Math Fun

[James_S_Saint]

@Flannel Jesus

I showed you why brown doesn’t work for 1 person.

Show me when you had only one person.

[Hobbes_Choice]

This is a simple enough problem.
There are two possible outcomes.

1. That the problem is coherent and has a solution within the parameters of the situation.
2. That the solution, the problem or the situation is badly conceived and the answer is bogus.

Now, as the problem/ solution invites us to believe that the logicians given no new information will (in the passage of time), find out what their own eye colour is without the help of ANY further information except the fact that they can see three groups; 100 of one colour; 99 of another and 1 green. ( Remember you are speaking for any of the logicians not the guru, and thus they all have exactly the same information.The possible permutations being B100, b99, plus their own unknown colour not necessarily B or b, OR B99, b100 plus their own unknown colour not necessarily B or b.)
Why would time provide more information?
This is completely counter intuitive. Given that it is counter intuitive it only remains to reverse engineer what the problem is with the ‘situation’.
As I could not think of a solution. I asked others for their’s. Thanks Flannel for trying. But for the reasons you are not coming to realise the solution assumes knowledge that the logicians cannot have.
The final outcome is 2) above.

[Flannel_Jesus]

@James S Saint

@Flannel Jesus

I showed you why brown doesn’t work for 1 person.

Show me when you had only one person.

If there is 1 brown-eyed person, he sees various other people, not with brown eyes, and he doesn’t know his own eye color, could be red, could be blue (until the other blue-eyeds leave), could be green, whatever, and so he doesn’t leave, because he doesn’t know.

[Flannel_Jesus]

You’ve asked me this goddamn question like 3 times now, literally. Brown doesn’t work for 1 person. Remember.

[Hobbes_Choice]

@Flannel Jesus

@Hobbes Choice

It only works if there are only two logicians; one brown eyed and one blue eyed.
As the blue eyed one can see the other has non-blue eyes he can conclude that his own eyes are blue.

Why only two?

What if there is one blue-eyed and 400 brown-eyed? The guru still says ‘I see someone with blue eyes,’ the blue-eyed person sees 400 people without blue eyes, and knows that he has to be the only one. So surely he can leave.

Granted.
But only he can leave.

[Flannel_Jesus]

You also seem to keep changing your mind…but never to the correct position. You change it from “it works for brown-eyed people too” to “it works for nobody”. What is it now? It’s hard to keep you straight.

[James_S_Saint]

@Flannel Jesus

@James S Saint

@Flannel Jesus

I showed you why brown doesn’t work for 1 person.

Show me when you had only one person.

If there is 1 brown-eyed person, he sees various other people, not with brown eyes, and he doesn’t know his own eye color, could be red, could be blue (until the other blue-eyeds leave), could be green, whatever, and so he doesn’t leave, because he doesn’t know.

YOU DO NOT HAVE ONLY ONE BROWN EYED PERSON ON THE GODDAMN ISLAND!!!

[James_S_Saint]

@Flannel Jesus

You also seem to keep changing your mind…but never to the correct position. You change it from “it works for brown-eyed people too” to “it works for nobody”. What is it now? It’s hard to keep you straight.

Go give me that BS.
I said that it worked for the browns also and that led to realizing the because of that and them being even in number, they get trapped and thus neither can decide in the end.

[Flannel_Jesus]

@Hobbes Choice

@Flannel Jesus

@Hobbes Choice

It only works if there are only two logicians; one brown eyed and one blue eyed.
As the blue eyed one can see the other has non-blue eyes he can conclude that his own eyes are blue.

Why only two?

What if there is one blue-eyed and 400 brown-eyed? The guru still says ‘I see someone with blue eyes,’ the blue-eyed person sees 400 people without blue eyes, and knows that he has to be the only one. So surely he can leave.

Granted.
But only he can leave.

Uh huh.

So, Conclusion 1 we’ll call this.
Logically Deducible Conclusion 1
LDC1
is
If there is only one blue-eyed person, he leaves on the first day.

Now, consider the case of two blue-eyed people.
I’m a person on the island, I see one person with blue eyes, right?
I see 100 people with brown eyes.
And the guru.
Now, there are only 2 possibilities:
a) I don’t have blue eyes, and so the person I see with blue eyes is the only blue-eyed person.
b) I have blue eyes

Let’s consider case a. If the person I see with blue eyes is the only blue eyed person, he MUST leave on the first day.
LDC1, remember? He must leave.
Right?

So, I wake up after the first night, and he’s still here.
a) must be false.
It MUST be.
You agreed to this. You agreed that if there is only one blue-eyed person, he MUST leave on the first day.
So, if I wake up after the first night and he’s still here, he MUST NOT be the only blue-eyed person.
b is true, I have blue eyes, I can leave the second night.

[Flannel_Jesus]

@James S Saint

YOU DO NOT HAVE ONLY ONE BROWN EYED PERSON ON THE GODDAMN ISLAND!!!

YOU JUST ASKED ME TO SHOW YOU WHEN I ‘HAD’ ONLY ONE PERSON!!!GODDAMN!!

Did you not want me to show you that? Why’d you ask for it? That’s a silly response to something you asked for bro.

[James_S_Saint]

@Flannel Jesus

@James S Saint

YOU DO NOT HAVE ONLY ONE BROWN EYED PERSON ON THE GODDAMN ISLAND!!!

YOU JUST ASKED ME TO SHOW YOU WHEN I ‘HAD’ ONLY ONE PERSON!!!GODDAMN!!

Did you not want me to show you that? Why’d you ask for it? That’s a silly response to something you asked for bro.

Because you NEVER only had one of any color… of any color.
The LOGIC still applies, regardless.
But if the groups just happen to be even, then it still leads to a stagnation.

[Flannel_Jesus]

The logic that ‘applies’ to one brown-eyed person is that he doesn’t get off the island. So if the logic still applies regardless, then no brown eyed people leave. Idk what logic 100 brown-eyed people could use to figure out that they have brown eyes. You haven’t explained it. You say the logic still applies, but you don’t specify what logic.

[phyllo]

Seems to be another case of messed up wording.

The ‘common knowledge’ formulation only allows two colors. Contrary to this statement :

So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

[James_S_Saint]

@Flannel Jesus

The logic that ‘applies’ to one brown-eyed person is that he doesn’t get off the island. So if the logic still applies regardless, then no brown eyed people leave. Idk what logic 100 brown-eyed people could use to figure out that they have brown eyes. You haven’t explained it. You say the logic still applies, but you don’t specify what logic.

Well, give me some time to figure out how to cause you to see it… if I even can.
Realize that Hobbes has not been able to see the logic that you present.
That is not a black or white situation.
There are degrees of logic wherein even an otherwise very logical person can’t see (yet) what was actually true.
So give me some time to come up with something to reveal the distinction between what you seem to be locked into versus what I clearly see as more rational.

[James_S_Saint]

FJ,

Can you agree that;

1. If there had been only one blue eyed, that person could finally deduce his eye color because of what the guru (supposedly always correct), had said?

2. Because there were actually more than one blue eyed, the guru didn’t say anything that wasn’t already known?

3. Any logic that actually applies can only apply to the actual situation, not hypothetical situations such as there only being one blue eyed?

That isn’t my entire argument. I just want to make sure that we are on the same page.

[Hobbes_Choice]

@Flannel Jesus

Uh huh.

So, Conclusion 1 we’ll call this.
Logically Deducible Conclusion 1
LDC1
is
If there is only one blue-eyed person, he leaves on the first day.

Now, consider the case of two blue-eyed people.
I’m a person on the island, I see one person with blue eyes, right?
I see 100 people with brown eyes.
And the guru.
Now, there are only 2 possibilities:
a) I don’t have blue eyes, and so the person I see with blue eyes is the only blue-eyed person.
b) I have blue eyes

Let’s consider case a. If the person I see with blue eyes is the only blue eyed person, he MUST leave on the first day.
LDC1, remember? He must leave.
Right?

So, I wake up after the first night, and he’s still here.
a) must be false.
It MUST be.
You agreed to this. You agreed that if there is only one blue-eyed person, he MUST leave on the first day.
So, if I wake up after the first night and he’s still here, he MUST NOT be the only blue-eyed person.
b is true, I have blue eyes, I can leave the second night.

When are you going to give up this nonsense. Why do I have to repeat myself.
The case is that there is more that one blue eyed person. LDC1 is not applicable.
CASE CLOSED!
The point is, that BECASUSE LDC1 is not applicable. NO ONE can use that conclusion; thus NO ONE can leave.
That is it!
THe problem is shit.

[Hobbes_Choice]

@James S Saint

@Flannel Jesus

The logic that ‘applies’ to one brown-eyed person is that he doesn’t get off the island. So if the logic still applies regardless, then no brown eyed people leave. Idk what logic 100 brown-eyed people could use to figure out that they have brown eyes. You haven’t explained it. You say the logic still applies, but you don’t specify what logic.

Well, give me some time to figure out how to cause you to see it… if I even can.
Realize that Hobbes has not been able to see the logic that you present.
That is not a black or white situation.
There are degrees of logic wherein even an otherwise very logical person can’t see (yet) what was actually true.
So give me some time to come up with something to reveal the distinction between what you seem to be locked into versus what I clearly see as more rational.

Maybe you cannot read.
I can see the logic.
Look back!

[Hobbes_Choice]

This is a simple enough problem.
There are two possible outcomes.

1. That the problem is coherent and has a solution within the parameters of the situation.
2. That the solution, the problem or the situation is badly conceived and the answer is bogus.

Now, as the problem/ solution invites us to believe that the logicians given no new information will (in the passage of time), find out what their own eye colour is without the help of ANY further information except the fact that they can see three groups; 100 of one colour; 99 of another and 1 green. ( Remember you are speaking for any of the logicians not the guru, and thus they all have exactly the same information.The possible permutations being B100, b99, plus their own unknown colour not necessarily B or b, OR B99, b100 plus their own unknown colour not necessarily B or b.)
Why would time provide more information?
This is completely counter intuitive. Given that it is counter intuitive it only remains to reverse engineer what the problem is with the ‘situation’.
As I could not think of a solution. I asked others for their’s. Thanks Flannel for trying. But for the reasons you are not coming to realise the solution assumes knowledge that the logicians cannot have.
The final outcome is 2) above.

[Hobbes_Choice]

@James S Saint

FJ,

Can you agree that;

1. If there had been only one blue eyed, that person could finally deduce his eye color because of what the guru (supposedly always correct), had said?

2. Because there were actually more than one blue eyed, the guru didn’t say anything that wasn’t already known?

3. Any logic that actually applies can only apply to the actual situation, not hypothetical situations such as there only being one blue eyed?

That isn’t my entire argument. I just want to make sure that we are on the same page.

Dah. It’s what I have been saying all along.
Do drop in won’t you?

[James_S_Saint]

@Hobbes Choice

When are you going to give up this nonsense. Why do I have to repeat myself.
The case is that there is more that one blue eyed person. LDC1 is not applicable.
CASE CLOSED!
The point is, that BECASUSE LDC1 is not applicable. NO ONE can use that conclusion; thus NO ONE can leave.
That is it!
THe problem is shit.

I often argue with FJ, but I certainly do not consider him an idiot.
I have to ask, why do you think both he and I such idiots that we cannot see even the simple minded “logic” that you present… but see a flaw that “impossibly”, you cannot see?

Why do you think yourself so very above us in the arena of logic?