# Tag Info

124

The knowledge of each islander consists of: the color of the eyes of every other islander; any past pronouncement from the guru; the history of who left the island on previous days (including their eye color), which provides knowledge about others' knowledge (that either they did or did not know their own eye color on previous days). At the beginning of ...

54

Every blue-eyed person sees 99 blue-eyed people. Since they don't know that they have blue eyes, they suspect it might be the case that every other blue-eyed person can only see 98 blue-eyed people, and if those people only see 98 blue-eyed people, they might think that each of them only see 97 blue-eyed people. And so it continues, until someone considers a ...

52

Let's continue the induction, since the jump to 99 blue eyes does seem weird. After all, everyone knows that someone has blue eyes. If there are 4 blue eyed-people, A will look at B,C,D, thinking : Now, the issue here is that I, being A, can see that B has blue eyes. Therefore I know that C sees at least D and B as having blue eyes. But this is the ...

35

The whole process is inductive, so it needs a starting point. If there were only one blue-eyed person, he would never know that there is "at least one person with blue eyes," so he would not go the first night. If there are only two, neither of them can know whether the other doesn't go the first night because he only sees brown eyes, so they don't know if ...

27

The only explanation I've seen that's sufficiently precise to be satisfying is this answer to the corresponding question on math.SE. The key fact that the "oracle" (guru) gives you, that you didn't have before, is that "(everybody knows)N there is at least one blue-eyed person" for any value of N. In particular, you need it to be true for N=100, but the "...

18

This might not qualify as a blue-eyes puzzle because it does not use common knowledge, but it involves chains of deductions based on nothing happening for a particular amount of time: $n$ villagers wear either black or white hats. They sit in a line, so that each villager can see all the hats in front of them, but not the hats behind them. If a villager ...

17

I think considering it backwards might actually be the easier way to understand it. A given blue-eyed person does not want to leave, so he hopes he has brown eyes and assumes he has brown eyes. He sees 99 blue-eyed people. Because he has assumed he does not have brown eyes himself, he must assume all of those other blue eyed people see 98 other blue eyed ...

15

The colour of the guru's eyes is not relevant. The guru is allowed to speak about eyes and nobody else is. If any blue eyed person said "I can see someone with blue eyes" where everyone on the island could hear it, the same thing would happen. Also if any brown eyed person did. The moment a blue-eyed person hears that someone else can see some blue eyes, and ...

15

For the muddy children variant of this problem, there are several earlier sources. For instance, A.A. Bennett (Problem No. 3734, American Mathematical Monthly 42, 1935, page 256) formulated the following version back in 1935: A car with $n>2$ passengers of different speeds of mental reaction passes through a tunnel, and each passenger acquires ...

13

As you did, let's reduce it to the case of three people for clarity's sake. Aaron, Bob, and Charlie have blue eyes. No guru says anything. Aaron thinks: If Bob sees only Charlie with blue eyes, then Bob knows after the first night, viz after Charlie doesn't leave, that Bob has blue eyes. Er, no. That'd be true if the guru said someone has blue eyes. But ...

13

I'm going to have to disagree with BianB BB's answer, which was formerly the accepted answer: For other values of n, the number of blue-eyed islanders:

12

One important difference between this problem and the blue-eyes problem is that here, even if there was one 50 and ninety-nine 51s, the person with the 50 would not know their own number (it could also be 52). This means that the crucial chain of deductions cannot start from there. However, there is a place to start this chain. If there was one 2 and ninety-...

11

Let's take the case where there are 3 blue eyed people. each blue eyed person sees two blue eyed people but that is not enough for him/her to realize they have blue eyes. for that fact to be inferred he needs to observe the two blue eyed people he sees not leaving after two days. and the only reason he would expect them to leave in two days is because he ...

10

The Guru's information makes the blue-eyed-people special. It is a bit easier to understand if you imagine the Guru saying "those with blue eyes may go". Then on day 1, you see nobody leaving, so you know nobody knows his own eyecolor, so you can conclude that at least 2 persons must have blue eyes. Then on day 2, you see nobody leaving, so you know ...

10

The earliest occurrence of the puzzle that I am aware of is from 1958. George Gamow and Marvin Stern: "Puzzle Math", Viking Press (February 7, 1958) Here is the Amazon page for this book: http://www.amazon.com/Puzzle-math-George-Gamow/dp/0670583359 Chapter 1 of the book contains several mathematical puzzle stories on the great Sultan Ibn-al-Kuz of ...

10

Let's start with a simpler question. Suppose there were three blue-eyed people - then what information do the three people learn on the first night? Before the first night, blue-eyed people see two blue-eyed people, and reason that there must be either two or three blue-eyed people total. These people can reason hypothetically about the two cases of their ...

10

You don't know that everyone is in the exact same situation with respect to who has blue eyes. Everything else is the same, but the configuration of blue eyes is not known to be symmetric. (And in many tellings of the riddle, it's not.) So, with 3 people, you can't say that "Person(1) knows that Person(2) knows that Person(3) sees someone with blue eyes&...

8

While JonTheMon and Emrakul's answers are entirely correct, here's a different way of explaining why. Everyone knows what they can see themselves, and can put themselves in everyone else's shoes to guess what they see. Sometimes that works: Scenario: blue blue blue blue blue blue - 1 concludes everyone else sees 4 or 5 blue eyes - decides it's ...

8

Does the Guru's statement bring any new information? The misleading thing here is that you might get tricked into the belief that the statement of the Guru just tells the people on the island that there is someone with blue eyes. But that is nothing new! The people already knew that by looking around. The statement of the Guru says something deeper. It not ...

8

A version of this puzzle, involving three ladies with dirty faces laughing at one another, appears in J. E. Littlewood's "A Mathematician's Miscellany," published in 1953. Littlewood writes that this is "a well-known puzzle that swept Europe a good many years ago and in one form or another has appeared in a number of books." The puzzle involves just three ...

7

I was able to more or less understand the solution only by imagining that this whole story is happening in Island 100 - our island, and there are another 99 islands in the ocean, each called Island 1, Island 2, Island 3, ..., Island 99, each of them named after the total number of people with blue eyes in them. The total number of people in each island is ...

7

The oracle disproves a nested hypothetical. I'll try to prove this from the top down without using induction. First, a definition: Person(n) is the n'th blue-eyed person. We number the blue-eyed people 1 to 100 without loss of generality, with each person being Person(1) from their own perspective. Those without blue eyes are not relevant to this proof and ...

7

I started writing my definitive explanation for how everyone is actually wrong about the necessity of the Oracle's proclamation and in the process finally explained to myself why, in fact, it's essential. Possibly not adding anything new to the list of answers (how ironic is that??) I'll throw in my explanation. This is highly unintuitive, but the way ...

7

Variant 1: The oracle saying "not everyone has blue eyes" does not add any information here. Consider the case where there are only two people on the island, one of whom has blue eyes and the other having brown eyes. Being told that "not everyone has blue eyes" gives the blue-eyed person no information - seeing a brown-eyed person, the statement could have ...

7

The answer is Proof Let's generalise the problem and say there are $n$ logicians on the island, all with blue eyes. Call them $L_1,L_2,\dots,L_n$. So our specific problem is where $n=100$. If $n=1$, then the single logician knows from the oracle's statement that the number of blue-eyed logicians on the island isn't $0$, so it must be $1$; he leaves on the ...

7

The same logic really does work inductively for every case based on the previous case. Let's say there are 4 people on the island. You're one of them, so you can see the other three all have blue eyes. IF you don't have blue eyes, then you know that, from the point of view of any of the others, they can see just two people with blue eyes. So they should ...

6

According to the puzzle definition, they can only leave if they know their own eye color, and they all want to leave as soon as possible. So the first logician honestly states "I see X people with blue eyes." If he has blue eyes, he says "I see 99 people with blue eyes." All the brown-eyed people see 100 people with blue eyes. Each of them knows that if he ...

6

If there were $51$ blue eyed people on the island, any blue eyed person would know from the statement that he had blue eyes. They would see $50$ blue eyed and brown eyed people and saying $51>50$ is the only way to rectify this. All blue eyed people would, therefore, leave on day $1$. If there are more than $51$ then on day $2$ they know there are more ...

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