I see that if I’m tripped up the ferry and getting on the boat is definitely contributing to it. I keep thinking well if they get on the boat at the same time or if one of em gets on the boat. It’s not about getting on the boat at all it’s about knowing your own eye color for sure (where your eye color is equal to x and not maybe x).
That’s the point they need to get too and I keep paying attention to the ferry. hehe
*First of all thank you Frylock for trying to help me one on one kind of but with forums it’s real hard to reply in a reasonable amount of time. But yes I can possibly have a yellow card according to Bob.
I don’t get this part of the stuff. from the perspective of any of the 100th blue eyed people. Everyone has one of the 99 blue eyed people they see is thinking 2 possible scenario’s
they each know that none of those blue eyed people see less than 98 because fact is there are at least 99 and they can each only be missing themselves.
that is your information independently
you are POSSIBLY the 100th blue eyed person.
If this is the case then the 99 you see are each staring at at least 98 blue eyed people and you may be the hundredth. They know and only they know whether you are your missing hundredth but can’t tell you.
that they the 99 they see are all of them. Which in that case the 99 are still there after 99 days and you yourself know they are the only 99 and know they are all blue and can leave but you can’t tell them that because you never could otherwise with your perfect logic would have already told them.
*When the 99 days pass and you’re all still there it’s not because they didn’t board the ferry it’s because they didn’t know their eye color (Which could still be 1 out of any assortment of colors as opposed to a binary 1 or 0 boarded the ferry or didn’t board the ferry?
I made a picture as well of the patch that I see as an option which leads to pure inconclusiveness for any of them. Not sure if it’s useful to any of you to point to me exactly where I’m wrong but I can’t drop this possible path as opposed to the bubble thought picture posted earlier and with this options I don’t see any sort of resolution to anyone’s independent conclusion to their own eye color. As long as I see this I see a possible path of repetitiveness and since as long as there is no evidence to prove beyond a reasonable doubt that their eyes are blue that means they don’t know for sure and if they don’t know for sure it means they’re guessing so speak. To me it sounds like when they board they ferry they say well I know my eyes are EITHER blue or something else and the ferry just happens to pick em up and all the other colored eyes may as well do the same exact thing thinking well what do I have to lose?. Am I crazy???..
Next question: Is it possible that, according to Bob, Carl is the only person with a green card? (Sorry if these first questions appear trivial–I just want to be certain we’re starting from the same baseline facts.)
True, those tow layers do not include that everyone knows that everyone knows that everyone knows about A liking cookies, so with only that rule it is possible for someone to think that someone else thinks himself to be the only one who knows that everyone knows.
And if we add another layer, so that everybody knows that nobody thinks themselves to be possibly the only person to know that everyone present knows A likes cookies, each person still might think themselves to be the sole possessor of that last bit of information, that everyone knows, and that the others might think they were unique in figuring that out.
It seems like it doesn’t describe reality very well, though, in much the same way as the “if you step half the distance to the wall each time, you’ll never reach the wall” thing: technically true, but at some point I’ll be close enough that it makes no difference.
It’s not even technically true. The conclusion that you’ll never reach the wall is false. It’s not that you’ll be “close enough that it makes no difference”–you actually do reach the wall. If there’s a hole in the wall, you can even pass beyond the wall.
Okay, I’m not afraid of math. I like math. I do algebra for fun.
However, my last formal math classes were decades ago, and the stuff in that wikipedia article uses symbols I don’t immediately recognize (and other symbols I do recognize but can’t remember what they mean), so could somebody dumb it down a bit for me?
It seems to be giving a formula (or actually, a couple of competing formulas) that would tell you how many layers you need for it to be “common knowledge”. So for a group of 201 (all the folks on the island), how many layers is effectively the same as the Guru speaking?
Or explain what the formulas mean. Please and thank you.
Is that the corrected question? to me the answer to that question is if Bob see’s that I have a green then No Carl can’t be the only one because I’d make the second but if I have anything other than green Carl might be the only one.
Well, I want to insist that it’s a yes/no question. Either it’s possible that according to Bob Carl might be the only green-card person, or it’s not possible that according to Bob Carl might be the only green-card person.
But to make that super extremely clear, let me back up a half-step.
Remember, here’s what we know:
So it might be that, according to Bob, you have a yellow card. We’ve stablished that. Here’s a new question (we’ll come back later to the one I asked in the previous post):
According to Bob, is it possible that Bob has a yellow card?
To make sure I understand you, you’re saying that if you happen to have a yellow card, then Bob thinks that both you and he might have yellow cards. Right?