you isolate n=x on day x, the dragons in the set don't simply see (x-1) other dragons (which is necessary for the logic to be run.. right?), they instead see 99 dragons
Date: October 8th, 2014 1:20 PM Author: motley lodge
i think it takes each day for someone not to change and to conclude that one was using someone else for the at least one. each dragon is seeing 99 who are seeing 99 etc so they are ruling them out one midnight at a time
Date: October 8th, 2014 1:59 PM Author: Flirting plum public bath
For the reasons I set out in this poast ( Dragon IQ problem: dragon thinking it may have green eyes stops ), every additional night that no dragons sparrow provides additional information concerning the maximum number of non-green eyes in the total population.
Date: October 8th, 2014 2:00 PM Author: Flirting plum public bath
I think most people understood the induction aspect of it, if not specifically then generally. What many of us struggled with was sorting out exactly what kicked of the process.