Need to apologise, the edit and multi-quote functions are a doddle to use and doesn't give the impression you're just posting to make as many posts as possible.
Problem in logic
- Thread starter Ha'kiv
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Need to apologise, the edit and multi-quote functions are a doddle to use and doesn't give the impression you're just posting to make as many posts as possible.
Ah, now I understand where you're coming from. You're right, if they all choose to solve for blue, they could do it without the guru. Equally, as you say, if they all choose to solve for brown, they could do it without the guru. In other words, if you could direct each islander to solve for the right thing from the start, you could solve it without the guru. In fact, you would be taking the place of the guru!
But it is not trivial for them all to do this by themselves. They can't communicate with each other. They're not telepaths either.
They can't KNOW that everyone else is choosing to solve for blue, so the scenario on day 100 is that they can't KNOW that they have blue eyes, because they can't KNOW that everyone else was doing the same thought experiment they were through the previous 99 days. Some annoying person may have been solving for brown, or even worse, not choosing to solve for anything at all!
It's only when the guru ENSURES that they ALL know that they ALL know that there is someone with blue eyes, that they ALL know that EVERYONE else will now opt to use blue eyes as the specific data point around which to construct the shared thought experiment.
This shared common knowledge is what makes the thought experiment solvable once enough days have passed. Which is why they attempt it. They won't attempt it until they KNOW that deductive logic will ENSURE an answer.
(Humans, not having to rely on deductive logic alone, WOULD try to attempt it anyway using precisely your line of thinking to justify the attempt... and may well get it right through a bit of good fortune, but they can't KNOW they would get it right on day 100.)
Explain how, in the scenario when there is only one blue on the island.
Okay, then explain how it can be solved without the guru in the case with two blues. (Hint: It can't be.)
I feel like I'm going round in circles here.
They can't by the rules of the riddle (ie using logic only - know you're right on eye colour)
They can if you break the rules of the riddle (ie allow a form of guessing correctly - believe you're right on eye colour)
I don't know how to phrase it more simply! I've been pretty consistent in saying this all along, I think.
Wow, what a thread. One thing that no one has addressed or mentioned is the total population of the island, which, to my understanding is 201 (100 blues, 100 browns, 1 green-eyed guru). However, going back to what boobatuba said about "98 blues, 100 browns, and self," that doesn't quite add up to 200 (excluding the green-eyed guru).
I'm speaking from a purely logical viewpoint. I'm not sure how guessing is even relevant to the question. I'm just trying to make sure everyone understands why the guru is necessary to the solution.
Ok, here's my problem with the base case induction and necessity of the guru statement. The assumption is that if the Guru is saying something about blue eyed people now, with 100 people, that the Guru would say the exact same thing if there were only one blue eyed person. That is completely unfounded... the Guru could in that case decide to talk about seeing a brown eyed person.
If you can make that leap, then you could also come to that same conclusion for any eye color where you see 2 or more people with that eye color (thus guaranteeing that EVERYONE can see 1 or more person with that eye color), and I imagine they could all leave on ferry #(eye_color + 1) if they have that eye color, and if not they would see that eye color leave on ferry #(eye_color) and thus never have the opportunity to leave.
Specifically for the situation at hand all the brown eyes would see 100 blues, and 99 browns. They would all come to the same plan, that is, if it's ferry 100, and all the browns they see haven't left, they will declare themselves brown and leave. If ferry 101 comes, and all the blues haven't left they will all declare themselves as blues and leave.
So, all the browns would leave on day 100 knowing they are browns, all the blues would leave on day 100 knowing they are blues.
The Guru would plan to leave on day 101, and see both groups leave and thus not get to leave.
If you can make that leap, then you could also come to that same conclusion for any eye color where you see 2 or more people with that eye color (thus guaranteeing that EVERYONE can see 1 or more person with that eye color), and I imagine they could all leave on ferry #(eye_color + 1) if they have that eye color, and if not they would see that eye color leave on ferry #(eye_color) and thus never have the opportunity to leave.
Specifically for the situation at hand all the brown eyes would see 100 blues, and 99 browns. They would all come to the same plan, that is, if it's ferry 100, and all the browns they see haven't left, they will declare themselves brown and leave. If ferry 101 comes, and all the blues haven't left they will all declare themselves as blues and leave.
So, all the browns would leave on day 100 knowing they are browns, all the blues would leave on day 100 knowing they are blues.
The Guru would plan to leave on day 101, and see both groups leave and thus not get to leave.
The guru would first have to set the condition in the same way as the blue-eyed people. In other words, he/she would've made an announcement at the beginning, saying "I see someone with brown eyes."
Hmm. Does the guru know his/her own eye color (green)?
The rules say:
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The Guru can't leave, as there's no information as to his own eye color. Actually, the only way he could leave is if HE also had blue eyes, and could leave on day 101 with the 100 blues.
The browns also can't leave, as they still have no information that they have brown eyes. They now know they DON'T have blue eyes, as they all just left, but for all they know, it's 99 browns, 1 green, and 1 purple. No decision they can logically make based on the Guru's initial statement.
The browns also can't leave, as they still have no information that they have brown eyes. They now know they DON'T have blue eyes, as they all just left, but for all they know, it's 99 browns, 1 green, and 1 purple. No decision they can logically make based on the Guru's initial statement.
For ease of explanation, we discuss the problem as if there could be different numbers of blues. However, there aren't----there are always 100, and the guru's statement is what it is.
When we refer to cases with fewer blues, we are not actually reducing the number of blues (creating a different scenario)----we're recursively following the non-blue hypothesis which needs to be disproven.
Let's look at it from the other end. We know there are 100 blues, but one of them chosen arbitrarily sees only 99. His eyes may or may not be blue. He assumes the not-blue hypothesis, thus analyzing a 99-blue situation. (Eventually, this assumption will lead to a contradiction, thus proving he has blue eyes. But that's later.)
This blue (A) who is assuming he is not blue is trying to figure out what the other 99 blues are thinking. Recursively, he knows that one of them (B, chosen at random) can see at least 98 other blues. In fact that one can see 99 including A, but since A is assuming he is not-blue, that isn't relevant here. So A assumes that B assumes that B is not blue, in which case B would start to think about a 98-blue situation.
And so on, and so on. It isn't actually the case that there are fewer blues----just that one particular blue, A, is following the logic chain WHAT-IF-I'M-NOT-BLUE, and on every level is assuming that every other randomly chosen blue he can see is also following the same branch of the tree. It's assuming about what others are assuming about what others are assuming about what others are assuming to a fairly mind-boggling level, which is why we need to assume they're perfect logicians.
A knows this is false for the others, but he also knows that they *don't* know this, so they'll explore that branch of the decision tree.Each level of the tree assumes logic at work which relies on fewer blues, until we get down to the 1-blue case. That case is a contradiction due to the guru's statement. At that point, each level of the tree triumphantly determines that I'M-NOT-BLUE is false, cascading all the way back up to A.
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