And again, you make no effort to know the truth of the puzzle, but merely focus on defending yourself.
Two problem with that:
- It is not a general solution since it doesn’t work for 1, 2, or 3 blues.
- There would have to be an agreement to start from 97 rather than 1. That adds an extra element of common knowledge which is not part of the puzzle definition.
No one asked for a general solution to work for all combinations of possible groups.
So no, that is NOT a problem.
The “agreement” is the fact that they can’t deny it.
So no. No problems.
It has to work for any number of blues. You don’t have different cases … in case A start counting from 1 … in case B start counting from 97.
Are you sure you guys understand the concept of ‘fun puzzle’?
:-"
Who made up THAT rule??
You start with the real situation.
The others are merely explanatory.
I discovered that wasn’t the case the second day this whole thread was started… too many ego sensitive people can’t enjoy being sometimes wrong.
Or: “there has to be an agreement to start with reality instead of non-reality”.
Yes indeed. You’re close to getting it.
And that is what’s deluding you chaps - you all would like to start with the hypothesis that there is in fact only one blue and that this one knows his eye color.
But as I keep saying, this is not the case and this is not how the hypothesis functions. It doesn’t work because it’s true, but because it’s not.
First the Facts of the situation are drawn.
Then the negative proof series is initiated to draw Fact from Fact.
There is Never Factually a moment when there is actually only one blue and the guru speaks that she sees one blue. And yet your deductions rely on this hypothesis. Why? Because what is usable of this hypothesis is that it is required to be untrue to explain the real situation.
Perhaps think of it this way;
If there was only one and he didn’t know there was at least one and the guru told him there was one, he would leave.
But because there was not only one and he did know there was at least one when the guru told him there was one, he wouldn’t leave.
This is still valid;
We don’t care about any lesser combinations.
The 4th day solution does not work, because it is not knowing that there is someone on the island with blue eyes (or that there are 97 people on the island with blue eyes) that makes the logic work. Both James and Fixed Cross have acknowledged as much in the 2 case and the 3 case
In the 2 blue case, everyone in this thread agrees: everyone on the island knows “there is at least one person on the island with blue eyes” before the guru utters those words.
This is true in the 3 case, the 4 case, every case greater than 1. Yet still, everyone in this thread agrees that when the guru says, “there is at least one person on the island with blue eyes,” in the 2 and 3 case the islanders learn important information, and leave on day 2 and 3, respectively, and no sooner. This tells us that content of the guru’s statement is not what’s important.
Another way to see this is to have her whisper the statement separately to each islander. If she does this, the 2 case doesn’t work: neither islander knows what she told the other (because he can’t rule out the possibility that she said “I see someone with blue eyes” to him, and “I see someone with red eyes” to the other), so they can’t deduce anything from the statement, no matter how long they wait.
Fixed Cross, you say that “The fourth hypothetical is impossible if there are as much as 4 blues,” but it’s no more impossible than the 3rd (after all, if you can see three blues, it can’t be the case that there are only two).
And note I say “no more impossible”, and not “also impossible”. Neither hypothetical is impossible. Why? Because it is not an islander making the hypothetical, but a hypothetical islander making the hypothetical, and the hypothetical islander has a different set of hypothetical facts.
YOUR logic doesn’t work.
But the reality is that the 4 Day does work.
No matter if you are a brown or a blue or a red, it still works.
Use a simple truth table and see.
You keep trying to make things work for cases that don’t exist on the island.
Carleas, you keep trying to work with what others on the island MIGHT think. That is misleading although not entirely wrong to do.
What you should be asking, as FJ and I discussed, is how does everyone know to use 97 without just assuming that they are blue themselves.
A simpler case to see is the 50 Day case.
In the 50 Day case, you simply divide the total number on the island by 4.
Again, because everyone is starting with the same number, after a certain amount of days, they all leave.
As long as everyone is starting with the same number, it will always work.
The trick is in figuring out what number to start with.
Perfect logicians are better than us, so that is something they can figure out.
Our game is to try to figure out how they knew.
I don’t think there is such a table. If you think so, show us.
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].
The problem enters at the same line in each:
Reads:
D) They all know [that they all know] [that if there were only 97 and (A) were true, they would leave the 1st day].
Should read:
D) They all know [that they all know] [that if there were only 97 and those 97 knew that (A) were true, they would leave the 1st day].
It’s not enough for A to be true outside the nested hypothetical, the hypothetical islanders need to know that A is true.
Going back to my X = not-blue, O = blue notation.
A blue, call him A, sees 99 blues. He doesn’t know what he is, but he know he might be not-blue:
A B C D E F . . . .
X O O O O O . . . .
He knows that every blue he can see can see at least 98 blues, but that they don’t know what color their eyes are. But they all know they might be not blue. So as far as A knows, for B the following is possible:
A B C D E F . . . .
X X O O O O . . . .
Note here: A knows that B is blue, but A also knows that B doesn’t know that.
A also knows that B knows what A’s eye color is, but since A doesn’t know it, he can’t count it. A knows that B knows that there are 98 blues.
A knows that B knows that for C, the following is possible:
A B C D E F . . . .
X X X O O O . . . .
Note here: A knows that B knows that C is blue, but A also knows that B knows that C doesn’t know that.
And A knows that B knows that C knows what B’s eye color is, but since B doesn’t know it, he can’t count it. A knows that B knows that C knows that there are 97 blues.
A knows that B knows that C knows that for D, the following is possible:
A B C D E F . . . .
X X X X O O . . . .
Note here: A knows that B knows that C knows that D is blue, but A also knows that B knows that C knows that D doesn’t know that.
And A knows that B knows that C knows that D knows what C’s eye color is, but since C doesn’t know it, he can’t count it. A knows that B knows that C knows that D knows that there are 96 blues.
A knows that B knows that C knows that D knows that for E, the following is possible:
A B C D E F . . . .
X X X X X O . . . .
Note here: A knows that B knows that C knows that D knows that E is blue, but A also knows that B knows that C knows that D knows that E doesn’t know that.
And A knows that B knows that C knows that D knows that E knows what D’s eye color is, but since D doesn’t know it, he can’t count it. A knows that B knows that C knows that D knows that E knows that there are 95 blues.
It continues on. The point is, as you go down the nested hypotheticals, you are hypothesizing down your total number of blues. They all know that they all know that there are at least 97 blues, but they don’t all know that they all know that there are at least 97 blues who know that there are at least 97 blues. Indeed, there aren’t.
No, that is what YOU keep throwing in. I already demonstrated that it works without that.
The 97 are already taken care of as “everyone”.
There are no 97 “others” outside of everyone.
Everyone knows that there are at least 98.
But everyone doesn’t know that there are more than 98.
But then again everyone does know there are more than 97.
It doesn’t, and you haven’t. You just keep saying it.
I don’t understand why it’s so unpalatable. You accept that in the 2 case, they need to know that they know, and that in the 3 case the need to know that they know that they know, but some how the 4 cases requirement that they know that they know that they know that they know is unacceptable, and the 5 case requirement that they know that they know that they know that they know that they know is not even worth talking about. What’s the principled distinction?
I showed you every possible case for everyone on the island (except the actual browns… trivial case).
How do you think it didn’t work??
Do you understand that if they all start with the same number, as long as it is less than 100, it will work for everyone?
If the guru had said, “I see 50”, they could have all started with 50?
I actually agree with this. And if she had said “I see 97,” they could have started with 97. But she didn’t. She said 1.
And notice: if she had whispered “I see 99” to each, they wouldn’t have been able to leave on day 1. They wouldn’t have been able to leave at all, for the same reason the two to whom she whispers “I see 1” can’t leave.
They actually would have left before she had a chance to even whisper.
But you didn’t answer the question;
Do you understand that as long as everyone starts with the same number, it will always work?
How did that 97 example NOT work, since everyone accurately discovered their eye color?
No, it won’t. It requires an assumption that isn’t justified: that the number know that they know that they know that they know…
What’s the difference between
- knowing that they know that they know, and
- knowing that they know that they know that they know,
such that the second can happen spontaneously without the introduction of the information necessary to make the first happen?