# Blue-eyes logic puzzle **URL:** \ **Category:** Factual Questions\ **Created:** [December 8, 2011, 5:31pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531 "2011-12-08T17:31:14Z")\ **Posts on this page:** 20\ **Page:** 6
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [December 9, 2011, 4:13pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/101 "2011-12-09T16:13:18Z")
> [@Zeriel](#): > > That, in a nutshell, is why you need the guru to speak to provide the logical bootstrap. Regardless of the fact I KNOW there are either 99 or 100 blue-eyed people, I cannot deduce what EVERYONE ELSE might know since they have **different sets** of limited information. You mean “because for all I know they might have different sets of limited information.” ---
**Author:** ![Zeriel](https://avatars.discourse-cdn.com/v4/letter/z/58956e/32.png) [@Zeriel](https://boards.straightdope.com/u/Zeriel)\ **Post date:** [December 9, 2011, 4:13pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/102 "2011-12-09T16:13:24Z")
> [@coremelt](#): > > You need to look at it from what the system knows, not what anyone one person can know, and if we can logically work out what the system as a whole knows (that the minimum number of blue eyes is 99) then so can your perfect logicians. A perfect logician can work out that if he can see 99 blues, someone else might only see 98, and that would color the decision-making process of that other person. Said logician has no way of knowing whether everyone else sees 99 or 98. > [@Frylock](#): > > You mean “because for all I know they might have different sets of limited information.” By the rules of the puzzle, they all DO have different sets–Each Person N lacks the information of “N’s eye color”. ---
**Author:** ![coremelt](https://avatars.discourse-cdn.com/v4/letter/c/4491bb/32.png) [@coremelt](https://boards.straightdope.com/u/coremelt)\ **Post date:** [December 9, 2011, 4:15pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/103 "2011-12-09T16:15:45Z")
> [@Frylock](#): > > **Coremelt,** tell me where you think this reasoning in the four-person case goes wrong. The reasoning is wrong because the islanders are only relying on this type of inductive reasoning and missing the fact that each person is only missing one piece of information (which you still haven’t acknowledged). The inductive logic would only follow if every player was not told that everyone else is only missing one piece of information, then they would have to reason as you say and would come to the incorrect conclusion that some people might believe there is no blue eyes. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [December 9, 2011, 4:30pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/104 "2011-12-09T16:30:13Z")
> [@coremelt](#): > > then they would have to reason as you say and would come to the incorrect conclusion that some people might believe there is no blue eyes. No-one is saying that they would come to that conclusion which, as you say, would be incorrect\*. We are saying that they would come to conclusions like “some people might believe that some people might believe there are no blue eyes”, which are quite different. I think your argument is “they’re reaching an absurd conclusion, so their reasoning must be wrong”, but you are talking about a different conclusion. - except in the trivial cases of two blue-eyed islanders or less. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [December 9, 2011, 4:33pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/105 "2011-12-09T16:33:17Z")
> [@coremelt](#): > > The reasoning is wrong because the islanders are only relying on this type of inductive reasoning and missing the fact that each person is only missing one piece of information (which you still haven’t acknowledged). > > The inductive logic would only follow if every player was not told that everyone else is only missing one piece of information, then they would have to reason as you say and would come to the incorrect conclusion that some people might believe there is no blue eyes. Did you read my post? At all? No, you skipped it because it was too long, didn’t you? Go back and read it. You did exactly what I accused you of doing. When asked “What does A think about what B thinks about C’s thoughts?” You answer with what A thinks about C’s thoughts. You don’t seem to understand that those “missing pieces of information” stack up when going through the logic chain. I even took the time to write it all out for you, each and every pattern. Now go tell me where I’m wrong. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [December 9, 2011, 4:45pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/106 "2011-12-09T16:45:26Z")
> [@coremelt](#): > > The reasoning is wrong because the islanders are only relying on this type of inductive reasoning and missing the fact that each person is only missing one piece of information (which you still haven’t acknowledged). There is no inductive reasoning (in either the formal or informal sense) in the code-block in my post. Can you please go back and re-read it step by step with that in mind? Each person is indeed missing just one relevant piece of information–the color of his own eyes. But the point is that this single missing piece of information leads to a cascade of beliefs about what is possible concerning what others believe about others’ beliefs and so on. And once the guru speaks, some of those beliefs about what is possible are eliminated. And what is left turns out to be enough to draw concrete conclusions after a number of days. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [December 9, 2011, 5:12pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/107 "2011-12-09T17:12:15Z")
> [@Chessic\_Sense](#): > > I even took the time to write it all out for you, each and every pattern. I meant to say that (with all due respect to the other perfectly good and fundamentally equivalent explanations) I did find your explanation especially clear, particularly how you highlighted where people might go wrong in thinking about the puzzle. ---
**Author:** ![Rain\_Soaked](https://avatars.discourse-cdn.com/v4/letter/r/e19adc/32.png) [@Rain\_Soaked](https://boards.straightdope.com/u/Rain_Soaked)\ **Post date:** [December 9, 2011, 5:28pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/108 "2011-12-09T17:28:33Z")
I want to thank **Frylock** , **Chessic Sense** and **Ximenean** and the rest for the effort they have put into explaining this logic puzzle. It has really helped me understand it and more importantly how to explain it to others. Thanks again. ---
**Author:** ![Quercus](https://avatars.discourse-cdn.com/v4/letter/q/7ab992/32.png) [@Quercus](https://boards.straightdope.com/u/Quercus)\ **Post date:** [December 9, 2011, 6:25pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/109 "2011-12-09T18:25:36Z")
> [@coremelt](#): > > The reasoning is wrong because the islanders are only relying on this type of inductive reasoning and missing the fact that each person is only missing one piece of information (which you still haven’t acknowledged). > > The inductive logic would only follow if every player was not told that everyone else is only missing one piece of information, then they would have to reason as you say and would come to the incorrect conclusion that some people might believe there is no blue eyes. Back up for me. I’m not sure what (if anything) you’re objecting to. Do you think the given solution is wrong, in that perfect logicians would not follow the reasoning and have all the blue-eyed islanders leave together on day 100? If you believe the solution is wrong, what would they do? It may help to realize that the Guru does two things: they provide one small bit of information (relevant in the small n cases) but they also provide a common ‘Starting Now!’ point, which is important whenever there is more than one blue-eyed person. And, I want to point out, it’s the exact same puzzle whether or not there’s an agree-upon strategy beforehand, as long as the strategy is logical. Remember in the original puzzle they’re all perfect logicians, so they could deduce the optimal strategy without having to discuss it among everyone, and they would all know that the others are all also following the optimal strategy. But it’s easier to think about if we imagine people agreeing on a strategy beforehand. Or maybe I’m wrong about this; if so, I invite you to give me an agreed upon strategy that’s better than this logical deduction. ---
**Author:** ![coremelt](https://avatars.discourse-cdn.com/v4/letter/c/4491bb/32.png) [@coremelt](https://boards.straightdope.com/u/coremelt)\ **Post date:** [December 9, 2011, 7:02pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/110 "2011-12-09T19:02:47Z")
The best I can put my objection in formal terms is that there is an implied paradox in the description of the situation and that the formal logician islanders can only come to the right conclusion by ignoring an obvious paradox when it conveniently suits them. Namely they have to believe that 1 other islander can somehow think there is no blue eyed islanders when they themselves can see 99 or 100 blue eyed islanders AND in addition they know everyone else can see the same as them + or - one. I can’t spell it out in formal logic terms, but I think the paradox is there. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [December 9, 2011, 7:06pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/111 "2011-12-09T19:06:33Z")
> [@coremelt](#): > > Namely they have to believe that 1 other islander can somehow think there is no blue eyed islanders when they themselves can see 99 or 100 blue eyed islanders AND in addition they know everyone else can see the same as them + or - one. Nope. That’s not what they’re believing at all. You keep trying to interpret the question of “What does A think about what B thinks…Z thinks” as “What does A thinks Z thinks?” I’ll stick around for as long as necessary to get you to get this. But you have to promise me that you’ll play along with my reindeer games. I’m going to put you through a similar exercise and you can see for yourself how the thought process goes. Deal? ---
**Author:** ![Enderw24](https://avatars.discourse-cdn.com/v4/letter/e/ba9def/32.png) [@Enderw24](https://boards.straightdope.com/u/Enderw24)\ **Post date:** [December 9, 2011, 7:18pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/112 "2011-12-09T19:18:21Z")
If we take a ten person island where five people have blue eyes. One of the blue eyed people looks around and sees 4 people with blue eyes. He can then imagine one of those 4 blue eyed people looking around and seeing 3 people with blue eyes (because he, himself, doesn’t realize he has blue eyes too). But that’s it. That as far as it goes. In reality, as there are 5 blue eyed people, every blue eyed person knows for certain there are at least 4 blue eyed people. And every non blue eyed person knows there are at least 5 blue eyed people on the island. There is no twist of logic, no “everybody knows that everybody knows that everybody knows” where you end up with zero blue eyed people. There can’t be because if you find yourself falling down that rabbit hole you can defeat it by just looking at your damn neighbors and saying “oh yeah blue eyes.” Here’s what everybody knows. There are blue eyed people on the island. Everyone knows this, with or without the guru’s interference. ---
**Author:** ![Zeriel](https://avatars.discourse-cdn.com/v4/letter/z/58956e/32.png) [@Zeriel](https://boards.straightdope.com/u/Zeriel)\ **Post date:** [December 9, 2011, 7:23pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/113 "2011-12-09T19:23:38Z")
> [@coremelt](#): > > The best I can put my objection in formal terms is that there is an implied paradox in the description of the situation and that the formal logician islanders can only come to the right conclusion by ignoring an obvious paradox when it conveniently suits them. > > Namely they have to believe that 1 other islander can somehow think there is no blue eyed islanders when they themselves can see 99 or 100 blue eyed islanders AND in addition they know everyone else can see the same as them + or - one. > > I can’t spell it out in formal logic terms, but I think the paradox is there. > [@Enderw24](#): > > If we take a ten person island where five people have blue eyes. One of the blue eyed people looks around and sees 4 people with blue eyes. He can then imagine one of those 4 blue eyed people looking around and seeing 3 people with blue eyes (because he, himself, doesn’t realize he has blue eyes too). > > But that’s it. That as far as it goes. In reality, as there are 5 blue eyed people, every blue eyed person knows for certain there are at least 4 blue eyed people. And every non blue eyed person knows there are at least 5 blue eyed people on the island. > > There is no twist of logic, no “everybody knows that everybody knows that everybody knows” where you end up with zero blue eyed people. There can’t be because if you find yourself falling down that rabbit hole you can defeat it by just looking at your damn neighbors and saying “oh yeah blue eyes.” > > Here’s what everybody knows. There are blue eyed people on the island. Everyone knows this, with or without the guru’s interference. I think you might actually be conflating two things here–the actuality of the situation and the theoretical constructs one can build when reasoning about it. Obviously, no one within the puzzle believes that there are actually people who can’t see any blue-eyed people for any case where N\>3. What instead they believe is that there is a theoretical universe where it’s possible that given a certain set of knowns and unknowns, there are scenarios where a person \*could \*see no blue-eyed islanders. I want to say it’s almost like every iterative step is saying “Okay, what if I wasn’t here. What conclusions could Person B reach about Person C’s behavior? Now what if Person B wasn’t here either, what can C say about D’s behavior?..” This, again, is why the guru’s pronouncement is important. Without it, you have no way whatsoever of figuring out for yourself whether there are 99 or 100 blue-eyed islanders. With it, you have both a time zero where a theoretical lone blue-eyed islander could escape, and a theoretical construct by which everyone can figure it out for N blue-eyed islanders–and because that theoretical construct bootstraps itself (even if the lower-order versions are literally unbelievable because they’re provably wrong about the number of blue-eyed islanders), it works. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [December 9, 2011, 7:28pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/114 "2011-12-09T19:28:25Z")
> [@Enderw24](#): > > If we take a ten person island where five people have blue eyes. One of the blue eyed people looks around and sees 4 people with blue eyes. He can then imagine one of those 4 blue eyed people looking around and seeing 3 people with blue eyes (because he, himself, doesn’t realize he has blue eyes too). > > But that’s it. That as far as it goes. There is no reason to stop there. It’s not relevant that I know there are at least four people with blue eyes–I don’t have any reason to think that everyone else believes that. You already explained, for example, why I can rationally think it possible for someone to think there are only three blue-eyed people. You can absolutely keep going with this. It is also rational for me to think it possible for someone to think _that it’s possible for someone to think_ that there are two blue-eyed people. Please read that carefully. Notice what I did _not_ say. I did _not_ say it’s rational for me to think it possible for someone to think that there are only two blue-eyed people. What I _did_ say, rather, is that it’s possible for someone to think that _it’s possible for someone to think_ that there are only two blue eyed people. Take me and four others–persons B, C, D and E. I see that all four of them have blue eyes. Let’s represent this as follows: What I see: B – Blue C – Blue D – Blue E – Blue. Now I ask, “what does B see?” Since I don’t know what color my own eyes are, here’s what I’ve got: What B sees: A – ? C – Blue D – Blue E – Blue. That question mark means the color is unknown. In other words, for all I know, what B sees is a non-blue pair of eyes. So: What B _might_ see: A – non-blue C – Blue D – Blue E – Blue Okay so far so good? Now that was me asking what B might see. Now let me ask what B might think C might see. So I put myself in B’s mind, but I’m still me so I’m only privy to the information _I_ have. What B might think: What C sees: A – ? (this question mark is here because _I_ don’t know what color my eyes are. B – ? (this question mark is here because I know _B_ doesn’t know what color _B’s_ eyes are. D – Blue E – blue So then, the B I’m imagining (because remember, we’re not talking about what B might think C thinks but what _I_ think B might think C thinks), _for all I know_, might be like this: What B might think: What C _might_ see: A – not-blue B – not-blue D – Blue E – Blue I have to go but and so on and so on. I’ll maybe try to draw a full on diagram with thought balloons and stuff. B – not-blue In reality, as there are 5 blue eyed people, every blue eyed person knows for certain there are at least 4 blue eyed people. And every non blue eyed person knows there are at least 5 blue eyed people on the island. There is no twist of logic, no “everybody knows that everybody knows that everybody knows” where you end up with zero blue eyed people. There can’t be because if you find yourself falling down that rabbit hole you can defeat it by just looking at your damn neighbors and saying “oh yeah blue eyes.” Here’s what everybody knows. There are blue eyed people on the island. Everyone knows this, with or without the guru’s interference. [/QUOTE] ---
**Author:** ![Zeriel](https://avatars.discourse-cdn.com/v4/letter/z/58956e/32.png) [@Zeriel](https://boards.straightdope.com/u/Zeriel)\ **Post date:** [December 9, 2011, 7:31pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/115 "2011-12-09T19:31:23Z")
The main thing I keep having trouble deciding if I understand or not is the idea that they have to wait N days. It seems to me, riffing on what some of the dissenters are saying, that regardless of the theoretical constructs, we can deduce pretty easily that everyone on the island is going to actually see between N-2 and N blue-eyed people (if we presume we as a blue-eyed person can see N-1 blue-eyed people). So why can’t the count start as though it were day N-2? I think it’s because if we assume we have non-blue eyes, and thus Person B actually sees N-2 people, then Person B has to consider that Person C might only see N-3 people–and we’re in the bootstrap chain again. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [December 9, 2011, 7:37pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/116 "2011-12-09T19:37:27Z")
Ah, this will probably not help but I did just want to give the last diagram. ```auto For all I know (remember, I am person A): Here is what B might think: Here is what C might think: Here is what D might think: Here is what E might think: A -- unknown (because unknown to me) B -- unknown (because unknown to B) C -- unknown (because unknown to C) D -- unknown (because unknown to D) E -- unknown (because unknown to E) ``` In other words, I have no idea what B might think C might think D might think E might think. For that reason ```auto For all I know: For all B knows: For all C knows: For all D knows: E sees: A -- non-blue B -- non-blue C -- non-blue D -- non-blue E -- unknown ``` Yeah on second thought that’s probably just confusing… ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [December 9, 2011, 7:48pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/117 "2011-12-09T19:48:53Z")
> [@Enderw24](#): > > But that’s it. That as far as it goes. In reality, as there are 5 blue eyed people, every blue eyed person knows for certain there are at least 4 blue eyed people. And every non blue eyed person knows there are at least 5 blue eyed people on the island. > … > Here’s what everybody knows. There are blue eyed people on the island. Everyone knows this, with or without the guru’s interference. That’s not as far as it goes. That’s just your natural limit. You keep wanting to ‘return to the surface’ and your brain won’t let you dive deeper. That’s why you called it a rabbit hole. But it’s a mental block, not a logical one. Sure, everyone knows this. But they don’t know that everyone else knows that everyone else knows that everyone else knows that everyone else knows…that everyone else knows that. We’ve shown you multiple times in multiple ways how that works. Like I said, I’m willing to put you through a virtual classroom exercise so you can see it for yourself. You just have to agree to play. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [December 9, 2011, 7:56pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/118 "2011-12-09T19:56:18Z")
> [@Frylock](#): > > There is no twist of logic, no “everybody knows that everybody knows that everybody knows” where you end up with zero blue eyed people. There can’t be because if you find yourself falling down that rabbit hole you can defeat it by just looking at your damn neighbors and saying “oh yeah blue eyes.” Defeat _what_? The notion that there are blue-eyed people? No shit. No one thinks otherwise. And everyone knows that no one thinks otherwise. But again, that’s not the conjecture at all, and that’s what you guys keep forgetting. No islander, at any time, concludes that 0000000 is a possible pattern for the islanders. What they _do_ conclude, however, is that everyone’s errors can add up. Every person is ignorant of their own eye color, and that error carries backward through everyone else’s assessment of their thoughts. No one says “Well, the guy ahead of me thinks his eyes might be brown, so he could be right.” What they’re saying is that “The guy ahead of me thinks his eyes are brown, and _even though that’s wrong_, he still thinks it.” Again, at no time does anyone consider any of the brown-thinkers to be correct. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [December 9, 2011, 7:58pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/119 "2011-12-09T19:58:27Z")
> [@](#): > > Quoth **Enderw24** : > If we take a ten person island where five people have blue eyes. One of the blue eyed people looks around and sees 4 people with blue eyes. He can then imagine one of those 4 blue eyed people looking around and seeing 3 people with blue eyes (because he, himself, doesn’t realize he has blue eyes too). > > But that’s it. That as far as it goes. In reality, as there are 5 blue eyed people, every blue eyed person knows for certain there are at least 4 blue eyed people. And every non blue eyed person knows there are at least 5 blue eyed people on the island. > > There is no twist of logic, no “everybody knows that everybody knows that everybody knows” where you end up with zero blue eyed people. There can’t be because if you find yourself falling down that rabbit hole you can defeat it by just looking at your damn neighbors and saying “oh yeah blue eyes.” > > Here’s what everybody knows. There are blue eyed people on the island. Everyone knows this, with or without the guru’s interference. No, you can’t leave the rabbit hole by just looking at your neighbors, because while you can look at all of your neighbors, your neighbors can’t look at all of your neighbors. ---
**Author:** ![Enderw24](https://avatars.discourse-cdn.com/v4/letter/e/ba9def/32.png) [@Enderw24](https://boards.straightdope.com/u/Enderw24)\ **Post date:** [December 9, 2011, 8:01pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/120 "2011-12-09T20:01:01Z")
> [@Frylock](#): > > There is no reason to stop there. But there is. I mean, I get the logic of you thinking what someone else is thinking that someone else is thinking that someone else is thinking. This whole thing sounds like a Godel, Esher, Bach conversation, really. The point is that it can really only go down two levels here. There are five blue eyed people on the island. I look around and see four. I can imagine one of those four looking around and seeing three. But I can’t imagine one of those three looking around and seeing two. OK, no, that’s not right. I can imagine it happening but I know that, in reality, it never would happen. Because I know for certain that every one of those four people sees at least three other blue eyed people. Every one. So my logic twisting games of inception level thinking have no basis in reality. [Previous page](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531.md?page=5) [Next page](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531.md?page=7)