A group of people with assorted eye colors live on an island. They are all perfect logicians – if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph. On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes. The Guru is allowed to speak once (let’s say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following: “I can see someone who has blue eyes.” Who leaves the island, and on what night? There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn’t depend on tricky wording or anyone lying or guessing, and it doesn’t involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she’s simply saying “I count at least one blue-eyed person on this island who isn’t me.” And lastly, the answer is not “no one leaves.”
I think I have the end case figured out, but I not yet how to expand it to the base condition. Remember when BS used to Google the answer and then pretend he came up with it himself? Good times.
I haven’t thought about it since then, but if you have only one Blue remaining, then it’s obvious that he will know he’s the one.
If you have two Blues remaining, then you will solve the game in 2 rounds. The reason is that Blue 1 will not know if he is the one or if Blue 2 is the one. So they both don’t leave in Round 1. Since they both observe no one left in Round 1, they know that both of them are Blues.
If you have three Blues remaining, then Blues 1, 2 and 3 will not know if they are a Blue. Like above, no one will leave in Round 1. However, additionally, no one will leave in Round 2 either. So, in Round 3, all three Blues will know they are Blue.
And then, presumably, you can expand that to further levels. I’m more of a 3d chess person though, so have trouble doing this in my head. This needs more of a 4d person I think.
It seems that, based on the iterative approach, all the Blues will wait for 100 rounds, then simultaneously leave. Then the Browns will simultaneously leave. I might haven’t really drawn it out though, so maybe something changes at higher iterative orders.
“Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate.”
They can’t communicate, so they can’t set up a system to go through rounds. Even if they could set up rounds, you’re always going to know that there are 99 or 100 people on the island with blue eyes. The Guru really didn’t give any new information. For example, let’s assume I have blue eyes and ohai has brown eyes and we look at each other night 1. I’m not going to leave after that round or any round, because I already know there are 99 people with blue eyes. Ohai won’t leave either because he knows there are 100 people with blue eyes.
They don’t need to communicate a system because they’re all operating through logic per the provided background. While the first rounds seem superfluous they are necessary to adhere to the logic train. It does not become crucuial until the 99th round in which noone goes which signals to each other that each blue eyed person also sees 99 other blue eyes which means that if you’re only counting 99, you must be the 100th. For the brown eyed people they are simply observing because they can each tell from the onset that this will occur. They will then begin their own process.
They can communicate, but only in one way - by leaving or not leaving. This is the only action the islanders can make. Therefore, it is the only way that new information can be added to the scenario. No one can leave on day 1. Therefore, assuming the question has a solution, the fact that no one leaves on any day must be adding information.
They know there are either 99 or 100 people with blue eyes. Furthermore, they know the number of rounds of inaction. You will not get to 100 rounds of inaction if only 99 people have blue eyes. Therefore, if they pass 99 rounds, then they know that there must have been 100 people with blue eyes, including themselves. Therefore, they know they have blue eyes.