Same way it expands to two: When there are three blue eyes, then each of them guesses they might have brown or something and there could be only two blue on the island, in which case as described those two would have left on the second night.
I don’t think that’s right.
Let’s try it out:
Basic case: 1 brown, 1 blue. Day 1. Guru says I see someone with blue eyes, blue eye person immediately leaves. End
Next: 2 brown, 2 blue.
Day 1; Guru speaks. It doesn’t help anyone immediately because everyone can see a blue eyed person, so no one leaves first night.
Day 2; The next night, everyone knows this, that everyone else can see a blue eyed person. Which tells the blue eyed people that their eyes are not brown. (They now know no-one is looking at all brown eyes). So the 2 blue eyed people who now realise their eyes aren’t brown leave that night on day 2. The end
Next case: 3 brown, 3 blue (I’m arbitrarily making brown = blue, I don’t think it actually matters).
Day 1, guru speaks, no-one leaves.
Day 2 everyone now knows no-one is looking at all brown. So if anyone could see only 1 other person with blue eyes at this point, they would conclude they themselves have blue. I suppose if you were one of the three blue eyed people you wouldn’t know if the other blue eyed people were looking at 1 blue or more. No-one leaves that night.
Day 3 I suppose now everyone can conclude that no-one was looking at only 1 blue, everyone can see at least two blue. So if the other blue eyed people can see 2 blues that means you must have blue eyes. So all blue eyed people leave Day 3?
Hmm. Maybe I’ve talked my way round to it. Maybe this keeps going on, each day without departure eliminating anyone seeing that many blue eyes until you get to 100.
It just seems so utterly counterintuitive that everyone sits there for 99 nights unable to conclude anything?
I was able to think this through by an inductive argument. Here’s how I thought about it:
If you are the only person with blue eyes, you leave night 1 when you see no one else with blue eyes. I’ll call this the “Algorithm for 1 Blue-Eyed Person”.
Say that you have blue eyes and you see only one other person with blue eyes. You expect them to use the “Algorithm for 1 Blue-Eyed Person”, but they don’t leave night 1. The only way the algorithm would fail is if you yourself also has blue eyes, so you would both leave night 2. This is now the “Algorithm for 2 Blue-Eyed People”.
Now say there are three blue-eyed people, including you. You look at the other two blue-eyed people and expect them to do the “Algorithm for 2 Blue-Eyed People” and leave on night 2. But they don’t use the algorithm, which would only happen if you had blue eyes as well. So this becomes the next algorithm… etc.
In a group with 100 blue eyes, each of them see 99 other blue eyes and expect them to do the “Algorithm for 99 Blue-Eyed People” and leave by night 99, which doesn’t happen so you all leave on night 100.
Though the hardest part is getting an intuition about why the “algorithm 1” “algorithm 2” thinking happens at all when they’s a group of 100 people and everyone can see at least 99 blue eyed people. I get the induction, but why does anyone think ‘well algorithm 1 people would leave first night’ when there obviously can’t be anyone in this group. The only immediate question on everyone’s mind is “are there 99 blue eyed people (what I see) or 100 (me included)?”
Yeah. It does seem counterintuitive, but it’s a result of the uncertainty that what they see is what others do. So they have to communicate a number, and the only way they can is leaving or not each night to count up to it.
I thought about it more and concluded that if the guru had said “I see only blue and brown eyed people” then everyone (but her) could leave the island using the same logic, regardless of how many of each color there was (greater than zero of course because otherwise she wouldn’t see that color). Same for any number of colors too as long as she lists them all and makes it clear that’s all of them and doesn’t include herself.
I don’t think that’s right.
Let’s try it out:
Basic case: 1 brown, 1 blue. Day 1. Guru says I see someone with blue eyes, blue eye person immediately leaves. End
Next: 2 brown, 2 blue.
Day 1; Guru speaks. It doesn’t help anyone immediately because everyone can see a blue eyed person, so no one leaves first night.
Day 2; The next night, everyone knows this, that everyone else can see a blue eyed person. Which tells the blue eyed people that their eyes are not brown. (They now know no-one is looking at all brown eyes). So the 2 blue eyed people who now realise their eyes aren’t brown leave that night on day 2. The end
Next case: 3 brown, 3 blue (I’m arbitrarily making brown = blue, I don’t think it actually matters).
Day 1, guru speaks, no-one leaves.
Day 2 everyone now knows no-one is looking at all brown. So if anyone could see only 1 other person with blue eyes at this point, they would conclude they themselves have blue. I suppose if you were one of the three blue eyed people you wouldn’t know if the other blue eyed people were looking at 1 blue or more. No-one leaves that night.
Day 3 I suppose now everyone can conclude that no-one was looking at only 1 blue, everyone can see at least two blue. So if the other blue eyed people can see 2 blues that means you must have blue eyes. So all blue eyed people leave Day 3?
Hmm. Maybe I’ve talked my way round to it. Maybe this keeps going on, each day without departure eliminating anyone seeing that many blue eyes until you get to 100.
It just seems so utterly counterintuitive that everyone sits there for 99 nights unable to conclude anything?
I was able to think this through by an inductive argument. Here’s how I thought about it:
If you are the only person with blue eyes, you leave night 1 when you see no one else with blue eyes. I’ll call this the “Algorithm for 1 Blue-Eyed Person”.
Say that you have blue eyes and you see only one other person with blue eyes. You expect them to use the “Algorithm for 1 Blue-Eyed Person”, but they don’t leave night 1. The only way the algorithm would fail is if you yourself also has blue eyes, so you would both leave night 2. This is now the “Algorithm for 2 Blue-Eyed People”.
Now say there are three blue-eyed people, including you. You look at the other two blue-eyed people and expect them to do the “Algorithm for 2 Blue-Eyed People” and leave on night 2. But they don’t use the algorithm, which would only happen if you had blue eyes as well. So this becomes the next algorithm… etc.
In a group with 100 blue eyes, each of them see 99 other blue eyes and expect them to do the “Algorithm for 99 Blue-Eyed People” and leave by night 99, which doesn’t happen so you all leave on night 100.
Thanks. Yes, I’m managing to absorb it now.
Though the hardest part is getting an intuition about why the “algorithm 1” “algorithm 2” thinking happens at all when they’s a group of 100 people and everyone can see at least 99 blue eyed people. I get the induction, but why does anyone think ‘well algorithm 1 people would leave first night’ when there obviously can’t be anyone in this group. The only immediate question on everyone’s mind is “are there 99 blue eyed people (what I see) or 100 (me included)?”
Yeah. It does seem counterintuitive, but it’s a result of the uncertainty that what they see is what others do. So they have to communicate a number, and the only way they can is leaving or not each night to count up to it.
I thought about it more and concluded that if the guru had said “I see only blue and brown eyed people” then everyone (but her) could leave the island using the same logic, regardless of how many of each color there was (greater than zero of course because otherwise she wouldn’t see that color). Same for any number of colors too as long as she lists them all and makes it clear that’s all of them and doesn’t include herself.