Blue Eyes – Explained (I think)

Here’s my attempt to explain the answer to the famous Blue Eyes puzzle.  Bear with me.  Also, this may all be wrong.  It’s quite confusing!

The answer to the puzzle is that all 100 blue-eyed people will leave on the 100th night.  Why?  If there was only one person with blue eyes, they would leave the first night, knowing  from simple observation that they were the person the guru had mentioned.  Two blue-eyed  people will leave on the second night, because each knows that if the single blue eyed person they could see was the only one they would have left on the first night.  When they don’t leave they can assume that the blue-eyed person they see must also see a blue-eyed person, and seeing no other blue-eyed people, the know it must be themself.  So the night of leaving is the number of blue-eyed people each blue-eyed person can see, plus one.

The brown eyed people and the guru can never leave.

However, whenever this puzzle comes up, lots of people seem to get the answer but have a lot of difficulty understanding how the guru’s words can be important.  “She says she sees a blue eyed-person?  Well, duh, there are loads of them.  Everyone on the island can see a blue-eyed person.  What’s she telling them that they don’t already know?”

It’s a good question, and had me stumped for a while.  But she is telling them something very important:  she’s telling them how much confidence they can have in the assumptions of their peers.

The easiest way to understand this is to consider what will happen if there is only one blue-eyed person on the island.  In this case, the guru’s words are clearly very important.  If she announces she has seen at least one blue-eyed person, that blue-eyed individual can deduce the colour of his eyes because he can observe no other blue-eyed people in the population.  So he leaves the island on the first night, and the rest of the population are stranded.

In the case of two blue-eyed people the guru’s words are, again, very clearly important.  If individuals A and B are both blue-eyed they can deduce their eye-colour through the principle demonstrated in the one-blue-eyed-individual example above.  That is to say that A observes B as having blue-eyes and realises that, after hearing the guru’s words, if B had observed no other blue-eyed people he would have left the island.  Therefore when B does not leave, and A observes no one else in the population with blue eyes, A can deduce that he has blue eyes.  Exactly the same logic applies in reverse (from person B to person A), so they both leave the island on the second night.

Now consider the example of three blue-eyed people on the island: A, B and C.  Each of these three individuals can see two blue eyed people, so the guru’s announcement is not news to them.

We’ll look at the three-blue-eyed-person problem from A’s perspective but it’s very important to remember that the same thing is going on in the minds of B and C.

Okay, here goes: A knows that B cannot know the colour of his own eyes.  So A’s assumption about B is that, if A’s eyes are non-blue, B can only see one person with blue eyes: C.  Further to this (and this is where it starts to get a bit brain-knotty!) A can also assume that B is assuming that C can see no-one at all with blue-eyes (because A’s baseline assumption is that his own eyes are non-blue and he’s assuming that B mistakenly thinks his own eyes are non-blue).  If B believes C to be seeing no-one with blue eyes (he doesn’t believe that, but that is A’s logical assumption) then A cannot trust B to deduce his own eye-colour based on C’s behaviour.  And we end up with a logical stalemate.

However, when the guru announces that she can see at least one person with blue eyes  the stalemate is broken.  A need no longer worry that B thinks that C sees no blue-eyed individuals because A knows that each individual has been given the knowledge that there exists at least one person with blue eyes.  On this assumption, A can confidently observe the behaviour of B and C and compare it to what would happen in the two-blue-eyed-people example in paragraph seven (behaviour itself which is dictated by what would happen in the one-blue-eyed-person example).  If B and C don’t leave on the second night, A can deduce that he also has blue-eyes and all three, each observing no other blue-eyed-people, can leave on the third night.

It’s the same principle in the puzzle, but taken to mind-melting proportions.  Basically, everyone can see 99 blue-eyed people, but everyone only knows for sure that everyone else can see 98 blue eyed people, and everyone only knows for sure that everyone else only knows for sure that everyone else can see 97 blue eyed people.  And on and on and on, until you need the guru to break the stalemate and give each member of the population the confidence to trust that everybody knows that everybody else knows that there is at least one blue-eyed person.

QED, motherfuckers.

Criticisms encouraged.

Edit: I now no longer believe that this “bottom-up” way of looking at the problem accurately reflects the puzzle, but the principle is the same. The way I’ve worded it makes it look sort of like having blue eyes is a goal from the outset.

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3 Responses to “Blue Eyes – Explained (I think)”

  1. […] Reword themeless musings « Blue Eyes – Explained (I think) […]

  2. They all are oot there tits and have all got blue eyes bit are colour blind and canna be fucked with it all so they just make shit up and 2 have blue eyes. Ur due me a a note les lol x

  3. Hi Les! I’m not sure if people have lost interest in this in the BS.net thread or are still thinking about it… So anyway, the flaw in your solution is that it’s missing the explicit induction step proof. Once you put that in, the confusion about how exactly the guru’s statement breaks the symmetry goes away too.

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