Well, let’s stop beating around the bush. You say that, in the case of 1 brown-eyed person, he could figure out that he has brown eyes. So…how? How can he? You show me how, and I’ll change my mind. I pinky promise. It has to be logically deducible from the given information in the problem.
The logic fails for all logicians as none of them know whether or not they belong to the blue, brown, green or red groups.
When they all turn out to leave none of them can say with any certainty what their eye colour is.
The solution exploits the knowledge WE have of the true situation, it cannot be used to ascertain each logician’s eye colour.
The fact that the logic fails for people of brown eyes ought to alert you to this simple fact.
Please take the trouble to examine your own words:
The problem clearly states that NO logician knows what his own eye colour is, so much that it also state the possibility that red eye colour might exist. Thus he is unable to assess what group he belongs to. He cannot assume he is not brown eyed, thus by your won statement, he is unable to apply the logical solution to himself.
I know you get this. You have to know that the fault in logic has to apply to every logician as none know what there eye colour is.
You accuse me of not listening; not understanding, and have patiently re-iterated your points.
I have taken on those points and shown how they do not help the case.
Please have the curtesy to do the same for me.
You still haven’t responded to my recent key response towards you, Hobbes. Please do that. It was the one in which I provide a syllogism as to how you extrapolate that 2 people can leave from the premises that you already accepted about 1 person being able to leave.
On this island there are 100 blue-eyed people, 100 brown-eyed people
The logic isn’t about what is actually there. The logic merely shows how it could be done IF there were only one, or two, or three, and so on.
The logic applies to any size group over 1.
Ok, that’s your CLAIM, brother. Now, show me how it works for 1 brown eyed person. Please. Don’t claim it any more, please, little brother, just prove it. 1,2,3, go
It cannot work for any logician, unless they already know their eye colour.
It can’t work for the brown eyed. yes? Then how can it work for any of them; none of the know what their group is.
If you had to consider that case of 1 brown-eyed person that would be true. But you don’t have to consider it because there are several brown-eyed people. They can start playing the game.
In what way is the guru’s statement a clue? Why is blue-eyed group privileged? Everyone already knows that there exit blue-eyed people who are visible to the guru. Similarly, everyone already knows that there exit brown-eyed people who are visible to the guru.
Phyllo, I’ve made a response to you that you haven’t responded to yet. I’m in the middle of a different conversation with you. Please respond to that one.
I’m keeping each of my separate disagreements in order, because you all disagree for different reasons. If you want clarity, respond to the posts addressed to you. If you don’t want clarity, keep doing what you’re doing guys.
If there was merely one brown, you would have a problem, but there isn’t.
After the guru speaks and the game begins, the brown eyed people can still count 99 days just like the blues and for the same reasons. They never had the situation of having only 1 of a color. That step in the logic is merely mental. Everyone always saw more than one of each color.
Sadly the solution assumes that there are 100 of each, Sadly it also says that the logicians DO NOT KNOW the disposition of eye colour; their eyes could even be red.
Thus the solution is false.
Duplicate post
Yeah, but so what? They count 99 days and then what?
Hobbes, I’ve proved the case of 2 blue-eyed people from premises you accepted. Do you accept that proof or not? Please respond.
At 99 days, each brown eyed knows that the other brown eyed haven’t been leaving because of him. They all suddenly leave together with the blue eyed.
Can you prove it James? In a similarly thorough way to what I’ve started with the blue-eyes? I don’t see the proof. I don’t see how any brown-eyed person can deduce his own color of eyes, ever, given what the guru said.
Ooppss… wait.
Actually, because the groups are actually even, even at 99 days, NO ONE can deduce and no one leaves.
Dah.
As the dispositions of the eyes could be any of the following; they cannot know which day to leave on.
G= green, B brown, b = blue, R = red (or colour unknown, could be G)
Permutations.
G=1, B=100, b=99, R=1
G=1, B=101, b=99
G-1, B=99, b=100, R=1
G=1, B=99, b=101
At the end of the allotted time, any given logician is clueless about the colour of his eyes. There is no way that he can infer his own eye colour as the above permutations will never change regardless of the time passing.
Let me explain the thought-process that goes into a brown-eyed person in various situations:
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.
If there are 2 brown-eyed people, they each see 1 other brown eyed person, as well as various people with non-brown eyes. He doesn’t know his own color. Is there any information available for him to figure it out? No. Could be green, could be blue (until the other blue-eyeds leave), could be red, could be purple.
If there are 3 brown-eyed people…you get the picture.
So…you’re claiming that some deductive thought process could be used by a brown-eyed person to prove that he has brown eyes. What is it?
So you’re now changing your position to Hobbes position? Is that right? Not that you’re not allowed to – totally, if that’s what you now think – but I have some solid arguments against that that I hope you’ll be open to. Hobbes wasn’t open to them, idk why, he seems like a reasonable guy but he stopped responding to me after I proved the 2-blue case…