100 Blue Eyes Puzzle

Puzzle:

This puzzle is one of the hardest puzzle in the world. If you are able to solve this, believe me you are genius.

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.”

Puzzle Solution:

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 other’s knowledge (that either they did or did not know their own eye color on previous days).

Blue eyed people leave on the 100th night.

If you (the person) have blue eyes then you can see 99 blue eyed and 100 brown eyed people (and one green eyed, the Guru). If 99 blue eyed people don’t leave on the 99th night then you know you have blue eyes and you will leave on the 100th night knowing so.

Proof:

Imagine a simpler version of the puzzle in which, on day #1 the guru announces that she can see at least 1 blue-eyed person, on day #2 she announces that she can see at least 2 blue eyed people, and so on until the blue-eyed people leave.

So long as the guru’s count of blue-eyed people doesn’t exceed your own, then her announcement won’t prompt you to leave. But as soon as the guru announces having seen more blue-eyed people than you’ve seen yourself, then you’ll know your eyes must be blue too, so you’ll leave that night, as will all the other blue-eyed people. Hence our theorem obviously holds in this simpler puzzle.

But this “simpler” puzzle is actually perfectly equivalent to the original puzzle. If there were just one blue-eyed person, she would leave on the first night, so if nobody leaves on the first night, then everybody will know there are at least two blue-eyed people, so there’s no need for the guru to announce this on the second day. Similarly, if there were just two blue-eyed people, they’d then recognize this and leave on the second night, so if nobody leaves on the second night, then there must be a third blue-eyed person inspiring them to stay, so there’s no need for the guru to announce this on the third day. And so on… The guru’s announcements on the later days just tell people things they already could have figured out on their own.

4 Thoughts on “100 Blue Eyes Puzzle

  1. Andrew King on May 23, 2017 at 8:01 pm said:

    There is a huge problem with the wording of this puzzle which debunks the whole concept. In the puzzle, you fail to specify that the guru picks a different person or “new information” every single night, otherwise she could be looking at the same blue eyed person every time. Secondly, your answer of all the blue-eyed people leaving on the 100th night presumes that the guru said blue eyes every night prior when she could have spotted a brown eye unless it is presumed that this is part of her logic-solving genius plan. And if that were the case all the blue eyes could all leave on the 1st night because they would have figured out that the guru would be using the quickest way possible to get people off the island and when they realised that there were 99 other blue-eyes (according to their personal count) and she could have just picked on the odd one out which could have been themselves, with red eyes, they would have then realised what the guru was doing and determined that they must be 1 of the 100 blue eyes and been sent away. The 2nd night everyone would also realise again that they must all be the same, so when she said brown they could have all gone too and the guru would be left alone. BOOM

    • Working of the puzzle actually is corret, but could be more clear :
      “The Guru is allowed to speak once AND ONLY ONCE FOR HER ENTIRE LIFE” would be more precise.
      I rather think the solution is confusing, as it holds 2 elements
      a) A counter 1,2,3, … blue eyes and b) The daily announcement.

      Solution :
      Let’s reduce the problem to only blue-eyed people (call them B’s) and the green-eyed guru (call her G) just after G sais: “I can see someone who has blue eyes.”
      So let’s start with the smallest possable constalation of one B and one G and add addtional B’s ass we go on.
      - 2 people: 1 B and 1 G: B would leave at first chance on night 1.
      - 3 people: 2 B’s and 1 G: The B’s seeing each other must first assume the other B is the blue eyed person in question and (beeing conservative) assume that their own eye collor is NOT blue, thus expection the other B to leave on night 1. BUT if NO one leaves each B must conclude that he himself must also have blue eyes, thus both B’s will leave the island the next night (Note: night 2 and two B’s).
      - 4 people: 3 B’s and 1 G: Les’s pick one of the B’s and call him Bert. Bert sees 2 B’s and (beeing conservative) assume that his own eye collor is NOT blue. If this is true he will conclude that on night 2 the 2 other B’s will leave the island.
      BUT they don’t, thus Bert must assume his eyes are also blue and he’ll leave on the next night (Note: night 3 and 3 B’s). But he’ll not be alone as the other B’s would have come to the same solution.
      - 5 people: 4 B’s and 1 G: Now we can see the patteren. If Bert sees 3 B’s (and assumes he’s not blue eyed) he must expect the 3 B’s to leave on night 3. If this doesn’t happen he knows that he actually IS blue eyed and will leave on night 4 (with the other B’s).

      a.s.o.

  2. There needs to be more rules in regards to the ferry. It was not stated that the islanders are only allowed to guess their eye color once in their life time. From the stated riddle there is no penalty for trying to get on the ferry only you get denied that night. So my answer would have been everyone guesses they have blue eyes on the first night and 100 of them get off the island on the first night.

  3. Andrew 2 on April 16, 2021 at 5:06 am said:

    Andrew (other Andrew) – sorry, there do not need to be any other rules for what you’re talking about. It’s clearly stated that the islanders are all “perfect logicians” – that if a conclusion can be deduced they will do so (instantly). And: any islanders “who HAVE figured out” the color of their own eyes then leave the island (not “try to”).
    Taken together, this precludes “guessing”, or “trying out” solutions – they must KNOW. Also, “trying” to leave and getting kicked off the ferry would provide additional information which is not in the scope of the rules, which obviously aim to be comprehensive & exhaustive – i.e. complete.
    There is sufficient information and rules for a solution as-is; the only way your solution works is if they all guess that that’s the solution they’ll arrive at, but there are multiple arrangements of eye colors that could lead to a guess of that solution with them being wrong. With this arrangement, they can be right WITH CERTAINTY *ONLY* on night 100, and, again, “perfect logicians” would have to have certainty — not a guess. That constraint’s already baked in.

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