@Simon – It’s unreasonable to suspect that each successive person could imagine that the next one could be thinking that there is one fewer blue-eyed person than the person “above” him in the chain imagines at minimum. In the extreme example of 100 blue-eyed islander, a blue-eyed seeing 99 other blue-eyed eventually has to think that another islander thinks that another islander could think, etc., that there might just be one other person with blue eyes, in light of the fact that there’s demonstrably many other islanders that person will see with blue eyes, and the person “above” him will see, and the person “above” him will see.

@all who believe the blue-eyed islanders will eventually figure it out – please answer me two questions: (a) why does the Guru have to announce that at least one has blue eyes before they start counting off nights? (b) why can’t the brown-eyed islanders use the same inductive logic? After all, they all know that everyone knows that there is at least one brown-eyed prisoner, right?

CAPTCHA: metamphetamines jubilant (teehee)

One person left on the night of the guru’s announcement; the only blue-eyed person who didn’t wear sunglasses during the announcement.

If this answer has already been posted, I apologise, but all the cod-logic in the varios posts was killing me!

For example, a brown eyed man, seeing there are 99 brown eyed and 100 blue eyed people, would guess that he is brown eyed. If he got it right, he leaves that night. If not, assuming a person may try once a night, he guesses blue the next night.

Since they dont die if they guess wrong, why not go ahead?

The guru is for show, telling people what they can already see for themselves. Both brown eyed men and blue eyed men can see there are blue eyed people among them. all of them except for the guru may leave by the second night if they all tried.

the guru looks pretty content to be staying on the island seems like it..

And instead of saying “I am” it says “I A-M”.

It made me giggle.

With at least 200 people on the island, the fact that there exists a person with ‘blue’ eye colour is simply noise. If everyone is a perfect logician, then they all understand that everyone can see at least 99 people with blue eyes and at least 99 people with brown eyes. This isn’t new information. With the noise of a sample size of 200, there’s no reasonable logic to suggest you could figure out whether you’re blue eyed or brown eyed.

The only way to solve the problem is to introduce more information. Since everyone is on an island, the best thing to do is look at your reflection in the water. Oh yeah! My eyes are yellow!

Matt: So awesome.

Richard: No one else has given your prisoner question a shot, so here’s my idea. There’s good logic in the assertion that midnight on Jan 30th, he can’t be killed on Jan 31st, because there’s no days left. But he can’t extend that to midnight on Jan 29th, because there are two days left. Either day is fair game; he only can’t be killed on the 31st if the 30th has already passed! Until the 30th has gone by and he hasn’t been killed, the 31st isn’t safe, and either is the 30th. All he really knows is that he isn’t dead yet.

Although even as I say that I’m getting sucked into wondering why he couldn’t be certain on the 30th that he was getting killed that day. They CAN’T leave it one more day. And by extension they wouldn’t set it for the 30th, because he would know on the 30th! So if he survives until the 29th, he would know it’s the 29th…

Perhaps it’s that each day, he knows it has to be the day he’s going to die, because it can’t be any of the days afterwards. They screwed up by saying that, because every day he knows, and every day he’s wrong, and they simply lied to him. :P

@XKCD: I totally agree that you should clarify that nobody is colour-blind. My first thought was that everyone would assume if there was more than one blue-eyed person, then the Guru would have said so, so therefore there really is only one, and *everyone I’m looking at must actually have different coloured eyes – I must be colour blind – I must have blue eyes!!!* Then all the blue-eyed people would leave that night.

By the same token, I kept getting caught up in motivations. Why on earth didn’t the Guru just say “I see an equal number of blue-eyed and brown-eyed people”??? Then everybody could go home that night. Obviously she must have some special motivation for saying what she did – she wants to go home herself, so somehow what she said will result in her leaving eventually. (My original hypothesis, before I figured it out, was the Guru would leave on day 100).

There’s another version of this that says the not-talking-about-eyes-thing is a religious thing, which really makes a lot of sense, and that the “Guru” is really a tourist who slips up. (”Strange seeing people with my eye colour in this part of the world!”) Personally, I think this version helps a lot because it removes all question of motivation. (But I really do appreciate how much you’ve already removed ambiguity from the problem – it was very refreshing to have a problem that I KNEW didn’t have some stupid answer. Even though I kept going back to the colour-blind thing, I knew in my heart of hearts that this wasn’t the answer, so kept going.)

Also, thanks so much for continuing to provide us all with the best comic ever. :)

rómverskur riddari réðist inn í rómarborg(a roman knight invaded rome)
rændi þar og ruplaði radísum og rófum(stole and robbed radishes and onions)
hvað eru mörg err í því?(how many r’s are in that?)

Let’s take a simpler situation where there are five blue-eyed people (I’ll call them “blues”) and a number of “non-blues”. We’ll call the five blues B1, B2, B3, B4 and B5.

Before the Guru does his announcement, B1 is contemplating the number of blues on the island. Here’s what he is thinking:

B1: Ok, so I see four blues [B2, B3, B4, B5]. Either I’m blue, in which case there are five blues, or I’m not, in which case there are four. I wonder what B2 is thinking. If I’m blue, he’ll be seeing four blues as well, and he’ll be thinking the exact same thing as I am. If I’m not blue, he’ll be seeing three blues, and might be thinking something like this:

“B2: Ok, so I see three blues [B3, B4, B5]. Either I’m blue, in which case there are four blues, or I’m not, in which case there are three. I wonder what B3 is thinking. If I’m blue, he’ll be seeing three blues as well, and he’ll be thinking the exact same thing as I am. If I’m not blue, he’ll be seeing two blues, and might be thinking something like this:

“B3: Ok, so I see two blues [B4, B5]. Either I’m blue, in which case there are three blues, or I’m not, in which case there are two. I wonder what B4 is thinking. If I’m blue, he’ll be seeing two blues as well, and he’ll be thinking the exact same thing as I am. If I’m not blue, he’ll be seeing one blue, and might be thinking something like this:

“B4: Ok, so I see one blue [B5]. Either I’m blue, in which case there are two blues, or I’m not, in which case there is one. I wonder what B5 is thinking. If I’m blue, he’ll be seeing one blue as well, and he’ll be thinking the exact same thing as I am. If I’m not blue, he’ll be seeing no blues, and might be thinking something like this:

“B5: Ok, so I see no blues. There are either zero ore one blue on the island. I don’t know.”
”
”
”
”

So, before the Guru announcement, B1 knows P; B1 knows that B2, B3, B4 and B5 know P; he knows that B2 knows that B3, B4, and B5 know that P, etc. recursively, but he can NOT say that he knows that B2 knows that B3 knows that B4 knows that B5 knows P!!