No - it is very clear what the guru says. And there is nothing new here. No monk can tell from what he says what his own eye colour is. The guru only says what they already know.
Counting is not relevant. Nor are the number of days. They do not appear in the problem.
Tell me in your own words how any sing;e monk can know the colour of his own eyes from the information given.
OR tell me why this is not relevant to the solution.
Counting is implicit in the physical situation. The total number of blue-eyed people is the number you see or the number you see plus one if you yourself have blue eyes. The number of days which have passed, since the guru made the statement, gives you information about your own eye color.
If there is 1 blue-eye, then he leaves on night 1 obviously.
If there are 2 blue-eyed then, there is one night of doubt which prevents leaving on night 1. You see 1BE but you might also be 1BE, so you canât move. When the 1BE, that you see does not leave, then you know that he sees your blue eyes-otherwise he would have left. He gives you information and receives the same information in return.
The logic extends to any number of people. If you see ânâ blue-eye people, on the day after night ânâ you will have a count of how many blue-eyed people your blue-eye companions are seeing.
Hobbes, you are missing the point.
Since you only see 99, you now know it has been you all along holding up the show.
And all 100 deduce that of themselves at the same time. They all leave.
PHylo and James S SAint.
Iâm not missing any point.
The problem as presented to me does not make any account to days or time.
The guru speaks once and tells them nothing they do not already know.
See below.
I quote:
But as they are perfect logicians, they all know that they no one can tell what their eye colour is, and that the words of the guru say nothing more than they already know. And the passage of day is completely irrelevant, as none of them can tell which is their own eye colour. And that is why none of them can ever leave.
BECAUSE NO ONE KNOWS WHAT THEIR OWN EYE COLOUR IS. That is why they do not leave. Them not leaving cannot be used to conclude that you have blue eyes. Your eyes could be red or purple.
Itâs like you have not actually read the problem.
After 99 days all the blue eyed people still have no idea what their own eye colour is, all they know is that with frustration he can still see 99 other blue eyed persons, who could have left is only they had known, and 100 brown eyed persons who also could have left had they only known, and the 1 green eyed guru who is still clueless as to her own eye colour as no one is able to tell anyone how many and what colour is the total eye count on the island.
Iâm surprised and rather puzzled that so many of you are falling for that.
No. As no one knows their own eye colour you cannot infer from the fact that no one leaves because they do not leave as they are also ignorant of their eye colour as are you.
The number of passing days cannot change that fact. The solution claims a temporal solution for an eventual simultaneous action; bogus.
THis is similar to the surprise exam problem, except that it is poorly executed by a person that has not thought it through properly.
With the surprise exam problem the students conlcude that the exam has to be on Friday, but they can only conclude that due to the fact that they will all have the knowledge that the exam would not be a surprise for each day. They have a line of knowledge which is NOT available to the logicians throughout the time, as the fact that no one is leaving assumes an ignorance and knowledge that is unavailable to any of them at any time.
For the students at least they have the experience each day of knowing that no exam has been sprung on them.
But as with that problem we all know it is ridiculous as it this one, and the one with the monks.
At the time of speaking no one knows when the exam is set so - it is a surprise exam at the time of notice. The fact is that the exam could be on any day or surprise, surprise there is no exam until the following week.
With the logicians not seeing people leave is meaningless, when the problem state that your eyes might also be brown or even green.
What it the matter with you people?
Stop saying âcase closedâ please, itâs dismissive and disrespectful.
Now, let me try again:.
You say if there were 1 with blue eyes, he could look around, conclude he has to be the one, and leave the first night, right?
Let me stress this: if there is only 1 guy with blue eyes, it is logically necessary, given your reasoning, that he leaves the first night.
Is this acceptable?
Now, letâs say Iâm a guy, I donât know what color my eyes are, but I see only 1 other person with blue eyes.
There are 2 and only 2 options: either heâs the only person with blue eyes, or we both have blue eyes.
So, when I wake up after the first night, I can verify which of those two is true, definitively.
If he is gone, I know that he was the only one with blue eyes.
If he is still here, I must have blue eyes.
If I didnât have blue eyes, he would have left. You said this yourself. You agree with that. If I donât have blue eyes and he does, he looks around, sees no one else with blue eyes, and leaves, so therefore if he doesnât leave, he must see someone else with blue eyes. Since I see no one else with blue eyes but him, he must be seeing MY blue eyes. Thatâs the only option.
Now, try to let go, try to let go, I know youâve concluded that youâre right and Iâm wrong and itâs hard to back down now, but donât think about it as âbacking downâ, just think about it as a logical problem. Think about what Iâm saying. Think about the logic. Realize that if you see 1 person with blue eyes, and heâs the only person with blue eyes, he must leave on the first night. If he doesnât leave, the only explanation is that someone else has blue eyes. That person must be you. Donât worry about âbacking downâ, you donât need to save face here, in fact youâd look better if you just thought about it for a bit and realised the logic behind it. This is the real and actual solution, it has already been solved, soâŚthereâs no question of if itâs correct or not, the only question is do you get it?
1 more time, for perfect clarity:
In the case of 2 blue-eyed people:
You see only 1 blue-eyed person.
If and only if there is 1 blue-eyed person, he will leave on the first night. (y/n?)
If X, then Y.
X = there is 1 blue eyed person, Y = he leaves on the first night.
You donât see him leave on the first night.
Not Y.
Negating the consequent:
Not Y, therefore Not X.
There is NOT 1 blue eyed person.
But since you only see 1, the other, by deduction, must be yourself.
QED
Iâm trying to take it one step at a time. We donât need to bring up 99 yet. You first need to understand why 2 works before you can fathom why 100 works. So, donât bring up the âhe sees 99â please, I know. My argument for now is just about 2 people. 2 people with blue eyes. Not 100âŚnot yet. Weâll get to that later.
This is not about backing down or facing up.
The logic is faulty and I know when you think it through you will get it.
Letâs say that I know what your eye colour is and that you do not.
I tell you that someone has blue eyes.
When do you leave and what is your eye colour?
Use any logic you can.
When I read your post Iâll then tell you what your eye colour is.
For utter clarity.
200 logicians donât know what colour their eyes are.
The next day the 200 of them are still there.
Conclusion: each cannot determine the colour of his eyes.
Day 2 the facts remain the same.
Your case is based on the assumption that an inference can be made about eye colour when no one leaves. But they do not leave as they are ignorant.
Day 100 the case remains the same.
The fact is that at no time can any logician say what his own eye colour is, and thus has no warrant to leave.
I think you are running away with the idea that I am stupid, I understand the logic, but do not accept itâŚ
This does nothing more that prove that logic whilst it might appear convincing is not a warrant for truth or action.
I remain disappointed that you are so gullible to accept the conclusion.
It remains the case that given a poll, the leaving logicians would still be unable to articulate what their own eye colour was, and as such would be cheating on the rules. They might try to escape the Island with this bit of Flim flam but not if I were the gatekeeper.
Okay, that was the standard argument from the web site.
If there is one blue-eyed person, then the guru is giving an important piece of information. But since there are several blue-eye and several brown-eyed people, then there does not appear to be anything new disclosed. He could have said that he saw a brown-eyed person.
The browned-eyed people could start counting the days and all leave together on the critical night. That would be before the blue-eyed if there were fewer brown-eyed people.
So letâs say that there is you(Y), 1 guru, and 6 other people - you see 3 blue-eyed(X1,X2,X3) and 3 brown-eyed(X4,X5,X6). The guru makes the same statement.
What is seen:
Y(blue eyes) see 3 blue, 3 brown
X1 or X2 or X3 sees 3 blue, 3 brown
X4 or X5 or X6 sees 4 blue, 2 brown
Therefore, logically X4 should assume brown eyes and leave on night 3. So the smaller group of browns leaves first.
What if X4 is blue-eyed?
What is seen:
Y(blue eyes) sees 4 blue, 2 brown
X1 or X2 or X3 sees 4 blue, 2 brown
X4 sees 4 blue, 2 brown
X5 or X6 sees 5 blue, 1 brown
Therefore, X5 and X6 will leave on night 2.
Smaller group leaves first.
What if the groups are equal size? 3 blue (X1,X2,X3) and 3 brown(X4,X5,X6)
What is seen:
X1 or X2 or X3 sees 2 blue, 3 brown
X4 or X5 or X6 sees 3 blue, 2 brown
Therefore, everybody leaves on night 3.
Nope, there is no information as to the disposition of eye colours, and has such no logician would be able to tell what is own eye colour was a the end: red blue or brown. He could say that 99 were one colour and 100 were another, but would still be ignorant of his ownâŚ
Your example relies on more than the knowledge each logician has at his disposal, his eyes could still be pink, red or polkadot.
Each of your algorithms, relies on unavailable knowledge.
This is the case until the end of time.
As a matter of interest, are you also convinced by the Surprise Exam Problem?.
What about Buridenâs ass?
I have a good solution to that one.
Thatâs why you think that âMath Funâ is an oxymoron ⌠because you are not having fun with it.
But you JUST agreed, you JUST AGREED! That IF he was the only one with blue eyes, he WOULD HAVE been able to determine the color of his own eyes. So to say âHeâs still unable to determine the color of his own eyesâ by necessity means that heâs NOT the only one with blue eyes. Do you not see how that follows?
Premise 1: If he was the only one with blue eyes, he would be able to determine his own eye color â THIS YOU AGREE WITH
Premise 2: He was not able to determine his own eye color â THIS YOU AGREE WITH
Conclusion: He is not the only one with blue eyes.
So, we can logically conclude that heâs not the only one with blue as, as per the above proof which is both logically sound and with premises youâve already accepted. And since heâs not the only one, but you donât see anybody else with blue eyes, who does that leave? Only you.
If you have any problems with the argument, refer to the premises or conclusion and explicitly lay out which you agree with and which you donât, and why. Iâve explicitly and clearly laid out the argument, so that should be pretty easy for you to do.
Why doesnât the smaller group leave first FJ?
I donât know what that question means.
Everybody knows that there are many blue-eyed and many brown-eyed people there. So the guruâs statement that he sees a blue-eyed person doesnât add info for anyone. The brown-eyed people can start counting the days just like the blue-eyed people. I posted an example earlier. Logically the fewer people that belong to a group, the sooner the trigger date will be reached.
Thus far we have not done any maths as such.
But when we do, we often come across problems that only exist in maths, not in nature, or that are paradoxical in logic but not in nature or have solutions in maths and logic, but are bollocks in nature.
Such are the problems we have so far discussed, and most of the other so-called paradoxes.
For example. Xenoâs arrow, or Achileus and the Tortoise are solved in nature by shooting an arrow or running a race. The Monk/Logician solution is falsified by practical example; and Buridenâs Ass is falsified by rejecting the synchronic and thinking it through with the diachronic.
Maths and Logic are incomplete tools for the modelling of reality. Reality is not written in Maths or Logic, and they both have to comply with reality or show themselves false.
Further proof that you canât persuade with any logical argument to those who canât perceive logic.
âŚthus they got religion.