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Coin tossing, probability and some logic, kinda


Swan Red

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They know there's a curse.

They don't know that everyone knows there's a green-eyed dragon (in the 2 dragon scenario) until the human gives the info. In the 3 dragon scenario, they don't know that everyone knows that everyone knows there's a green-eyed dragon until the human gives the info and no one subsequently becomes a sparrow.

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They know there's a curse.

They don't know that everyone knows there's a green-eyed dragon (in the 2 dragon scenario) until the human gives the info. In the 3 dragon scenario, they don't know that everyone knows that everyone knows there's a green-eyed dragon until the human gives the info and no one subsequently becomes a sparrow.

 

Yep.

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They know there's a curse.

They don't know that everyone knows there's a green-eyed dragon (in the 2 dragon scenario) until the human gives the info. In the 3 dragon scenario, they don't know that everyone knows that everyone knows there's a green-eyed dragon until the human gives the info and no one subsequently becomes a sparrow.

 

This is where i think a perfectly logical creature would conclude that each of them would be aware they all know there is a green eyed dragon, making the assumptions you do that no dragon is blind, they all know that and know each of them recognises green, because " At least one of you has green eyes " doesn't answer any of those questions.

 

In my opinion, in that example above, the dragons aren't perfectly logical before the human speaks but are afterwards. Each dragon can see 99 green eyed dragons. " At least one of you has green eyes " confirms nothing.If a dragon reaches a conclusion after that sentence, he'd have reached it before

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No, you're missing a bit. I think. But I only "get it" up to 3. Haven't read SR's Marx Bros post properly yet though.

 

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

 

This is the key bit isn't it?

Understand this and all else follows.

 

Need help getting there though!

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No, you're missing a bit. I think. But I only "get it" up to 3. Haven't read SR's Marx Bros post properly yet though.

 

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

 

This is the key bit isn't it?

 

 

And how does what the human says confirm that ?

 

 

It's flawed. In your scenario

 

Chico :

 

" I can see 99 green eyed dragons. But how do i know Harpo does ? And how do i know Harpo knows Groucho does ? "

 

Human

 

" At least one of you have green eyes "

 

Chico still doesn't know if Harpo knows Groucho sees green eyed dragons because the above sentence doesn't answer those questions.

Edited by Earl Hafler
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This is where i think a perfectly logical creature would conclude that each of them would be aware they all know there is a green eyed dragon, making the assumptions you do that no dragon is blind, they all know that and know each of them recognises green, because " At least one of you has green eyes " doesn't answer any of those questions.

 

No.

 

In the case of 3 dragons,dragon A cannot logically deduce that dragon B knows that dragon C can see at least one dragon with green eyes. If dragon A thinks he could have red eyes (and he doesn't know if he does or not) then dragon A will think that dragon B could be sitting there thinking "dragon A has red eyes, maybe I do to? If that's the case then dragon C may not see any green eyes at all". It's only when it's confirmed as communal knowledge that dragon C must see at least one dragon with green eyes that the logic unfolds.

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And how does what the human says confirm that ?

 

 

It's flawed. In your scenario

 

Chico :

 

" I can see 99 green eyed dragons. But how do i know Harpo does ? And how do i know Harpo knows Groucho does ? "

 

Human

 

" At least one of you have green eyes "

 

Chico still doesn't know if Harpo knows Groucho sees green eyed dragons because the above sentence doesn't answer those questions.

 

Hold up on the 99, we're still dealing with 4.

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No.

 

In the case of 3 dragons,dragon A cannot logically deduce that dragon B knows that dragon C can see at least one dragon with green eyes. If dragon A thinks he could have red eyes (and he doesn't know if he does or not) then dragon A will think that dragon B could be sitting there thinking "dragon A has red eyes, maybe I do to? If that's the case then dragon C may not see any green eyes at all". It's only when it's confirmed as communal knowledge that dragon C must see at least one dragon with green eyes that the logic unfolds.

 

100 dragons, all with green eyes. All the dragons recognise the colour.

 

That's the scenario i'm talking about.

 

" At least one of you has green eyes " doesn't confirm what KAMF is saying it does. Elsewhere in the thread, someone said the dragon can now rule out the possibility that they don't have green eyes - that aint true either

 

Gerry said " Dragon A DOESN'T know whether dragon B knows that C knows at least one dragon has green eyes "

 

Well the humans statement doesn't answer the logical question some dragons have - do we all know what green is ? And if you're assuming the answer to that question is yes, then Gerry's sentence is incorrect. " At least one of you has green eyes, and you all know what green is " would need to be the sentence to resolve that.

 

yeah

 

SR, that sentence alone doesn't answer Chico's question.

 

Or are you saying that now Chico can deduce that even if Harpo doesn't know what green is, the fact he ( Harpo ) can see at least 98 other dragons with the same colour eyes, Harpo can now do the math ? ( being a perfectly logical dragon )

Edited by Earl Hafler
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How does the multi-day scenario come about? All dragons have green eyes and equal deductive powers and the knowledge imparted by the human, and none are aware of the colour of their own eyes. So whatever happens must happen to all of them at the same time. Without getting into any of the who worked out what about who, symmetry tells us they either all survived the first night or all become sparrows that night or, if surviving the first night as dragons is the key, the next night.

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How does the multi-day scenario come about? All dragons have green eyes and equal deductive powers and the knowledge imparted by the human, and none are aware of the colour of their own eyes. So whatever happens must happen to all of them at the same time. Without getting into any of the who worked out what about who, symmetry tells us they either all survived the first night or all become sparrows that night or, if surviving the first night as dragons is the key, the next night.

 

It's over on day 3.

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Why are we talking about whether the dragons know what green is? It's assumed they do.

 

And over on day 3? It's only over on day 3 when there are 3 dragons involved.

 

The problem is the scenario given earlier is flawed. Having read the example on the Harvard website My link it's a maths / physics thing which asks to ignore or assume certain things, treating perfectly logical creatures like numbers in an equation.

So it assumes A didn't know B knew there was at least one green eyed dragon but assumes that both dragons know the other knows what green is because the " new " info doesn't confirm that.

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Who said there are only two colours? There could be an infinite number of alternative colours for the eyes and it matters not a jot. I think you're going to have to fill in the gaps for "So on day three..." because I'm not sure we know what you're driving at.

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The problem is the scenario given earlier is flawed. Having read the example on the Harvard website My link it's a maths / physics thing which asks to ignore or assume certain things, treating perfectly logical creatures like numbers in an equation.

So it assumes A didn't know B knew there was at least one green eyed dragon but assumes that both dragons know the other knows what green is because the " new " info doesn't confirm that.

 

The presentation differs in a variety of cases, I've linked two alternates, the issue is one of logic, it assumes that all dragons know every other dragons eye colour, each dragon knows the curse and each knows what green means. None of this is contentious it's important but they are merely the assumptions the puzzle relies on.

 

Also the harvard article you linked to only provides the solution I don't see where the presentation differs and the response is consistent with those posed.

 

Consider the case N = 3. A knows that B green eyes, and he also knows that

B knows that there is at least one dragon with greens eyes (because A can see that

B can see C). So the two bits of information in the N = 2 case above are already

known before you speak. What new information is gained after you speak? Only

after you speak is it true that A knows that B knows that C knows that there is at

least one dragon with green eyes.

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The presentation differs in a variety of cases, I've linked two alternates, the issue is one of logic, it assumes that all dragons know every other dragons eye colour, each dragon knows the curse and each knows what green means. None of this is contentious it's important but they are merely the assumptions the puzzle relies on.

 

Also the harvard article you linked to only provides the solution I don't see where the presentation differs and the response is consistent with those posed.

 

It assumes those things and also that each dragon doesn't know for certain if other dragons knows what green is, yes ?

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It doesn't state it explicitly I guess

 

 

 

That said it's not something that's under question.

 

Ok.

 

So each dragon knows there are a maximum two colours involved but that a third must be considered - If Dragon A puts himself in the position of Dragon B, Dragon A could logically think that Dragon B might consider the green eyes Dragon A can see, Dragon A's own as yet unconfirmed colour and ( from Dragon B's viewpoint ) Dragon B's as yet unconfirmed colour.

 

Past that point, there's only one conclusion and it doesn't take 100 days to figure out.

Edited by Earl Hafler
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Ok.

 

So each dragon knows there are a maximum two colours involved but that a third must be considered - If Dragon A puts himself in the position of Dragon B, Dragon A could logically think that Dragon B might consider the green eyes Dragon A can see, Dragon A's own as yet unconfirmed colour and ( from Dragon B's viewpoint ) Dragon B's as yet unconfirmed colour.

 

Past that point, there's only one conclusion and it doesn't take 100 days to figure out.

 

The number of colours is irrelevant, the maximum of two is only a maximum when at least n dragons -1 have green eyes. In the case that there are 2 dragons of the 100 with green eyes there are 99 potential colours of eyes. All non greed eyed dragons are functionally equivalent. Consider what dragon A learns about what Dragon B knows about Dragon C......n

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It doesn't take 100 days when there are only 3 of them, certainly.

 

Fecks sake

 

I'm talking about 100 dragons. Each only has to consider an A, B or C

 

Each of the 100 dragons thinks the same " I might have red eyes. Each of the 99 green eyed dragons i see could then reasonably assume there are 98 green eyed dragons, one red eyed dragon and his own as yet unconfirmed colour "

In each case a dragon doesn't get past N=3, where the reasoning for N=2 is possible.

 

If N=10, the reasoning is still the same as N=3 because of the amount of possible colours / variables

 

The number of colours is not irrelevant

 

The number of colours is irrelevant, the maximum of two is only a maximum when at least n dragons -1 have green eyes. In the case that there are 2 dragons of the 100 with green eyes there are 99 potential colours of eyes. All non greed eyed dragons are functionally equivalent. Consider what dragon A learns about what Dragon B knows about Dragon C......n

 

Explain N=6 when all have green eyes.

Edited by Earl Hafler
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