Blue eyes

difficulty 5

riddle
Published

December 10, 2018

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. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. 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.”

Hover to show the answer.

All the 100 blue-eye people leave the island on the 100th night.

The following chart explains why using 4 people. The reflexion is the same, simply longer, for 100 persons. Click on the chart to open it in a new window.

The information “There is at least one blue-eyed person” is needed !
The information “There is at least one blue-eyed person” is needed !
If I am blue-eyed, he looks at 3 blue-eyed persons
If I am blue-eyed, he looks at 3 blue-eyed persons
If I am brown eyed, he looks at only 2 other blue-eyed
If I am brown eyed, he looks at only 2 other blue-eyed
When I look at 3 blue-eyed person, I wonder if I am myself blue eyed.
When I look at 3 blue-eyed person, I wonder if I am myself blue eyed.
I consider myself blue-eyed
I consider myself blue-eyed
I consider myself brown-eyed
I consider myself brown-eyed
ME
ME
I wonder if this guy thinks he is blue eyed or not (I wonder the same for the 2 other, the reflexion is the same)
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ME
ME
A
A
B
B
C
C
He considers himself blue-eyed
He considers himself blue-eyed
He considers himself brown-eyed
He considers himself brown-eyed
ME
ME
A
A
B
B
C
C
If A considers himself blue-eyed, there are 3 blue-eyed persons.
If A considers himself blue-eyed, there are 3 blue-eyed persons.
ME
ME
A
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B
B
C
C
If A doesn’t consider himself blue-eyed, he will try to understand what the 2 last one are going to do 
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This last guy could think he is brown-eyed
This last guy could think he is brown-eyed
This last guy could think he is blue-eyed
This last guy could think he is blue-eyed
ME
ME
A
A
B
B
C
C
Then there are 2 blue-eyed persons 
Then there are 2 blue-eyed persons 
ME
ME
A
A
B
B
C
C
Then he looks at a blue-eye guy that is alone
Then he looks at a blue-eye guy that is alone
ME
ME
A
A
B
B
C
C
If someone comes and say “There is at least one blue-eyed person”, than person B will see 3 brown-eyed and leave !
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If someone comes and say “There is at least one blue-eyed person”, Than, no one will dare to leave the first day !
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After one day, we know which one of those 2 supposed situations is the true one.

Either B left, and there was only 1 blue-eyed person.

Or B stays, and C will understand that he is blue after one day. 
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However, if after two days, B and C are not leaving…

A will understand that his supposition that he is brown might be false.

The third day, A, B and C are going to leave

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If the 3rd day no one leaves.. 

I should understand that the supposition that I am brown can not hold. 

Therefore, we are all blue-eyed.

The 4th day, we all leave.
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The problem is the same with 4 persons and with 100 persons.

This is the concept of common-knowledge. 
At first, I might think that it is common knowledge that there are at least 3 blue-eyed person. And I might think that the sentence “there is at least one blue-eyed person” is useless, since everyone sees at least two blue eyed person. 

But we don’t all have the same knowledge. If I am browm eyed, other people might only see 2 blue-eyed persons. Those persons will think that other could only see 1 blue-eyed person. And those persons that see only one blue-eyed person could think that someone sees no blue-eyed persons. 

This means that, until someone says “I see at least one blue eyed person” and 4 day passes, we don’t all have the same knowledge about the world.

The time it takes to reach common knowledge is the time the persons have to wait before leaving. 
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Yes this is comic sans ms :)
Yes this is comic sans ms :)<br>