This riddle, or rather puzzle or what you want to call it got famous from xkcd and you can read the challenge there, but myself and others got confused from the wording. Here I will present my own version, which I think is more simple and clear. The “offical” answer is very beautiful, but I will here present my alternative solution as well.

## My Wording of the Puzzle

On an island there is 100 blue eyed and 100 brown eyed men. The men are all very intelligent and to quote the xkcd wording “if a conclusion can be logically deduced, they will do it instantly”. They don’t know their own eye color and if they find out, they will magically be teleported to paradise at midnight, but they need to be 100% sure of their color. They can not communicate with each other in any form. They can just observe. One day an angel comes down from the sky and says infront of them all: “I see someone with blue eyes”, then the angel leaves and never returns.

Who will leave the island and on which day?

Here are some other pointers:

- They can not see the reflection of their eye color from the water or anything of the kind.
- It’s not relevant how long they have been on the island.
- There is no stupid answer. The answer is pure logic and really amazing.
- The angel is not relevant. They could just as well have found a note on the ground stating the same thing.
- Remember that every blue eyed person see 100 brown eyed and 99 blue eyed, and it’s tempting for every blue eyed to assume there is 100 of each and that way know their color, but there might be 101 brown and 99 blue eyed. And it’s of course vice versa.
- What do you mean by “which day”? Well, there is only one referance point.

## My Problems with the Original Wording

- Introducing a “guru” with green eye color. Useless information that pollutes the puzzle.
- Mentioning a ferry might indicate that there is a captain that they need to submit their eye color to. What if anyone is listening when a person submits his answer? And is there a restriction on how many times they can try to leave with the ferry? Can a person try “blue” one day and “brown” the next?

## The “Offical” Answer

You can read the answer on xkcd, but I will try with my own wording here.

If we start with only one blue eyed and one brown eyed, the answer is simple. After the angel speaks, the blue eyed will look at the brown eyed and instantly know that himself must have blue eyes and leave day 1.

Notice this answer applies also if there was 1000 brown eyed and 1 blue eyed.

If there was two blue eyed and 1 brown eyed on the island, the blue eyed can think like this “If I have brown eyes, then this dude with blue eyes will know for certain that he has blue eyes, because he sees two brown eyed, and will leave today. If he doesn’t leave today, then I must also have blue eyes”. They both think like this, because they are equal intelligent and they will both observe that the other guy didn’t leave on the first day and they will both leave on day 2. Notice also that we can introduce 1000 brown eyed and there will be no difference, so we can conclude that the brown eyed are irrelevant and will never leave.

If there was three blue eyed, then each of them see two other blue eyed and thinks “If I have brown eyes, these two blue eyed will leave on day 2”. Now, if the other two doesn’t leave on day 2, each person knows they have blue eyes and leaves day 3.

See a pattern now? The 100 blue eyed will leave the island on day 100. Every person must use the same logic and think “I see 99 blue eyed, then if no one leaves on day 99, then I must have blue eyes as well and leave the next day”.

If you think you get the answer you can try asking yourself these questions quoted from the xkcd answer:

- What is the quantified piece of information that the Guru/angel provides that each person did not already have?
- Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?
- Why do they have to wait 99 nights if, on the first 98 or so of these nights, they’re simply verifying something that they already know?

## My Alternative Answer

It starts with one guy just standing up. Then another guy will stand next beside him. Let’s say one has blue and one has brown eyes. They have now formed this row: **BL**/BR (Bold being blue and regular brown)

The third person will see the colors of these guys and he will place himself in the middle. It doesn’t matter which color his eyes are, but let’s say they are brown. Updated Row: **BL**/BR/BR

If the two initial guys have both the same color, the third guy will just move to one of either side.

The forth guy comes and he will place himself in the middle where blue and brown eyed are separated. His goal is to have one blue eyed and one brown eyed on each side. Let’s say he has blue eyes, which makes the row like this: **BL**/**BL**/BR/BR

The next guy will do the same and so on, until everyone is standing in the row, where the 100 men from the left is blue and the 100 men on the right is brown eyed.

Now everyone will know their eye color, just by looking around, except the two guys in the middle. They don’t know where the distinction between blue and brown is. They will see only blue eyes in one direction and only brown in the other, but they can’t know their own color. But for the other 198 men, they can leave the island on day 1.

I submited this answer to Randall (the creator of xkcd), but he claimed this answer was not valid because there was too much communication between the men :’(

xkcd (1) , blueeyes (1)