It’s not mind reading–it’s knowing what a perfect logician would think. Since they are all perfect logicians, it follows all know what the others are thinking.
it follows all know what the others are thinking to an extent. And that extent only takes you to everyone elses maximum possible count (which is not the same thing as the maximum possible count like we know it from an outside perspective and already got the answer). That one missing set of eyes could be added to any group colored eyes even if the guru said she saw one blue eyed set yours could still be green like hers in your world (so why not just leave when you saw all the green people still there after x amount of days? or even brown for that matter?
The problem is that you are taking these steps
- I can see the Guru and someone with blue eyes
- Therefore, the Guru can see a person with blue eyes.
- There are more than two people with blue eyes.
and replicating them at one remove with these steps
- Therefore, each other person on the Island can see the Guru and a person with blue eyes.
- Therefore, everyone on the island knows that the Guru can see a person with blue eyes.
(I omit step 5 because it wasn’t in the first three steps, and there is no need to repeat it).
So you have glossed over step 3. It should read like this:
- Therefore, each other person on the Island can see the Guru and a person with blue eyes.
- Therefore, everyone on the island knows that the Guru can see a person with blue eyes.
- Everyone knows that there are more than two people with blue eyes.
Now, if you keep repeating those three steps by tacking an “everyone knows that…” on the front, eventually you come to a point where the statement “everybody knows that … [yadda yadda yadda] … that there are more than two people with blue eyes” is not true.
Yes, the imaginary 97er could imagine a 96er, but that’s as far as it goes. Nobody imagines a 95er, because the 98er who is imagining the 97er knows that nobody can see fewer than 97. His imaginary 97er doesn’t know that, and so could imagine there being a 96er, but the imaginary 97er would know that nobody could see fewer than 96, so he can’t imagine that someone sees only 95.
And nobody thinks that there’s anybody who can see only one, or even any hypothetical person who thinks that somebody can see only one, so nobody thinks that somebody might see zero.
Correct. But nobody here is saying that.
If you are the hundredth blue eyed person you will see 99 for sure and at that point is ok to assume that they each only see 98. But any of those 99 people in this island and in this scenario in their real world see at least 99 including the you (since on the island that holds to be true). If they’re following logic they (each and every blue eyed person) would rule out any possibility of there being 98 or less blue eyed people. Leaving each single blue eyed person stuck only being sure of these things.
- that there are no less than 99 blue eyed people and no more than 100.
- that they are all still on the island and none have left.
- that days will pass with no change in counts or information of eye color or people leaving.
- that they will never know if they were the hundredth blue eyed person or the 101st brown eyed person or the secondgreen eyed person.
No, they don’t agree on anything. It just takes that many nights to reach the conclusion that they either are blue (and leave) or not blue (and stay). The “all leave together” part is just because they all reach the conclusion at the same time.
In a way, they communicate with each other every night by either staying or leaving. As I said, with five islanders, the thought chain A.B.C.D.E is 32 combinations, B.C.D.E. is 16, C.D.E. is 8, D.E. is 4, E is 2 and reality is 1. After the announcement, it’s A.B.C.D.E is 31 combinations, B.C.D.E. is 15, C.D.E. is 7, D.E. is 3, E is 1…possibly. Since E doesn’t leave the first night, now D knows which two possibilities E has the pattern narrowed down to. And on the second night, C figures it out. And on the third night, B has it. And on the final night, A has it all figured out. So A leaves.
Except everyone is reasoning as if they were A, i.e. in the first person. So they all conclude this at the same time after the fifth night.
Emoticorpse, if you logically accept that an island with 3 blue eyed people will leave on the 3rd night after hearing the guru say “I see at least one person with blue eyes” then that means you accept the fundamental logic for why 100 blue eyed people would leave on the 100th night. There is no difference between 3 and 100 for the purposes of the logic of this problem.
on this island no single person thinks that there less than the amount of eye colors than they count.
each single person knows that every other person has that same exact count except with a one number difference for a single set of eyes.
that second sentence applies all the same to every single person. the second sentence has a limit to how many different counts it the count offers only a possible total not a definite total so without out a count how the heck could someone even attempt to surely know the missing color
I still don’t see it playing out for a island with three blue eyed people and the guru making four 
Chassic Sense
“In a way, they communicate with each other every night by either staying or leaving”. That part right there by itself is what is bugging me. If that made sense I would move on the the other stuff but I don’t see how anyone can communicate “I’m still here so wait one more night and if the count gets to the amount you counted in total you’re next and we all leave tomorrow” to everyone else just by staying put. That’s the part that violates my understanding of solving this riddle. I feel like they can’t communicate they do through the act of staying put and showing the others by communicating.
Try really hard to understand the 3 person case, and the 100 person case will make sense. It’s been explained several times on this thread so far, but I’ll do my best to explain it again.
Three blue eyed people on the island hear the guru say “I see at least one person with blue eyes.”
Person A knows for sure than there are at least 2 people on the island with blue eyes, so he has to wait 3 days before he can leave, if ever. If both of the other people don’t leave on the 2nd night, then that means he must also have blue eyes, and he can leave on the 3rd with them.
All 3 people are thinking this of course, and all 3 people see 2 guys with blue eyes, so they all end up staying on the 2nd night, waiting to see if the others leave first, but they don’t, so they all leave on the 3rd night.
It would be impossible for them to stay longer than the 3rd night, because it would be obvious to them that on the 3rd night, if no one else had left, they must all have blue eyes.
Why don’t they all just leave on the first night? Well, that’s where the tricky part comes in. A can’t be sure that B can’t be sure that C sees anyone with blue eyes until the guru speaks. When the guru speaks, now A KNOWS that B KNOWS that C knows there is at least one person with blue eyes, but at first, A can’t know that B knows that C sees 2 blue eyed people, hence the waiting. As the days pass, the possibilities of what A thinks B thinks C sees diminish to only one possibility, that they all have blue eyes.
As I have said before, anyone who understands this better than me is welcome to correct any mistake in my reasoning.
guru says what she says I got that. person a knows for sure there are at LEAST 2 people on the island with blue eyes (you stated that he knew for sure there “are” 2 people with blue eyes" not sure if that’s what you meant?.
because for me the third person is where is becomes unsolvable.
because if he knows there is at least three then he knows that he may be the third but the other 2 are thinking the exact same thing leaving them in a kind of frozen catch22 kind of thing even after the wait. I don’t see the magic in waiting but I’m gonna leave this thread alone for a while maybe I’ll poof get it in the middle of the night or something thank you guys for all your help and i’m gonna keep tryin.
haha yes I did mean “at least 2” and went back and edited I think.
Think about it like this. 3 people on the island (and the guru but forget her). If only one has blue eyes, he sees 2 people with different eyes and leaves the first night. If two of those people have blue eyes, then the two people that have blue eyes look at the other and say “Ok, I might have blue eyes, and if I do, then that means that the other blue-eyed guy will have to wait until the 2nd night to leave, so if he doesn’t leave the first night, that means that I have blue eyes and we’ll both leave the 2nd night.” The 3rd guy is stuck forever with the guru.
Similarly, if all three of them have blue eyes, then all three of them simultaneously think, “Ok, there are 2 other guys on the island with blue eyes. If my eyes are brown, they will leave on the 2nd night, and it’s just like the case that I wasn’t here at all. But if my eyes are blue, then they WON’T leave on the 2nd night, because they will be waiting to see if I leave, therefore, we can conclude after the 2nd night that we all have blue eyes and we can leave on the 3rd night.”
I’ll lay it out line by line for you in a few minutes after the following bout of League of Legends.
One thing that might be tripping people up is that while “knows”, plausibly, interates, “everybody knows” doesn’t automatically interate. If I know that p, then I know that I know that p (and so on, as much as you like). But it is not the case that If “I know everbody knows p”, then “I know everbody knows everybody knows p”. Another angle at explaining the issue might be to explain how the “everybody knows” operator works. (Which I am not going to attempt to do.)
Ahem I do not agree but I shouldn’t hijack… 
I did say plausibly. In fact I don’t agree either, but that is neither here nor there with regards to people conceptualising the issue.
Re-jacking; I think that “everybody knows” doesn’t iterate should be uncontroversial. My thought was that giving people an explanation as to why it doesn’t iterate might help them conceptualise the issue. (And because it is uncontroversial, it might be possible to explain it in a way that helps people understand what is going on.)
Haha, I second this. For example, just a couple days ago, I knew that the name of the song I was thinking of was “Hold On” but I didn’t know that I knew it. It wasn’t until I looked up the lyrics and remembered the name of the song that I knew that I knew the name of the song. It was there in my brain, but I didn’t know that I knew it at the time!
Unless Frylock is talking about something else, haha.
There is an easy way to understand this puzzle. Apologies if it has already been posted.
If there is only 1 person with blue eyes, he leaves on day 1. Easy.
If there are 2 people with blue eyes, then when nobody leaves on day 1 they can eliminate the possibility of there only being 1 person with blue eyes, so they leave on day 2.
If there are 3 people with blue eyes, then when nobody leaves on day 2 they can eliminate the possibility of there only being 2 people with blue eyes, so they leave on day 3.
If there are n+1 people with blue eyes, then when nobody leaves on day n they can eliminate the possibility of there only being n people with blue eyes, so they leave on day n+1.
Bonus questions:
-
There are 100 people with blue eyes. One of them is struck dead by lightning directly after the announcement. What happens?
-
There are 100 people with blue eyes. One of them is struck dead by lightning on day 2. What happens?