Blue-eyes logic puzzle

No, not ad infinitum. Ad 99 or ad 98. The guru’s pronouncement makes it ad infinitum.

It isn’t directly giving you any new information about eye color distribution.

But it does reduce the possible uncertainty that you can deduce in other people’s logic. And that, as we have seen, leads eventually to all the blue-eyed islanders knowing their eye color.

The reason you can’t skip a day is you have no way to communicate to the others that you intend to skip a day.
You know that other people see a different number of blue eyed people than you, but you don’t know how many. Some of them could be seeing one fewer than you, or one more than you, and at best half of them see the same number as you, and they have the same uncertainty about how many YOU see, so they can’t logically deduce how many days you think it would be safe to skip, and you can’t deduce the same about them. Since you can’t communicate, you can’t decide how many days to skip, so the logical answer is: zero.

Gurus in logic puzzles march to their own (or possibly the puzzle creator’s) drummer.

I realise that. But the rules do not say that only one person can leave per day, so waiting 100 days would tell them no more than waiting five days.

Hi, folks, I’m up to speed now.

I get what the Guru told everybody. Now let’s see if I can say it clearly:

Everybody can see N blue eyed people. Everybody knows it is possible that the blue eyed people they see can only see N-1. So when they try to imagine what the others know, each person imagines that each blue eyed person thinks the others he can see are only seeing (N-1)-1. Therefore, even if there are a billion blue-eyed people on the island, eventually somebody thinks that maybe that guy thinks that that other guy can only see 2, and that he thinks maybe they can only see each other, and therefore each of them wonders if the other one can see any blue eyed people at all.

So I can see 99 blue eyed people, which means there are either 99 or 100 of them. But each of them might only see 98, which would tell him that the number is 98 or 99. And he thinks each of the ones he can see might only see 97, and think the number was 97 or 98.

Here’s the thing: I KNOW that nobody only sees 97. And the guy who only sees 98 of them knows that nobody only sees 96. So while we are quite fuzzy on how many there might be, we all know that there is at least one, and we all know that we all know that.
If the correct answer is 99, then nobody can see fewer than 98, and nobody thinks that anybody can see fewer than 97.

It isn’t waiting, it’s counting.

They have logically deduced a rule: Start with the day the Guru made his announcement as “Day One”. Each day, if the blue eyed people did not all get on the ferry, it is because the total number of blue eyed people is greater than the number of the day.

As a blue eyed person, you have no way of finding out your own eye color except by observing the actions of the others. When nobody got on the ferry on day 99, you know there must be at least 100 blue eyed people. But since you can only see 99, the 100th one must be you. All 100 of you learn that at the same time, by the same method. And now that all of you know what color your eyes are, you get on the next ferry, which is the one on day 100.

You haven’t gone down enough levels. The thing is, if we have the same eye color, then we see the same thing. If we have different eye colors, then you see one less of your own color than everyone else.

So I can see 99 blues and look at you, a blue-eye, and say to myself “Perhaps we have different eye colors and he only sees 98.” Now if that were the case, he’d look at a fellow blue-eye and think “Perhaps we have different eye colors and you only see 97”. And here’s where your brain rebels. You know it to be false that they have different eye colors. You can see them both and know they’re both blue. So your brain wants to stop you there.

But you’ve got to recall that now you’re pondering from the point of view of 98 about 97’s mind. And that’s totally legit because 98 is unaware of their matching eye color. So there’s no contradiction. Despite you knowing that there are no 97s in existence, you can’t rule out that a 98er might think so.

Now the 98er can think to himself “This poor 97er chap. He’s talking to another 97er chap, and they both think the other guy could be a 96! They’re wrong, but they don’t know it!”

I’d like to point out that there are no actual 98s in existence either, but that didn’t stop you, a 99er, from imagining it, did it? So why can’t a 98er imagine a 97er, even though neither actually exist? And then can’t the imaginary 97er make the conjecture that there’s a 96er standing beside him, even though there isn’t one?

That’s the crux of it. If it’s possible to suppose that blue-eyed guy #99 incorrectly thinks his eyes are not blue (making the total number of BEGs off by one), then it’s possible to suppose that #99 supposes that BEG #98 incorrectly thinks his eyes are not blue (making the total number of BEGs off by two), and then it’s possible to suppose that #99 supposes that #98 supposes that #97 incorrectly thinks his eyes are not blue (making the total number of BEGs off by three), etc.

Having said that, I had more success imagining it from the ground up.

If there’s 1 BEG on the island, he says “oh, the guru must be talking about me” and he leaves on night 1.

If there’s 2 BEGs on the island, #1 thinks “if I don’t have blue eyes, #2 will act according to the case where there’s 1 BEG on the island; if he doesn’t leave on night 1, I must have blue eyes too, so I’ll leave on night 2”.

If there’s 3 BEGs on the island, #1 thinks “if I don’t have blue eyes, #2 and #3 will act according to the case where there’s 2 BEGs on the island; if they don’t leave on night 2, I must have blue eyes too, so I’ll leave on night 3”.

If there’s 4 BEGs on the island, #1 thinks “if I don’t have blue eyes, #2, #3 and #4 will act according to the case where there’s 3 BEGs on the island; if they don’t leave on night 3, I must have blue eyes too, so I’ll leave on night 4”.

It’s easy to generalize to the case of 100 BEGs using the same logic.

I think I’m going to have to concede that I just don’t get it then. :frowning:

I reject the answer given and say that no one leaves. My answer hinges on the condition that everyone must KNOW their eye color before they can leave.

The problem is, on the 100th day, it is equally likely for all the blue-eyed and brown eyed inhabitants to think that they are the 100th blue-eyed inhabitant. So nobody leaves. On the 101th day, they would be equally clueless.

No. The blue eyed person sees 99 people with blue eyes. The brown eyed person sees 100 people with blue eyes. The brown eyed people would all leave on the 101st day (thinking they had blue eyes) if all the true blue eyed people hadn’t left the previous night.

Okay…I may have been hasty.

Yes, and it’s the same exact reasoning, but if this thread is any indication, it just causes the crowd to shout back “But they already knew that!” :smack: and “So the guru just gives them a coordinated starting date?” :smack::smack:

Here’s another version of the same logic, which I came up with this morning.

Suppose Joe’s in a bar, hitting on Sara. Joe thinks “Man, I want to do her tonight.” Sara’s thinking “Gee, I hope he comes back to my place tonight.” But both are too timid to make the first move- Joe thinks he’ll get slapped and Sara doesn’t want Joe to call her a slut. So they both want laid, but they’re afraid to say anything.

Then Joe’s buddy Mike walks over and blurts out “Looks like somebody wants laid tonight!” and walks off. Now, Joe already knew that. Sara knew that, too. So Joe didn’t really tell them anything they didn’t already know.

But Joe looks at Sara, waiting for her to realize that “the somebody” is him and slap him. Sara looks at Joe and waits for him to call her a slut and leave. But they both sit there, waiting. Then they realize that since no name-calling or slapping has gone on, they were both unsure of who “the somebody” was, and realize they’re both horny.

They make a b-line for the door, arm in arm. Even though Mike didn’t directly add any new info.

What part are you hung up on, exactly? Why it takes time for the departure to occur?

:stuck_out_tongue:

Funny that many of Chessic’s analogies involve people screwing :smiley:

I think that the “new information” theme is just a straw man to divert from the solution and confuses people who are still trying to grasp the solution. The point is not that the Guru information was new or not, but that it was relevant to the islanders’ situation and their inferences. Surely, what the Guru said was relevant information on day 1, 2, 3, … N-1 and the fact that no one has left the island on those days. While it may be interesting to ask what new information is provided by the Guru, recognize that it’s just a side discussion to the solution.

This reminds me of a (flawed) line of logic that my father presents in his math classes:

And so on, continuing with the conclusion that it is not actually possible for the professor to give them a pop quiz on any day of the semester.

I suppose there are some similarities with the Surprise Test/Unexpected Hanging paradox in that they are both about meta-knowledge, although I think that one is a little deeper. You can actually boil that paradox down to the following statement, as **ultrafilter **noted in an earlier thread about it:

“You can’t know that this statement is true.”

That is an unusual statement in that it actually is true - I know it, and everybody else except you can easily verify that it’s true. But you can’t.