Math Fun

[Silhouette]

Isn't it interesting that Sil, FJ, and Carleas all don't understand the words, "start with", "counting days", and "pick".
"But I don't understand".. "but I don't see the evidence"...

Yeah, those words are way too complex. What the hell is counting?
If you don't think we understand them then feel free to continue living in your bubble, but it won't get you anywhere or prove anything to anyone else.

In other words, you haven't been reading anything but your own posts for many pages now.
.. I suspected that.

When I read what you quoted the first time, it made no more sense than it did when you quoted it. I've kept up with this thread as best as I can, but the stuff that makes no sense, or isn't clear enough to me is just as much that way as I re-read it as it was the first time I read it.

I'm not the only one to have come up against this problem - which I know from having read the last many pages - Carleas has asked on many occasions for you to make clearer what you are saying, saying things like "communication is a two way street".

You do need to learn how to communicate.

Please start now by clarifying what the hell you're trying to get at, rather than pretending to want to engage, but running back off into the bushes of obscurity and self-proclaimed clarity at the first opportunity.

I have re-iterated my own presentations of the correct solution several times now, each time improved in clarity from the last. When people fail to understand you, I suggest you learn from this example.

[James S Saint]

Yeah picking a number between 0-99 is a really tough task to comprehend.
But somehow I managed it and gave it as an example displaying all possible combinations of results.
Yet somehow it magically escapes you?

Picking merely the number 97, we get this;

The only 3 possibilities;

I am blue out of 100 blues;
A) They all know [that they all know] [that there are at least 97 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 99 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 100 and (A) were true, we leave the 4th day].

I am brown out of 99 blues;
A) They all know [that they all know] [that there are at least 97 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 99 and (A) were true, they leave the 3rd day].

I am red out of 99 blues;
A) They all know [that they all know] [that there are at least 97 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 99 and (A) were true, they leave the 3rd day].

Now can you pick a number that would fail to work that way?

[Silhouette]

The only 3 possibilities;

I am blue out of 100 blues;
A) They all know [that they all know] [that there are at least 97 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 99 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 100 and (A) were true, we leave the 4th day].

I am brown out of 99 blues;
A) They all know [that they all know] [that there are at least 97 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 99 and (A) were true, they leave the 3rd day].

I am red out of 99 blues;
A) They all know [that they all know] [that there are at least 97 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 99 and (A) were true, they leave the 3rd day].

I "picked" out the flaws for you. And they go for each of your "only 3" colours that exist:

Consider first the incompatibility between (A) and (D).

If, as in (D), "there were only 97 blues", (A) would not be true. How would knowing that they all know that there were at least 97 blues be known by anyone when each of them does not even know their own eye colour (they only even see 96, never mind know that anyone else knows this)? Contradiction there methinks.

Consider now, (E). If there were only 98 blues, all islanders would only see that there were at least 97 blues (no knowing of knowing of seeing as in (A)).
(F). If there were only 99 blues, the blues would only know that other blues saw at least 97 blues (and not, as in (A), know that they all know that at least 97 blues were seen). Any browns/reds/misc-non-blues would see the 99, know that others see 98 and know that others know that others see 97, so (A) is valid for them.
(G) - only seen in the case of the blues, this is the only other case where anyone knows that they all know that there are at least 97 blues. (A) is also valid for them.

So they just leave because they each know that everyone knows that everyone sees 97 in their own particular circumstance? No they don't, because none of that fictitious progression you were attempting backs it up. They just notice something that tells them nothing about their own eye colour. Where's a good Guru when you need one, eh?

Essentially all you've done is pick out a few patterns that are incompatible with one another, and stick 'em together with a meaningless progression of numbers (upon proper examination of them) to make it look like there's a process of logic.
There is, in fact, a great deal of omissions in your logic that would only be noticeable if you actually worked it through thoroughly, rather than just trying to fit some numbers together that you can make look like they fit together.

Also, what does "I am a brown/red out of x blues" even mean? Are browns part of the set of blues?! (I know what you meant to say).

So basically:

Now can you pick a number that would fail to work that way?

Yeah.

All of them.

[Carleas]

James, when you say "pick", it seems like what you really mean "identify a number such that that number of blue eyed islanders know that there are that many blue eyed islanders". The distinction is important. The statement "I see 10 pairs of blue eyes" and the statement "10" contain different information. The statement "10" doesn't contain any information; it's simply yelling out an abstract property unattached to anything else. There is no relationship expressed. The statement "I see 10 pairs of blue eyes" contains information. The abstract property "10" is applied to a group of things, "pairs of blue eyes". "10" is related to "blue eyes the guru sees".

When you use the vague terms "pick", "start with", and "use", that is not clear. As you yourself said, clarity is vital. Be precise. When we're talking about the distinction between knowing that someone knows that someone knows and knowing that someone knows that someone knows that someone knows, we need significantly more precision that those terms provide. Specify what the numbers refer to, what the people are counting.

why don't you just state a specific number that wouldn't work for them and show how it would fail?

Carleas seems to think it's not credible to have a position that he finds difficult to challenge.

Let's be clear here: you're asking me to disprove a syllogism that you have not provided. Your claim is that simply having a number in mind is enough to prove eye color. My evidence for rejecting that is that the canonical solution is rigorously logical (something that has not been refuted), which entails that if there's another solution, the problem itself contains an inherent contradiction. You have not identified that contradiction, and you have not rigorously demonstrated your solution. As far as I'm concerned, I'm being charitable in trying to refute a solution that you refuse to expound in any detail, based on the scanty tidbits about it that I've gleaned over these few dozen pages.

Now, one thing I'm sure that you're saying is that everyone having the same number in mind at the same time is sufficient. It isn't.

FC, you say "the thing works easily with 4 blues and 4 browns," could you show that? I believe the mistake you are making is substituting knowing that they know that they know for knowing that they know that they know that they know, which breaks the logic that allows the canonical solution to work, and which requires the guru's statement. If you could spell out the syllogism, it would further the discussion whether we're able to refute it or not.

[James S Saint]

Wow..
Ask a simple question that any 3rd grader could understand, and what do you get...

Now can you pick a number that would fail to work that way?

You two want to jump into your future objections, "how do they know what number to pick".

Can't you calm down and just go one step at a time?
The question is pretty damn simple. It doesn't take a PhD in logic or mathematics.
If you can't answer such a simple question, I am really not interested in your excuses.
If you can't handle arithmetic because of your religion, I am not going to try to explain the calculus.

On the Physics Forum site, you get banned for even mentioning that there might be any alternate understanding.
So I still admire Carleas for being far more tolerant than many online, but honestly, how hard is it to temporarily (as in a hypothesis) accept a notion long enough to prove it wrong? Is it really that threatening?

[Carleas]

James, I have answered that question. Repeatedly. Silhouette has answered it. The answer is that for no number does "that way" work. "That" is not a "way" to solve the problem. Why? Because there is no valid deductive process that gets you from "this is a number" to "my eye are blue."

[James S Saint]

"That" is not a "way" to solve the problem.

"That is not the WAY!!!"

Answering simple questions is NOT the WAY. The WAY is to repeat what you're TOLD!!
Never ask questions. Never answer even simple questions, else the Devil will take your soul to Hell to be in torment forever more.
Glory be the WAY.
All hail the WAY.
Praise the WAY.


Wow. Are you guys really that mind clogged and blocked?
Are they going to come and take away your pension, your kids, or raise your alimony?
The Secular priests are at your door coming to get you?

Show me a number as I showed you, but one that would lead to the wrong deduction of eye color.

[Flannel Jesus]

Your method itself produces a wrong 'deduction'.

You already agreed, in a prior conversation about this same 'way', that whatever number of blues they actually start with, they leave on the 4th day. Remember that conversation? The one in which you insisted that if there were actually 99, they would leave on the third day, but if there were 90 they would leave on the 4th day? And you ended up agreeing that that didn't make sense, and that no matter what number there are (as long as it's above 4 or 3 or something), they would leave on day 4?

If that's agreed, then we agree that if there were 101 blue-eyed people, they would also leave on the 4th day, correct?
You take your logic and add one to everything:

I am blue out of 101 blues;
A) They all know [that they all know] [that there are at least 98 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 99 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 100 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 101 and (A) were true, we leave the 4th day].

Right?

So, the logic above can be summed up as, if you see 100 blues and you see that they don't leave on the third day, you know you're blue eyed.

And it becomes clear what's wrong with this.
You see, in the actual case, the brown eyed people ALL DO see 100 blues, and they ALL DO see that they don't leave on the third day. So if the logic works...then the brown eyed people...deduce that they're blue eyed.

I did it the other way in my first explanation -- instead of looking at 100 blue-eyed people, we looked at 99 -- but now we're going to 101 to show that the mistake works on all sides.

[James S Saint]

Your method itself produces a wrong 'deduction'.

You already agreed, in a prior conversation about this same 'way', that whatever number of blues they actually start with, they leave on the 4th day. Remember that conversation? The one in which you insisted that if there were actually 99, they would leave on the third day, but if there were 90 they would leave on the 4th day? And you ended up agreeing that that didn't make sense, and that no matter what number there are (as long as it's above 4 or 3 or something), they would leave on day 4?

And no matter how many times I tell you, you don't remember the actual issue involved.
"IF they ALL pick the SAME number."

If that's agreed, then we agree that if there were 101 blue-eyed people, they would also leave on the 4th day, correct?

It is NOT agreed.

You take your logic and add one to everything:

I am blue out of 101 blues;
A) They all know [that they all know] [that there are at least 98 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 99 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 100 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 101 and (A) were true, we leave the 4th day].

Right?

That certainly seemed to have worked, to me.

So, the logic above can be summed up as, if you see 100 blues and you see that they don't leave on the third day, you know you're blue eyed.

And it becomes clear what's wrong with this.
You see, in the actual case, the brown eyed people ALL DO see 100 blues, and they ALL DO see that they don't leave on the third day. So if the logic works...then the brown eyed people...deduce that they're blue eyed.

Except you forgot something (again.. you really should have your memory checked);;

You take your logic and add one to everything:

I am blue out of 101 blues;

In your example there actually ARE 101 blues and the browns see all 101 of them... leave on the 4th day.

[James S Saint]

Guys, let me propose a different scenario that has a similar issue involved, also from a classic puzzle.

A "perfect logician" is trapped on an island with limited food supplies. He has two identical containers and inadvertently places the exact same amount of food in each container. The next day, he goes to get a little food and realizes that both containers are exactly identical. How does he decide which container to open first?

So far, you guys would suggest that he will starve to death because being "perfect", he cannot choose which to open and thus can't open either, that is unless a guru or angel or someone comes along and opens one for him. And his problem is that he is perfect??

Would you starve from such a situation? Yet you would conclude that a perfect logician, better than you or I, would actually starve to death?

I propose that such a perfect logician knows a logic trap when he sees one and can logically deduce how to solve it. He doesn't need a guru, an angel, nor a woman to guide his little fingers. And he doesn't need to be imperfect either. Perfect logicians can get themselves out of perfect logical traps, else they are not as perfect as proposed. How logical is it really to starve to death due to such a situation?

So far, you guys are saying that perfect logicians should think like you have been and thus be helplessly trapped until saved by a guru woman (and since I have never heard of such a thing, I suspect that could be quite a while). You defend how they MUST think exactly like you suggest and thus are helpless. Whereas I suggest that they are smarter than that and know to take a different logical path.

"If they CAN deduce the solution, they WILL do so instantly."

[Silhouette]

"That is not the WAY!!!"

Answering simple questions is NOT the WAY. The WAY is to repeat what you're TOLD!!
Never ask questions. Never answer even simple questions, else the Devil will take your soul to Hell to be in torment forever more.
Glory be the WAY.
All hail the WAY.
Praise the WAY.

There's a wonderful irony here.

James is throwing a paddy because we are not accepting his assumptions and going along with his methodology. To him, his assumptions and methodology are THE WAY that we must follow. It is the way he is telling us to follow, and we should not ask questions because his mind is too clogged and blocked to comprehend that we might choose a different way that does not have flawed assumptions.

How dare we.

Just to clarify, James. The "way" that we are choosing is as a result of trying many other ways, including yours, though finding them to be unacceptable. At least I have discovered (and communicated) the reasons why no other ways work than the only one that does work - though I suspect that the others (apart from poor you and FC) at least intuitively know this and perhaps have also logically worked it out too (and maybe even presented it and I haven't seen it).
It doesn't take a PhD in logic or mathematics to understand that there is only one solution in this case.

Anyway...

A "perfect logician" is trapped on an island with limited food supplies. He has two identical containers and inadvertently places the exact same amount of food in each container. The next day, he goes to get a little food and realizes that both containers are exactly identical. How does he decide which container to open first?

Would you starve from such a situation? Yet you would conclude that a perfect logician, better than you or I, would actually starve to death?

This one is pretty simple.

You just irrationally/illogically choose one. There is no logic that will immediately just come into the mind of a perfect logician. It's a 50/50 with no reason at all to choose one or the other, just so long as one is chosen if the perfect logician values his life. To be honest, there isn't even a logical decision to choose one or the other if they were unequally filled with food. If the aim is to eat, then all that matters is that you choose a container that has food in it.

In enough instances of the same scenario, monitoring which box is chosen will almost inevitably yield results that are nigh on half and half. Unless one box is consistently put in a stupid place, or there's some kind of unnoticed bias, that is. Even perfect logicians would still sometimes go for one and sometimes for for the other.

Only if one is restricted only to logic, will they starve in this situation. But because people are built from the ground up NOT on logic and reason, but simply on circumstance and trial and error, they will make a choice anyway. They will not starve.

Seems to me like you load "logic" and "reason" with values (more assumption). I think you want/need logic and reason to underlie everything, and you get frustrated when people who are more in touch with reality question this. It's a shame for all of us that you will probably never understand this, but hey, we tolerate you and seem to never give up hope.

[James S Saint]

No. You are just refusing to answer a simple question.
Instead you run off to defend something that is NOT the question because you can't separate one issue from another.

It doesn't take a PhD in logic or mathematics to understand that there is only one solution in this case.

It doesn't take a PhD for you to ASSUME there is only one.
And it doesn't take a PhD to realize that there might be more an one.
Until the lack of alternatives is proven (which it hasn't been), nothing has been proven.

The question that you seem to fear to the bone;
"IF everyone chooses the exact same number to use in counting days before deducing that they are the remaining blue (or not), is there any number between 0-99 that would cause them to deduce improperly?"

[Silhouette]

No. You are just refusing to answer a simple question.
Instead you run off to defend something that is NOT the question because you can't separate one issue from another.

Let me explain to you how this works, since you seem to be confused.

If someone says "answer this question for me" and you don't question the question, you are accepting the assumptions behind it.
What I have done is look at your question and say "hang on, this question doesn't fly". I proceed to unravel the question and clearly show why it is a duff question, and to go along with it would lead to a duff answer. I don't like duff answers. I did, however, do you the courtesy of giving you an answer that was related to your question, but fundamentally based on my criticism of it, that in a way actually answers your question just fine!

Thus, can I pick a number that would fail to work in the same way as you were demonstrating in your example of picking 97?
Yes, I can pick any number that would fail to work in that way, as shown by my criticism of your whole process in dealing with any given number.
Question answered.
Is there any number between 0-99 that would cause them to deduce improperly? Yes, all of them ^ question answered. Zero fearing on my part.

What more do you want?

It doesn't take a PhD for you to ASSUME there is only one.
And it doesn't take a PhD to realize that there might be more an one.
Until the lack of alternatives is proven (which it hasn't been), nothing has been proven.

Yes, a PhD is completely irrelevant to either assuming there is only one solution in this case, or understanding that there is only one. That was my point: your hyperbole about PhDs was irrelevant.
I already did prove there were no alternatives (without a math or logic PhD) - I suspected that you hadn't been reading anything but your own posts for many pages now. Or maybe you just didn't understand the concept of dealing with all and only the certainties from any given islander's own perceptions, and only when that fails to lead to a solution on its own, adding in the Guru's communication and seeing if that led to a solution... which it just so happens to do - and considering ALL the certainties that arise from a combination of both of those factors, we get one single solution. This, by default, eliminates any other possibilities, because by process of elimination, they would necessarily rely on something other than certainty/definite knowledge and flawless deduction.

Case closed. Get over it.

[James S Saint]

More rhetoric.
Can't answer a simple question.

"IF everyone chooses the exact same number to use in counting days before deducing that they are the remaining blue (or not), is there any number between 0-99 that would cause them to deduce improperly?"

Show me one number that doesn't work and show it not working.

I showed you one that did work. It is your turn.

[Silhouette]

.......

what?

I... just answered your question. Right.... there....

Q: IF everyone chooses the exact same number to use in counting days before deducing that they are the remaining blue (or not), is there any number between 0-99 that would cause them to deduce improperly?
A: Yes. All of them.

How is that.... "not answering a simple question"?

Show me one number that doesn't work and show it not working.

Okaaay...
Here:

I am blue out of 100 blues;
A) They all know [that they all know] [that there are at least 97 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 99 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 100 and (A) were true, we leave the 4th day].

This is an example of one number that doesn't work.

(D) is the best example of why it doesn't work. (D) doesn't work because if there were only 97, (A) would not be true. Yet (D) claims both to be true.
In (D), "if there were only 97", each one would see 96 other blues. They would not know that they all know that there were at least 97 blues because they can only see 96 (which is less than 97, not "at least 97") blues because each do not know their own eye colour and they are not seeing blues that are not there. Not even seeing 97 blues means they do not "all know [that they all know] [that there are at least 97 blues]", which is (A). (D) says that they do.

I.... cannot be any more clear here. This doesn't work. It is "one number that doesn't work". I just showed it not working. Just like you asked.

You're telling me a clear answer to your question is "just rhetoric"? Really?

[James S Saint]

I am blue out of 100 blues;
A) They all know [that they all know] [that there are at least 97 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 98 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 99 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 100 and (A) were true, we leave the 4th day].

This is an example of one number that doesn't work.

Sure looks like it works to me. They all correctly discovered their own eye color.
So no. You didn't show anything not working.
"Not working" means they make a mistake in their color.

(D) is the best example of why it doesn't work. (D) doesn't work because if there were only 97, (A) would not be true. Yet (D) claims both to be true.

Was there ever only one person on the island when the guru said there was one?
Noooo.
It is irrelevant.

The issue is the ability to know your color. They each learned of their color.
And they will every time.... IF they ALL choose the SAME number.

[Carleas]

Again, James, what you're saying suggests that you don't understand the logic behind the canonical solution.

In the cases where 1, 2, or 3 islanders learn their eye color, they all know that there are at least two islanders on the island with blue eyes. But the counterfactual, "if there were only 1 he would leave on day 1", is still necessary to reach the conclusion. You seem to understand and accept this.

For some reason, though, when we get to 4, you and Fixed Cross discount the value of the same information, and suppose that the 1 would just spontaneously know that he has blue eyes. What is the difference between the case of 3 blue islanders and the case of 4 blue islanders such that the guru is necessary in the former but not in the latter?

If we use your example for the case of 4 blue eyed islanders, I think what Silhouette is saying becomes even clearer:

I am blue out of 100 blues;
A) They all know [that they all know] [that there are at least 1 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 1 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 2 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 3 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 4 and (A) were true, we leave the 4th day].

It should be obvious that while they all know that they all know that there is at least 1 blue, there is no 1 who knows that he has blue eyes. We are given that no one knows their eye color, and D requires that there is at least 1 who does. The contradiction, as Silhouette has repeatedly explained, is that you're taking the knowledge out of context. 4 know there is at least 1. But if there were only 1, he would not know that there is at least 1 - until the guru speaks.

[Silhouette]

Was there ever only one person on the island when the guru said there was one?
Noooo.
It is irrelevant.

The Guru never said that there was only one person on the island.

She said she can see someone who has blue eyes. Not only one someone.

Sure looks like it works to me. They all correctly discovered their own eye color.
So no. You didn't show anything not working.
"Not working" means they make a mistake in their color.

The issue is the ability to know your color. They each learned of their color.
And they will every time.... IF they ALL choose the SAME number.

Sounds to me like you'd be happy with anything that looks like "a working out" of one's eye colour just so long as the end step "says" that they definitely worked out their own eye colour. I'm not disputing that your solution looks like it works to you.
At the end of your process, it DOES say they all leave (and therefore know their own eye colour, as this is the sole stated condition on which their leaving depends).
It's just that the process doesn't deductively lead to that end through only logic, observation and certainty.

This issue raises an epistemological point about what constitutes knowledge.
Is it knowledge when someone uses uncertain assumptions in order to determine a fact?
How about if someone uses uncertain assumptions in order to come to the same conclusion as someone who has used certain deductions?

I do not class these two things as the same, though perhaps you do, and perhaps you are vouching for a utilitarian "ends justify the means" approach to the puzzle. You do say that the ability to know their own eye colour is all that matters, and you say nothing about how they arrive at this information. Perhaps your point, all along, has been that as long as you think you know your own eye colour, even if through flawed logic, information counter to observation, and assumption, that's good enough?

[James S Saint]

You have only the choices of logical deduction, spontaneity, and pot luck.
The only one of those that always works is logical deduction.

You can't think of a single number that they could all choose that would not work because any number they all choose will work every time. As logicians, they know what always works because they can deduce what always works. I didn't try every number, I deduced that any number would work. So if I can deduce what always works, why can't perfect logicians?

Was there ever only one person on the island when the guru said there was one?
Noooo.
It is irrelevant.

The Guru never said that there was only one person on the island.

Who said anything about the guru saying that there was ONLY one??

Again, James, what you're saying suggests that you don't understand the logic behind the canonical solution.

What it suggests to me is that You do not understand logic.

But the counterfactual, "if there were only 1 he would leave on day 1", is still necessary to reach the conclusion.

You should explain that to Sil.

For some reason, though, when we get to 4, you and Fixed Cross discount the value of the same information, and suppose that the 1 would just spontaneously know that he has blue eyes. What is the difference between the case of 3 blue islanders and the case of 4 blue islanders such that the guru is necessary in the former but not in the latter?

What FC and I know is that it is not spontaneous, else it could not work every time they do it. Why don't you know that?

If we use your example for the case of 4 blue eyed islanders, I think what Silhouette is saying becomes even clearer:

I am blue out of 100 blues;
A) They all know [that they all know] [that there are at least 1 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 1 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 2 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 3 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 4 and (A) were true, they leave the 4th day].
.
.
N) They all know [that they all know] [that if there were only 100 and (A) were true, we leave the 100th day].

You have trouble with patterns, I take it..?

We are given that no one knows their eye color, and D requires that there is at least 1 who does.

(D) does NOT require that anyone know their own eye color. It makes the exact same hypothesis as the canonical version, "IF there were only one and he knew there was one."

The contradiction, as Silhouette has repeatedly explained, is that you're taking the knowledge out of context. 4 know there is at least 1. But if there were only 1, he would not know that there is at least 1 - until the guru speaks.

I haven't taken any more out of context than you;
"If there was only one when the guru said that was one" (which never took place).




But the question, still incorrectly answered was simply;
"IF everyone chose the SAME number to start counting the days before deducing their color, would it always lead to an accurate deduction of their color?"

So far, you have each said "no", yet you cannot come up with a single number that doesn't lead to an accurate deduction of the color, every time they do it.

It is a yes or no question with a request for what number that would be if your answer is "no".
Let me guess, you guys took an oath to ban yourselves if you turn out to be wrong...??
Seems it has to be something like that for you to go to such extreme lengths to avoid such a simple question.

[Silhouette]

You have only the choices of logical deduction, spontaneity, and pot luck.
The only one of those that always works is logical deduction.

Then why didn't you choose it?

Q: IF everyone chooses the exact same number to use in counting days before deducing that they are the remaining blue (or not), is there any number between 0-99 that would cause them to deduce improperly?
A: Yes. All of them.

^

You can't think of a single number that they could all choose that would not work because any number they all choose will work every time.

^ Clearly

So if I can deduce what always works, why can't perfect logicians?

Oh, I can think of a reason.

Was there ever only one person on the island when the guru said there was one?

The Guru never said that there was only one person on the island.

Who said anything about the guru saying that there was ONLY one??

What's the point in talking to you?

[James S Saint]

What's the point in talking to you?

I seem to remember telling You that there isn't much point in You talking to me, because you merely attempt to be contrary to anything I say.. and that usually leaves you being seriously wrong, and in this case, looking a bit foolish.

[Carleas]

Code: Select all

I am blue out of 4 blues;
A) They all know [that they all know] [that there are at least 1 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 1 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 2 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 3 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 4 and (A) were true, we leave the 4th day].

Compare with

Code: Select all

I am blue out of 3 blues;
A) They all know [that they all know] [that there are at least 1 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 1 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 2 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 3 and (A) were true, we leave the 3rd day].

What is the difference here? Every blue knows that every other blue knows that there is at least 1 blue in both the three and the 4 example. Why don't the three leave the island on the first day?
The canonical answer is that they can't leave because they need to know [that they know [that they know [that there are at least 1 blue]]], and that the 4 islanders need to know [that they know [that they know [that they know [that there is at least 1 blue]]]]. Your answer seems to reject the nested knowledge requirement, so I'm curious why you see these cases as requiring a different logical process.

What is the difference between an island with 3 blues and an island with 4 such that the four can divine their own eye color just by counting the days?

[James S Saint]

Code: Select all

I am blue out of 4 blues;
A) They all know [that they all know] [that there are at least 1 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 1 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 2 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 3 and (A) were true, they would leave the 3rd day].
G) They all know [that they all know] [that if there were only 4 and (A) were true, we leave the 4th day].

Compare with

Code: Select all

I am blue out of 3 blues;
A) They all know [that they all know] [that there are at least 1 blues].
B) They all know [that they all know] [that they don't know their own color].
C) They all know [that they all know] [that they will leave if they could deduce their color].
D) They all know [that they all know] [that if there were only 1 and (A) were true, they would leave the 1st day].
E) They all know [that they all know] [that if there were only 2 and (A) were true, they would leave the 2nd day].
F) They all know [that they all know] [that if there were only 3 and (A) were true, we leave the 3rd day].

What is the difference here? Every blue knows that every other blue knows that there is at least 1 blue in both the three and the 4 example. Why don't the three leave the island on the first day?

That seems like a silly question.
They don't leave the 1st day because they see another uncounted blue.
That is the same reason they don't leave in the canonical version.

The canonical answer is that they can't leave because they need to know [that they know [that they know [that there are at least 1 blue]]], and that the 4 islanders need to know [that they know [that they know [that they know [that there is at least 1 blue]]]]. Your answer seems to reject the nested knowledge requirement, so I'm curious why you see these cases as requiring a different logical process.

You are STILL trying to argue about HOW they know what number to choose.
THAT is NOT my question.
The question is a mere hypothetical, "IF they ALL choose the SAME number..."

What is the difference between an island with 3 blues and an island with 4 such that the four can divine their own eye color just by counting the days?

IF they ALL choose the SAME number to count with/from (within each group), there would be no difference in the result.
They would both accurately deduce their own eye color.
What makes you think that it wouldn't work for 3 or 2?

[Carleas]

What makes you think that it wouldn't work for 3 or 2?

Just to be clear, are you saying that your method would work for an island on which there are only two people, both of whom have blue eyes?

[James S Saint]

What makes you think that it wouldn't work for 3 or 2?

Just to be clear, are you saying that your method would work for an island on which there are only two people, both of whom have blue eyes?

"Method"??

I have asked you a question. Why is that so hard to comprehend???
If you had only 2 on the island and you used that same "counting and deducing" procedure, given that everyone on the island is starting with the same count, would they still deduce their proper color? The obvious answer is "yes", but it seems you are hell bound to never say it.
Hell, any 3rd grader could figure that much out.