Blue Eyes Logic Puzzle

"A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?


There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves."
"

http://xkcd.com/blue_eyes.html

It's a good puzzle. Unfortunetly, I thought that I misunderstood the puzzle and looked up the solution, but there is really enough information to solve the puzzle.  

 

This riddle is hard to understand, let me try to clarify.

1) There are 100 people with blue eyes and 100 people with brown eyes
2) When a person finds out the color of their eyes, they leave at night
3) They know that at least one person has blue eyes, the Guru said it at noon
4) All the people on the island have perfect information about all the other people and are perfect logicians

This is really enough information to solve the riddle.

 

 

Nice try, but the people have to be 100% sure about the eye color they have in order to leave; they can't just leave if they only think their eye color is so and so.

I have heard a variation to this riddle. Same set-up, except, the situation is a bit different.

 

There are people stranded on an empty island. They find and read a message that states "At least one person amongst the stranded has red eyes." The islanders now all know that at least one person DOES have red eyes. They can see the eyes of others, but not their own. They cannot communicate whatsoever. No talking, writing, sign language, nothing. There are no reflective surfaces. They have an opportunity to leave each night. The guru doesn't exist. You can only leave if you know you have red eyes. Who leaves the island, and on what night? There's more to the answer than you may think. It's not a trick answer, and not a simple one. You may be correct, but with the wrong reasoning. The cleverness of this riddle lies within the correct logic and reasoning. The point of this riddle is not to get around the "no communication" or "no reflective surfaces" part. Just accept that, it's to guide you towards the right answer using internal contemplation and logic.

 

I've put much thought into this to ask and answer this riddle properly. 

Solution below... 

 

Solution: 

 Put yourself in the shoes of the person with red eyes, because at least one DOES exist. You cannot leave unless you are certain you have red eyes.

The solution to this riddle is an "If....then....." answer. 

 

 If only one person has red eyes, 1 leaves on the first night. Here's the description of the thought process of the one who leaves:

You look around, and you think to yourself, "I don't see anyone else having red eyes, and someone DOES have it, so I must have it. I leave on the first night."

 

 If two people have red eyes, 2 people leave on the second night. Here's the description of the thought process of the 2 who leave, they both are thinking the same thing, this is what one of them would think:

You look around, and you think to yourself, "I see one guy with red eyes, he should be leaving tonight." The next day, he's still there, so you think to yourself, "Why is he still here? He should've left last night because he has red eyes. He must not be certain that he has red eyes. If I were him, I'd see that no one else has red eyes, and I'd leave. The only reason he's still here is because he has to see someone else with red eyes. But I don't see anyone else besides him with red eyes. I must have red eyes too. We both leave on the second night."

 

The same principle is applied as the "If....then...." number increases.

In the example with Red Eyes, from the perspective of any Red Eyed Person:

For 1 Red Eyed Person:
I know there is at least one person with Red Eyes.
I see no other people with Red Eyes.
Therefore, the Red Eyed person must be the one I cannot see (myself).
I leave on the first night.



For 2 Red Eyed People:
I know there is at least one person with Red Eyes.
I see one person with Red Eyes.
If he is the only person with Red Eyes, he will leave on the first night.

On the second day, I see that the other Red Eyed Person has not left.
Since he did not leave, he must not be sure that he has Red Eyes.
He could only be unsure of this if he could see someone with Red Eyes I do not.

I see no other people with Red Eyes, besides the one I know about.
Therefore, the other Red Eyed person must be the one I cannot see (myself).
We both leave on the second night.



For 3 Red Eyed People:
I know there is at least one person with Red Eyes.
I see two people with Red Eyes.
If they are the only people with Red Eyes, they will leave on the second night.

On the second day, I see that the Red Eyed People have not left.
This is expected.

On the third day, I see that the Red Eyed People still have not left.
Since they did not leave, they must not be sure that they have Red Eyes.
They could only be unsure of this if they could see someone with Red Eyes I do not.

I see no other people with Red Eyes, besides the two I know about.
Therefore, the last Red Eyed person must be the one I cannot see (myself).
We all leave on the third night.



In other words, for a group of N people with the listed eye color, it will take them N days for each individual to figure out that they themselves have this color of eyes, on logic alone (without communication, mirrors, etc.), and thus they leave on the N'th night.

It's possible to prove, by mathematical induction, that this applies for all N:
• If there is only one person with the given eye color, he leaves on the first night.
• If you see N people with the given eye color, and they aren't gone by the N+1'th day, you leave with them on the N+1'th night.