Still avoiding the question.
If there were only 100 and you were one of them and starting a count from 100;
Go ahead, say it again, "But the canonical solution says…"
I don’t even know why there’s still an argument about the ‘counting from 97’ thing. It was already demonstrated that it doesn’t work.
The solution basically works like this:
You see 99 blues, you see that they don’t leave after 3 days, and you conclude that there must be 100 and you’re blue-eyed.
But James already agreed that if there were in fact only 99, they wouldn’t leave after 3 days anyway, so a brown-eyed guy would, as above, see that there are 99 and see that they don’t leave after 3 days. So clearly seeing 99 blue-eyed people and seeing that they don’t leave after 3 days is not enough to conclude that there are 100 and you’re blue-eyed, since…that’s exactly what a brown-eyed person would see if there were 99.
I explained it once, and James agreed once. It’s pretty clear to see why it doesn’t work.
Gyahd, after all of those explanations, you STILL can’t see YOUR mistake in that… Geeezz!!!
I certainly can’t see how you call it a mistake after explicitly agreeing with it…
You agreed that if there were only 99, they wouldn’t leave after 3 days.
If there were 100, they also wouldn’t leave after 3 days.
It doesn’t take much brain power, then, to realize, “Hey, them not leaving after 3 days is completely unrelated to the quantity of blue-eyeds there are, and therefore, I can’t use that information to deduce how many there are.”
You can’t use a fact that would be true regardless of the number of blue eyes, to determine the number of blue eyes.
You can’t use a fact that would be true regardless of your eye color to determine your own eye color.
Only if “starting a count from X” means “know it to be common knowledge that there are X blues”. So, for 100, if the blues know there are 100 blues from the get-go (e.g. the guru tells them “there are 100 blues”), they will all leave. But for 99 blues, it’s not enough that they know that there are 99 blues, they need to know that (the 99 blues know that there are 99 blues). And they don’t know that: they can see 99, they don’t know their eye color, so they don’t know if the 99 see 98 or 99. They can only leave on the 2nd day if they “start counting from 99”, i.e. “know it to be common knowledge that there are 99 blues.”
You are using a vague phrase, “start counting from”, which you’ve refused to define precisely. But the way that you’re reasoning from it is to treat it as them having common knowledge, i.e. for any hypothetical set of islanders, they know that there are X blue eyes. If they get the knowledge by looking at the other islanders, it doesn’t translate the same way: even if we see 99 blues, it’s clear that we can say that if there were only 3 blues, they couldn’t know that there are at least 99 blues.
NO!!
It does NOT!!
It means that they start counting from the same number REGARDLESS OF WHY!!!
You keep denying the hypothesis.
“[size=150]IF[/size] they ALL start with the SAME number,…”
That is all they have to do.
WHY they do it or HOW is NOT the question being asked.
Just answer the question that is asked.
Can we agree that the following chain of reasoning doesn’t work:
I see 99 blues.
If there are 100 blues, I must be one of them.
100.
Therefore, my eyes are blue.
Is it not clear that this doesn’t work?
Using your same reasoning (if we can call it that), “yes, it is not clear. It does work”.
So this must also work:
I am the only person on this island. I don’t see anyone.
If there were 1 person on the island with blue eyes, it would be me.
1
Therefore, I have blue eyes.
…
Sounds valid to me.
OK. I am satisfied that you are forced to accept that those are valid proofs in order to maintain that your solution is also valid.
So I take it that you didn’t get the whole, “I AM NOT TALKING ABOUT A SOLUTION, BUT MERELY ASKING A QUESTION” bit, huh.
Carleas, let me ask a different question;
In the canonized solution, it is proposed that a blue (you) is thinking that another blue is thinking that he might be brown, “if I am brown, then…”. Why would a blue be thinking that he is brown? He has no way to know that he is brown, especially since he really isn’t.
First, we need to be precise in our language, since this has gotten us into trouble before. It’s not what he “is thinking”, but what he logically could think.
Second, while he could assume he is brown, the relevant assumption to complete the modus tollens is that he is ‘not blue’. That is different from being brown. Being brown is one way to be ‘not blue’, but it is reasoning on the possibility of being ‘not blue’ that ultimately allows him to conclude that he’s blue.
Finally, he’s able to contemplate and reason on these assumptions because they are logically permitted. There are a million logical deductions he could make, most of which aren’t relevant to the solution of the problem, but which the givens state that every person on the island knows immediately; everything that is logically deducible. He doesn’t know his eye color, so he knows that he could be blue, not-blue, brown, not-brown, red, etc. Making a deduction that includes his being any of these does not contradict anything he knows. And he makes them all. One of those deductions is "if I am not blue, then those blues see 98 blues, and they might think, "If I am not blue, then those blues see 97 blues and they might think, "If I am not blue then those blues see 96 blues and they might think, “…” " " "
Anybody ready to try a different one?
Don’t worry, it’s far less ambiguous (I think so anyway…).
Again, no tricks/loopholes:
You’re talking to a woman, she says she has 2 children, you ask her, “Is at least one of your children a boy?” She says yes. What is the probability that the other child is a boy?
Just to be perfectly clear about no tricks/loopholes:
Assume she’s telling the truth.
Assume there’s no ambiguity about sex, everyone is either a boy or a girl in this universe.
Assume that each time a woman gives birth, it’s 50/50 boy/girl.
Assume that each child’s sex is independent of the other (no statistical anomalies caused by twins).
If you must cheat and look up the answer, at least put it in a [ tab ] thingy.
[tab]1/3[/tab]
yurp
A 1-metre stick is broken into two pieces at random. What is the length of the shorter piece, on average?
[tab].25 meters?
My reasoning is this: It will always be between .5 and 0 meters long, and the average of all the lengths between .5 and 0 should be .25[/tab]