They are signaling that to each other through their behavior. If each blue eyed person sees 99 OTHER blue eyed people none of them would have a reason to move on the 99th day because they would either A) assume they were not blue eyed and the count was 99 or B) assume they might be blue eyed and thus the count would be 100. In the circumstance of A, each blue eyed person would see 98 blue eyed people (browns would see 101) and on the 99th day would leave together while in circumstance B they would each see 99 and thus all leave on 100th day because by not going on the 99th day they are each communicating that they all see 99.
^ But they can see everyone else at all times. As a blue eyed person, I am ALWAYS seeing 99 blue eyed people. I saw 99 blue eyed people before the Guru said anything.
The eyes of all the islanders have no colour as colour is a merely a subjective perception of different wavelengths of light
They don’t have to see everyone else at all times. It’s a small island with 201 inhabitants and they’re all gathered together, counting would be simple.
What good does counting do? As stated in the original post, I could have red eyes, so it doesn’t matter if I see 100 blue eyed people or 99 blue eyed people. It also says that I, and everyone else, sees everyone else at all times. So, I have always known that I share the island with 99 blue eyed people, 100 brown eyed people, and one green eyed Guru. It doesn’t matter how many nights you put a different person in front of me or what their eye color is, I will never leave the island, and neither will anyone else, because I know and have always known that there are blue eyed people on the island.
Correct, good job!
not to derail the topic, but what makes the red man red?
because hes not the method man, hes the red man
So what new information do I get from an individual meeting? Let’s say all the blue eyed people are numbers 1-50, all the brown eyed people are 51-100, and I’m one of the blue eyed people. On night 1, I meet with 51. I see that 51 has brown eyes (something I already knew because I can see everyone one at all times). I know the Guru wasn’t talking about 51 because 51 has brown eyes, but again I already knew that being able to see everyone at all times. 51 sees my blue eyes (which 51 already knew) and thinks the Guru could have been talking about me, him, or people 2-50 since 51 already knows that 2-50 also have blue eyes. Since I don’t know if the Guru was talking about me or people 2-50, I don’t leave. Since 51 doesn’t know if the Guru was talking about him, me, or people 2-50, he doesn’t leave either. On night 2, I meet with person 2. I see her blue eyes (again, something I already knew) and she sees my blue eyes (again, something she already knew). I don’t leave because the Guru could have been talking about me, person 2, or 3-50, all of which I already knew. Person 2 doesn’t leave beause the Guru could have been talking about me, her, or 3-50, which she already knew.
I think the only way the iterative process works is if I don’t know anyone’s eye color until I meet them one-on-one. In that situation, when I meet with 51, I know the Guru wasn’t talking about them (new information). He could have been talking about me, but he also could have been talking about anyone else except 51, so I don’t leave. 51 doesn’t leave because he knows the Guru could have been talking about me, him, or anyone else.
Higgs, I don’t know what individual meeting you’re referring to, I didn’t reference one. Regardless, I get it, Ohai gets it, apparently the philosophy guy gets it and many math PhDs have weighed in that this is correct. It’s very likely that I’m doing a poor job of explaining it but much more unlikely that everyone is universally wrong on this logically self evident result, it reminds me of the invariable protests when people first encounter the Monty Hall problem. There are a number of pages dedicated to explaining it online that may do a better job than I did.
Help me out by answering these questions then, since you get it:
What is a “round” if there aren’t individual meetings (ohai used the word round in his answer which was deemed correct)?
What new information did the Guru give everyone by saying he sees someone with blue eyes?
What new information does everyone get each night that no one leaves?
Here’s how I see it. I’m on the island. I see 100 people with brown eyes and 99 people with blue eyes. The green eyed lady steps out of the kitchen long enough to tell me she sees someone with blue eyes. I, and everyone else know there are either 99 or 100 people with blue eyes. Fast forward through 99 nights of inaction and we get to the 100th day. everyone is still here, which means I have blue eyes. If I had brown eyes, everyone would have been raptured the day before.
i wrote this post since you simpletons/plebs cant figure it out. i am very smart. my iq is very big like gyna.
The guru exists only to remove the possibility that no one has Blue eyes. If the scenario said “at least one person has Blue eyes”, then (I think) there would be no need for the guru. The guru might be needed to tell the Brown people when the Blues have F’ed off, but I need to think about that - the Browns might be able to tell through logic.
Higgs you are thinking too much. There is no need for individual meetings. All islanders are continually aware of the whole scenario state at any time. There are only two things to keep track of: 1) The number of Blues any islander observes (there are only two variations, based on the eye color of the observer. It can be generalized beyond that), and 2) the number of rounds of inaction that have elapsed. A round is just defined as a period where every single islander makes a choice whether to stay or leave.
Actually, I think this whole logical process would be the same even without any Brown people. The Brown people are there to confuse you.
It’s also interesting that in the 100/100 scenario, the Blues and Browns are differentiated in their observation by only one unit of Blues. So, the Browns are just all one step behind the Blues from assuming they are Blue and leaving.
Basically what Ohai said, the rounds are the midnight Ferry arrivals that occur each day. The Guru offers the base rule that is required as the missing logical assumption to start a measured countdown via ferry arrivals. It’s the funkiest part of the riddle but still holds up.
The new information is the maximum number of X color. It’s a robust logic rule that would work under any blue brown ratio but for it to work you must go through this form of communication. If they could simply speak they could simply each say how many they’re seeing, but lacking that, this is the sole method. Given this 100/100 scenario it is the 99th day that holds the information for all blues as they are basically communicating whether they each count 98 blues (they are brown) or 99 (they are each one of the blue) among themselves. If they somehow screwed up and waited an additional day, their would be confusion as all of the browns that count 100 blues would try to join the exodus only to realize their had been a binary jump and be forever stuck on the island in confusion.
To reiterate, clearly I am doing a terrible job of explaining this, you should consult the internet and meditate on it. The only thing we’re each going to convince one another of successfully given by poor teaching skills is that I am not the person for this job.
He is consulting the internet.
Consult a different internet.
its all gucci. less thinking more efficient
i personally think the answer is no one leaves since no one knows what color their eyes are still.