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Coin tossing, probability and some logic, kinda


Swan Red

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No, it's possible that neither has green eyes. We have no information about who placed the curse or why.

 

A ) Why ? Why would there be a curse when there isn't a green eyed dragon ?

B ) " he says something that the dragons already know "

 

He's a starting gun. That's all he is.

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A) No idea, could be dozens of things, could be there just in case there are green eyed dragons in the future for instance.

 

B) What they already know is that there's at least one green eyed dragon. If there were only two dragons, they'd each know that but they wouldn't know whether the other one knew.

 

This is clearly a tricky one, that will take a lot of working out, and I don't think you can confidently assert "that's all he is" just yet.

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I know.

 

As I said,

That's how it works with 2 but I don't yet see how it works with more.

 

Consider the case of three dragons. All three dragons have green eyes, all dragons know that each dragon sees at least 1 dragon with green eyes. What they do not know is that every other dragon knows every dragon sees a green eyed dragon. (I know it wrecked my head as well and if I could have explained it better to myself I'm sure I'd have understood it quicker)

 

Consider Curly Larry and Mo. Curly does not know the colour of his own eyes, he knows that Larry sees Mo has green eyes and vice versa but he also knows they do not know the colour of their own eyes. Therefore Curly does not know that Larry knows Mo sees a green eyed dragon.

 

If Curly has blue eyes and Larry does't know his own eye colour Curly can't know that Larry knows Mo can see a green eyed dragon. When the stranger speaks he knows this, it becomes common knowledge.

 

I found the problem with getting my head around it is there has be be a regress. Each dragon knows every other dragons sees at least n-2 green eyed dragons but they don't know each dragon knows this. This is why it helps trying to understand it with much smaller numbers, once you get 3 it's a case of iteratively n+1.

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So Curly thinks "If I don't have green eyes, then Larry can't be sure that Mo knows that at least one of us has green eyes. So if I don't have green eyes then Larry thinks this may be new info to Mo. But if I do have green eyes then Larry knows this isn't new info to Mo. So if I don't have green eyes then Larry is waiting to see whether Mo turns into a sparrow."

What's next?

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Does anyone who cares still not get the 2 dragon version?

 

 

Consider Curly Larry and Mo. Curly does not know the colour of his own eyes, he knows that Larry sees Mo has green eyes and vice versa but he also knows they do not know the colour of their own eyes. Therefore Curly does not know that Larry knows Mo sees a green eyed dragon.

 

That's the best way to think about it

 

I think

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So Curly thinks "If I don't have green eyes, then Larry can't be sure that Mo knows that at least one of us has green eyes. So if I don't have green eyes then Larry thinks this may be new info to Mo. But if I do have green eyes then Larry knows this isn't new info to Mo. So if I don't have green eyes then Larry is waiting to see whether Mo turns into a sparrow."

What's next?

 

What is new information to Larry is that he now knows that you also know Mo sees a green eyed dragon and if his own eyes aren't green you will deduce that you have green eyes on day 2. So when you don't you all go piff paff poof day 3 (with a nod to the Great Soprendo).

 

It's why the dragons need to get the information simultaneously the information introduced is n level knowledge of the other dragons knowledge.

 

And I apologise for the particularly cackhanded way I'm explaining this.

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Doesn't more than 3 dragons make the human info irrelevant? With 3, the regress that A doesnt know that B doesn't know if C can see any green eyed dragons, etc. ...sparrows. But with more than 3, every dragon knows that the 'at least one of them has green eyes' condition has been met, making their own eye colour indeterminate even by the regress above?

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Seems that way to me JRC, but it took me a while to understand 2, I'm slowly coming around to 3, so I won't rule out 4-99 just yet.

 

Pretty sure the n=3 logic doesn't work past n=4. No way to construct a situation where 1 dragon can't be aware that their own eye colour doesn't give a clue to another dragon, when there are obviously 98 other green eyed dragons around.

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Just how big does the number of dragons present have to get before any dragon observing that they all have the same eye colour must reasonably conclude that it looks certain that gentically he must also have eyes that colour?

 

Red herring that.

They need to know for certain that they have green eyes. It being v v likely is irrelevant.

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You'll have to talk me through the deduction on day 2, and the further deduction on day 3, from the point of view of one dragon. Soz.

 

No probs I'll do my best.

 

Curly, Larry and Mo are three green eyed dragons who know each others eye colour but not their own. Before the stranger spoke Curly did not know that Larry knew Mo saw a green eyed dragon.

 

On day one Curly does not expect any dragon to turn into a sparrow knowing there are at least two green eyed dragons. What he is able to deduce is that both Larry and Mo now know there is more than 1 green eyed dragon on the island. If there had been one green eyed dragon it wouldn't have seen any green eyed dragons and thus would have known it was him. If there are two green eyed dragons on the island Curly knows they will have deduced that they have green eyes when neither turned into a Sparrow on day 1, thus if he does not have green eyes then both Larry and Mo will turn into sparrows on day 2.

 

He is able to deduce this because if he does not have green eyes and both Larry and Mo will know there is more than 1 green eyed dragon they will both learn they have green eyes.

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What is new information to Larry is that he now knows that you also know Mo sees a green eyed dragon and if his own eyes aren't green you will deduce that you have green eyes on day 2. So when you don't you all go piff paff poof day 3 (with a nod to the Great Soprendo).

 

It's why the dragons need to get the information simultaneously the information introduced is n level knowledge of the other dragons knowledge.

 

And I apologise for the particularly cackhanded way I'm explaining this.

 

I think that's not perfectly logical, which the dragons are.

 

The human or guru doesn't give them new info.

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Got it.

Thanks.

 

So I understand that the human makes a difference with 3. Significant progress. Once I understand 4 and 5 I will accept 100.

 

 

 

It is perfectly logical, and yes he does.

 

No, i mean that it's not perfectly logical for the dragons to need the human to say anything.

 

I understand the deduction between the x amount of dragons

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Got it.

Thanks.

 

So I understand that the human makes a difference with 3. Significant progress. Once I understand 4 and 5 I will accept 100.

 

I'll try.

 

4 Dragons Chico, Harpo, Groucho, Zeppo, all have green eyes all know the colour of each others eyes none know the colour of their own.

 

Chico knows every other dragon sees at least two green eyed dragons. Chico also knows that every other dragon knows each dragon will know there is at least 1 green eyed dragon.

 

Chico does not know that Harpo knows Groucho knows Zeppo sees at least one eyed dragon.

 

On day two Chico knows that Harpo knows there are more than two green eyed dragons. Had there been two green eyed dragons they would each have seen one green eyed dragon and two non green eyed dragons. They would have deduced after the one dragon they could see did not turn into a sparrow on day one that they would have been the second green eyed dragon and both would have turned into sparrows on day 2. Hence even though Chico does not expect any activity before day 3 on day two he knows that the others will have deduced they have green eyes if Chico doesn’t.

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It is, unless you're still on the "why would there be a curse unless there was a green eyed dragon" thing?

 

And your explanation is just " what if "

 

There's a curse. There's at least one green eyed dragon. Perfectly logical for a dragon to come to the deductions we've discussed without a human speaking before them, which some think needs to happen for Dragon A to know that Dragon B is also aware of the curse.

 

We've discussed different variations in the thread but the one i'm talking about is where it says the human is giving them no new information. In every day life, a person might think it's possible that they know of the curse or a law ( let's say under age drinking ) but the other possibly doesn't but i'd expect two perfectly logical creatures / humans not to think that.

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"what if" is enough because they need to be certain.

 

"there's at least one" does not logically follow from "there's a curse"

 

 

Excellent - so what does the human tell them ? He doesn't say publicly there's a curse ( for all we know ) He mentions eye colour. Why then, in your opinion, does Dragon A immediately realise Dragon B now knows of the curse if he didn't before ? Surely there's enough ' what if ' in that sentence.

 

" What if he's just mentioning eye colour ? " " What if it's a harmless comment based on a conversation not related to the curse "

 

The human says nothing they didn't know already and, in the reasoning given later in this thread, doesn't say enough to be the decisive factor.

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They know there is a curse, they also know that there is more than 1 green eyed dragon. It is an iterative process he sets the clock at t0 from which t1....tn

 

I think he sets the clock for us, as a point of reference, but he doesn't tell the dragons anything which could change them going from happily walking around the island to suddenly realising that every dragon knows of the curse.

A perfectly logical creature could have done that already if he's doing it then.

 

 

There are a hundred dragons in an island. All of them have green eyes. There is a curse on the island where if a dragon realizes by itself that it has green eyes, it will turn into a sparrow at midnight. Because of this curse, none of the dragons are allowed to talk about eye color and none of them can look at their own reflection. One day a human comes along for a vacation to the island and he has a lot of fun with the dragons. While leaving at a farewell party where all the 100 dragons are present, he says something that the dragons already know "At least one of you have green eyes". The dragons know what the color "green" means and they are perfectly logical creatures.

 

KAMF says " If I'm a dragon, I know he's not giving new info to me. But I don't know whether he's giving new info to the other dragon "

 

- He's not giving any new info because he's merely commenting on eye colour, which the dragons can already see. And that comment itself doesn't prove to Dragon A that the other now knows about the curse and can recognise the others with green eyes ...if Dragon A didn't know or assume that anyway.

 

They can see green eyes, they know what the colour is and what it represents. And it's eye colour they can't talk about - doesn't stop one from saying " You know about the curse, right ? "

 

Swan Red " Curly, Larry and Mo are three green eyed dragons who know each others eye colour but not their own. Before the stranger spoke Curly did not know that Larry knew Mo saw a green eyed dragon " which isn't true.

 

"At least one of you have green eyes" doesn't tell Curly that.

Edited by Earl Hafler
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