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This question already has an answer here:

In the solution to the Blue Eyes logic puzzle, there is a follow-up question:

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?

I have found an answer to that question here:

Blue-eyed people can't see their own faces, so blue-eyed people can see one less blue-eyed face than non-blue-eyed people can. Even though I can see that there are at least 99 blue-eyed people, I don't know that they can see that, so I need to imagine people who see only 98, who would base their actions in part by imagining people who can see only 97 who would base their actions in part by imagining people who can see only 96, and so on...

This answer doesn't make sense to me. Why are they basing their logic on some hypothetical islander who can only see 96 blue-eyed people when they know such an islander cannot possibly exist and - moreover - know that everyone else knows it too?

Taking a smaller example with 6 islanders for illustrative purposes:

A (blue), B (blue), C (blue), D (brown), E (brown), F (brown)

No matter what A's eye color is, he knows:

  • B will see at least one other blue (C)
  • C will see at least one other blue (B)
  • D, E and F will see at least two other blues (B & C)

Everyone can see at least 1 other blue-eyed islander, and everyone else knows it.

What value does Day 1 serve? Why not skip ahead to Day 2?

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marked as duplicate by March Ho, Gamow, KSmarts, Michael Rize, Haobin Feb 24 '15 at 16:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The question might be a bit different but the answers are the same. Both question basically are "Why is this the solution to this blue-eyes puzzle" in a way. By your logic you could skip 99 days because Everyone can see at least 99 other blue-eyed islanders. The other question goes in depth of wha the importance of day one is. Reading through all answers and comments there should give answer to you understanding the solution $\endgroup$ – Ivo Beckers Feb 18 '15 at 22:27
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    $\begingroup$ @IvoBeckers Thank you for the suggestion but I have already reviewed in depth the answers there and also on a couple "Blue Eyes" questions on the Math SE site. If there's a particular answer there you feel is directly relevant, perhaps you could highlight it. $\endgroup$ – Lynn Feb 18 '15 at 23:47
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    $\begingroup$ It seems to me that Gilles's answer there, in particular "Now suppose that only Alice and Bill have blue eyes. Before day 0, Bill already knew that there was someone with blue eyes, but he did not know that Alice knew. If Bill had had green eyes, Alice would have been the only blue-eyed person and would not have known. On the first night after the guru, Alice doesn't leave; this tells Bill that Alice did not know the color of her eyes, so learn that she was the only blue-eyed person." should resolve your question. Does it not? $\endgroup$ – Josh Caswell Feb 19 '15 at 0:01
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    $\begingroup$ @JoshCaswell - I'm not asking about whether they can skip the Guru's proclamation. Clearly they cannot, as you pointed out. What I'm asking is that once the Guru has spoken, why can they not collectively reach a conclusion about which days provide them all no useful information. $\endgroup$ – Lynn Feb 19 '15 at 0:03
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    $\begingroup$ @Lynn I've faced a smiliar problem, as seen in the comments to that question. Not satisfied with the explanations yet, either. $\endgroup$ – No. 7892142 Feb 19 '15 at 8:31
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After talking with you for a bit in chat, I see better what the problem is and where you're heading down the wrong path.

Here's the principal issue: knowledge is only useful if everyone on the island can make the same conclusion. Why? Let's say $C$ has a piece of knowledge that lets them skip a day, but can't verify that everyone else also has this piece of knowledge. If $C$ skips ahead a day, $C$ could easily be leaving people behind who haven't made the same conclusion.


Let's say there are three people: $A$, $B$, and $C$. They all have blue eyes.

Your premise starts off with the conclusion that $C$ sees two people with blue eyes, and knows nobody will leave the island on day 1. This conclusion is entirely correct. However, $C$ can't just go ahead and skip day 1 without first checking to make sure $A$ and $B$ will do the same. Otherwise, $C$ would skip ahead and leave $A$ and $B$ confused.

Here's where the logical trap is. While $C$ can conclude that nobody will leave the island, $C$ can't conclude that $B$ or $A$ will say the same. $C$ has two cases to explore: if they have blue eyes, and if they have brown eyes. They don't know which it is. Let's say $C$ supposes they have brown eyes (even though it's not true) - then, what does $C$ conclude about what $B$ knows?

In this thought experiment of $C$'s, $B$ sees only one person with blue eyes. But $B$ doesn't know their own eye color either! In $B$'s mind, their eye color could theoretically be brown. As a result, $A$ would leave the island on day 1!

Now, obviously, this isn't true. The point is that it's possible for $C$ to conceive of a situation in which neither $A$ nor $B$ could conclude nobody will leave the island. As a result, the knowledge that nobody will leave the island on day 1 isn't useful to $C$. $C$ may know this, but there's a chance that neither $A$ nor $B$ do, and in that case, $C$ would be skipping ahead.


To continue this thought experiment through, let's say day 1 passes. Nobody leaves the island. Now what's $C$ thinking?

Well, $C$ ponders what $B$ knows. In the case where $C$ has brown eyes, $B$ has seen nobody leave the island and has concluded that they have blue eyes. This is a correct conclusion. In the case where $C$ has blue eyes, $B$ still can't tell whether they have blue or brown eyes, but $B$ knows that everyone else is conducting the same thought experiments.

As a result, from the perspective of $C$, either $B$ and $A$ will leave the island tonight (if $C$ has brown eyes), or $A$, $B$, and $C$ will leave the island tomorrow night (if $C$ has blue eyes). The first night is critical, though, because it establishes the common knowledge that in everyone's thought experiments, there are at least two people with blue eyes.

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While JonTheMon and Emrakul's answers are entirely correct, here's a different way of explaining why.

Everyone knows what they can see themselves, and can put themselves in everyone else's shoes to guess what they see.

Sometimes that works:

Scenario: blue  blue  blue  blue  blue  blue 
- 1 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4
- 2 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4
- 3 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4
- 4 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4
- 5 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4
- 6 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4

Everyone correctly deduced that days 1 - 4 are pointless and skips ahead to day 5.

But sometimes it doesn't:

Scenario: blue  blue  blue  blue  blue  brown 
- 1 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3
- 2 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3
- 3 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3
- 4 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3
- 5 concludes everyone else sees 3 or 4 or 5 blue eyes - decides it's safe to skip 3
- 6 concludes everyone else sees 4 or 5 blue eyes - decides it's safe to skip 4

Ooops. Islander #6 decided something different than everyone else.

Skipping ahead only works if everyone's on the same page and skips ahead the same number of days.

So going back to the original 200-islander problem: Nobody's under the illusion that there's only 1 blue-eyed person on the island, or 2, or even 97. They can see with their own eyes that that's not the case, and they know that everyone else knows it too.

But some might think it's safe to skip 97 days, others 98. Since they can't agree on how many days to skip, they just have to go through the tedium of days 1-97 to be safe -- even though the outcome of those days surprises nobody.

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  • $\begingroup$ But couldn't 6 also conclude that the others could conclude something different, and, following that, conclude the same, i.e. he knows that everyone can see 4-5 blue eyes, but he concludes that everyone knows that everyone knows that everyone can see at least 3? $\endgroup$ – No. 7892142 Feb 23 '15 at 8:16
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    $\begingroup$ @No.7892142 - Everyone has to use the same system for deciding how many to skip. So yes, #6 can say: "Hmm, I think it's safe to skip 4, but everyone else might conclude 3, so I'll skip 3 instead." The trouble is that #1-#5 will also think the same thing: "Hmm, I think it's safe to skip 3, but I might be wrong so I'll skip 2 instead." They still end up skipping 2 different numbers. Nobody has enough information to conclude that they're the odd man out. $\endgroup$ – Lynn Feb 23 '15 at 8:25
  • $\begingroup$ Ah, that's a good point. (And actually probably resolves the issues I still had with accepting the solution.) $\endgroup$ – No. 7892142 Feb 23 '15 at 8:26
  • $\begingroup$ @No.7892142 - Incidentally, my example shows the exercise with #6 imagining "what would #1-5 see?" to reach his conclusion. Even if you do the exercise with #6 imagining: "what would #1-5 see? Now what would #1 guess that #2-4 see? What would #2 guess that the others see?" you still end up with folks guessing 2 different numbers (I wrote a program to prove it to myself because I had such a hard time believing it :)). There just doesn't seem to be enough clues for everyone to end up on the same page. $\endgroup$ – Lynn Feb 23 '15 at 8:31
  • $\begingroup$ I'm just glad someone finally convinced me of the solution! Thanks for that! $\endgroup$ – No. 7892142 Feb 23 '15 at 8:33
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Day one serves to handle the case that there is only 1 person with blue eyes. They would see no-one else with blue eyes, but since at least 1 person has blue eyes, they know it's them then.

Back to why they don't skip days, let's take the scenario where you have 4 blues and 2 brown.

  • Each person with blue will see 3 other blues.
  • But they don't know if they have blue eyes, so someone they see with blue might only see 2 other blue sets.
  • So, do you skip to day 3, or day 4 (a brown eye person would see 4 blues, so they assume a blue eye could only see 3)?

They can't communicate with each other.

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  • $\begingroup$ But we know from the puzzle setup that there isn't only 1 person with blue eyes, and it would seem that the islanders can deduce that also. You would have to skip ahead to the lowest day you could be sure of everyone ruling out. $\endgroup$ – Lynn Feb 18 '15 at 21:44
  • $\begingroup$ I think the biggest issue is coordinating what the "lowest day you could be sure of everyone ruling out" is. They can't communicate with each other. What if they're off by 1? $\endgroup$ – JonTheMon Feb 18 '15 at 21:51
  • $\begingroup$ True it would be a chore, but these are supposed to be perfect logicians per the puzzle setup. $\endgroup$ – Lynn Feb 18 '15 at 21:58
  • $\begingroup$ True, but they have to communicate somehow. Who leaves and who doesn't is the only way they communicate. $\endgroup$ – JonTheMon Feb 18 '15 at 21:59
  • $\begingroup$ I don't think that's true. In my example no nobody needed to communicate to deduce with certainty that everyone else could see at least 1/blue. $\endgroup$ – Lynn Feb 18 '15 at 22:06

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