# 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:** 8
**Author:** ![matt\_mcl](https://avatars.discourse-cdn.com/v4/letter/m/7ab992/32.png) [@matt\_mcl](https://boards.straightdope.com/u/matt_mcl)\ **Post date:** [December 9, 2011, 9:33pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/141 "2011-12-09T21:33:36Z")
> [@Chronos](#): > > Let’s just back up and re-state the original reasoning. Anyone who has a problem with the original reasoning, tell me at what step it fails. > > A: If there were exactly one blue-eyed person, we can logically deduce that he would leave on Night 1. > B: We were able to logically deduce this, and all islanders are perfectly logical, so every islander is also able to deduce this, and so all islanders know A. > C: If there were exactly two blue-eyed people, then on the first night, each one might stay put, thinking that the other one might be the only one. So nobody would leave on Night 1. > D: But everyone knows that if there were only one blue-eyed person, he would leave on Night 1. Given that nobody left, everyone knows that there isn’t only one blue-eyed person. Each of the blues would then look around and see only one blue, and realize that they must be the other blue. So all blues would then leave on Night 2. > E: But this is a logical deduction, which all the islanders are capable of. So all islanders know that if there were only two blues, then they’d both leave on Night 2. > F: If there were exactly 3 blues, then on the 2nd night, they would all stay put, since they would each think that the other two blues might be the only blues. > G: But once night 2 passes with nobody leaving, each of the blues would know that there are not exactly 2 blues, because if there were, they know they would have left. So after that point, each of the blues knows there must be at least 3 blues, and they must be one of them. > H: So if there are 3 blues, then they will all leave on the 3rd night. > I: But this is a logical deduction, and so the logical islanders will deduce it. So all islanders know that if there are exactly 3 blues, then they will all leave on the 3rd night. > J: So if there were 4 blues, they’d all stay put on Night 3, because they might think that the other 3 blues are the only ones, and are going to leave > K: But after Night 3 passes without anyone leaving, all the blues would know that there are at least 4 blues, because if there were only 3, then they would have left. > L: So now all four of the blues must know that they themselves are blue-eyed. > M: So on the 4th night, if there were exactly 4 blues, all of them would leave. > N: But this is a logical deduction, and so all of the islanders, being logical, will know that if there were 4 blues, they’d all leave on the 4th night. > > I can keep going with this, but I don’t think it’d be productive to carry it any further; the pattern is the same. So, if there’s a flaw in the reasoning, which step is it in? I won’t risk calling it a logical flaw as such, but the “common-sense problem” (as I call it) is here: > [@](#): > > Given that nobody left, everyone knows that there isn’t only one blue-eyed person. Everyone knows that there isn’t only one blue-eyed person _anyway_, and likewise all the way on up – because they can all _see_ at least 99 of them, and knows that everyone else can see at least 98 of them. So it’s difficult (again, intuitively as opposed to logically) to see how any of this helps. (I’m not really backsliding – the explanations really are helpful! I’m just restating the difficult bit to swallow.) ---
**Author:** ![aldiboronti](https://avatars.discourse-cdn.com/v4/letter/a/9fc348/32.png) [@aldiboronti](https://boards.straightdope.com/u/aldiboronti)\ **Post date:** [December 9, 2011, 9:41pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/142 "2011-12-09T21:41:41Z")
I was having a major problem with this but I can’t fault **Chronos** ’ chain of reasoning and I think I get it now. It’s just so damned counter-intuitive, that the Guru gives them the solution to escaping the island by telling them something which they already know,ie there is at least one blue-eyed person on the island. Great puzzle! ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [December 9, 2011, 9:46pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/143 "2011-12-09T21:46:45Z")
It’s not the fact that the islanders learn “There is a blue-eyed islander” that does anything. If the Guru went and whispered this privately into each islanders’ ear, nothing would happen. It’s the fact that the Guru states it publicly which has an effect. Because the Guru states it publicly, everyone knows that the Guru stated it, and everyone knows that everyone knows that the Guru stated it, and so on. That brings about a change; it makes it so that everyone knows that everyone knows that … everyone knows that there is someone with blue eyes. This wasn’t the case before. It’s as a result of stating it _publicly_ that this becomes the case. The islanders do not gain information from learning “There is someone with blue eyes”. Rather, there is a shift brought about in what the islanders know about what other islanders know about what other islanders know about … [to the Nth degree], by this fact being stated publicly, even though everyone already knew it privately. ---
**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, 9:48pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/144 "2011-12-09T21:48:10Z")
> [@matt\_mcl](#): > > I won’t risk calling it a logical flaw as such, but the “common-sense problem” (as I call it) is here: > > Everyone knows that there isn’t only one blue-eyed person _anyway_, and likewise all the way on up – because they can all _see_ at least 99 of them, and knows that everyone else can see at least 98 of them. So it’s difficult (again, intuitively as opposed to logically) to see how any of this helps. At the step you refer to, the argument is in the middle of a supposition–in other words, the argument is discussing a scenario in which there are only two people with blue eyes. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [December 9, 2011, 9:52pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/145 "2011-12-09T21:52:44Z")
Out of curiosity, is there anyone who has difficulty understanding what happens in the islands-and-bridges problem I gave in post [#27](http://boards.straightdope.com/sdmb/showpost.php?p=14545715&postcount=27)? Regardless of whether you think it is relevant to this problem or not, is there anyone who has difficulty understanding that problem on its own terms? ---
**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, 10:01pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/146 "2011-12-09T22:01:41Z")
> [@Galileo](#): > > This logic puzzle seems like the converse of the Monty Hall problem, where the solution is more obvious if you increase the number of doors. In the “Blue eyes” puzzle, the solution is more obvious if you decrease the number of islanders, and even more so if you decrease the number of islanders who have blue eyes. Yes, it’s quite similar to earlier old chestnuts such as the various Three Hats puzzles, some of which can be scaled up to mind-boggling complexity by adding a few hats. For some reason, this particular chestnut is always presented with a large number of islanders. If it were three or five islanders in the standard version, I think people would get it more readily. I’m not saying that the five-person version is trivial, because as we have seen it isn’t, but I still think that making it a large number like 100 makes the logical deduction seem unfathomable. ---
**Author:** ![hogarth](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hogarth/32/1773_2.png) [@hogarth](https://boards.straightdope.com/u/hogarth)\ **Post date:** [December 9, 2011, 10:04pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/147 "2011-12-09T22:04:10Z")
> [@Chronos](#): > > I can keep going with this, but I don’t think it’d be productive to carry it any further; the pattern is the same. So, if there’s a flaw in the reasoning, which step is it in? Once you get to four blue-eyed people (call them A, B, C and D), every possible subgroup of people would mutually agree that A sees at least one blue-eyed person (e.g. B & C and anyone else know that he sees D, B & D and anyone else know that he sees C, C & D and anyone else know that he sees B), that B sees at least one blue-eyed person, that C sees at least one blue-eyed person, that D sees at least one blue-eyed person, etc. Therefore, everyone knows that no one will leave on Night 1. There’s no further inference that can be made on Night 2, because everyone agrees that everyone else knows that nothing would happen on Night 1 (or every night, for that matter – elucidating the fact that there’s at least one blue-eyed person adds nothing, since everyone knows that everyone else knows that that’s true). ---
**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, 10:14pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/148 "2011-12-09T22:14:01Z")
> [@hogarth](#): > > since everyone knows that everyone else knows that that’s true). Yes, but everyone doesn’t know that everyone knows that everyone knows that everyone knows. And that’s the difference. We’ve already explained this. Saying “Everyone knows that everyone knows” isn’t far enough. ---
**Author:** ![sachertorte](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/sachertorte/32/3727_2.png) [@sachertorte](https://boards.straightdope.com/u/sachertorte)\ **Post date:** [December 9, 2011, 10:15pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/149 "2011-12-09T22:15:59Z")
Does it help if you think about the Guru as simply a coordination mechanism? In other words, The Guru sets a clock by which all the islanders operate? It isn’t strictly true, but the problem and solution do have a bit of a temporal aspect to it and ‘coordination’ or synchronization of some sort is necessary. Also, each day affords a singular opportunity for the islanders to share information. They aren’t allowed to speak to each other, but they communicate information through actions (or the lack of action – e.g., not leaving the island). On Day N, all islanders NOT leaving the island are communicating that “I see more than N blue eyes”. ---
**Author:** ![aldiboronti](https://avatars.discourse-cdn.com/v4/letter/a/9fc348/32.png) [@aldiboronti](https://boards.straightdope.com/u/aldiboronti)\ **Post date:** [December 9, 2011, 10:17pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/150 "2011-12-09T22:17:12Z")
The brown-eyed logicians, hearing the Guru’s announcement, will no doubt instantly realize the logical opportunity for escape that it offers those with blue eyes. They will surely also go on to reason that if the Guru had substituted brown for blue the escape route would have been theirs. It also might occur to them that if they act as if the Guru had said brown then they can use the same chain of reasoning as the blue eyes to leave the island. Would that work for them? ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [December 9, 2011, 10:24pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/151 "2011-12-09T22:24:54Z")
> [@sachertorte](#): > > Does it help if you think about the Guru as simply a coordination mechanism? In other words, The Guru sets a clock by which all the islanders operate? > > It isn’t strictly true, but the problem and solution do have a bit of a temporal aspect to it and ‘coordination’ or synchronization of some sort is necessary. > Also, each day affords a singular opportunity for the islanders to share information. They aren’t allowed to speak to each other, but they communicate information through actions (or the lack of action – e.g., not leaving the island). On Day N, all islanders NOT leaving the island are communicating that “I see more than N blue eyes”. I think this is the heart of it. The Guru is violating the communication rule. There has to be something to create the common knowledge, which in the description of the problem was insufficient to determine that anyone knew anything at any particular time. I think the point of the Guru’s statement is that the simple communication was the minimum violation of the non-communication rule necessary to get the process going. Kind of like hitting enter on a command line that everybody knows about, but no one else can start executing. ---
**Author:** ![Galileo](https://avatars.discourse-cdn.com/v4/letter/g/85e7bf/32.png) [@Galileo](https://boards.straightdope.com/u/Galileo)\ **Post date:** [December 9, 2011, 10:55pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/152 "2011-12-09T22:55:20Z")
> [@aldiboronti](#): > > The brown-eyed logicians, hearing the Guru’s announcement, will no doubt instantly realize the logical opportunity for escape that it offers those with blue eyes. They will surely also go on to reason that if the Guru had substituted brown for blue the escape route would have been theirs. It also might occur to them that if they act as if the Guru had said brown then they can use the same chain of reasoning as the blue eyes to leave the island. Would that work for them? How does the brown-eyed person know that his eyes are not blue, or green, or purple? Remember, they can’t communicate with each other. ---
**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, 11:04pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/153 "2011-12-09T23:04:27Z")
> [@Galileo](#): > > How does the brown-eyed person know that his eyes are not blue, or green, or purple? Remember, they can’t communicate with each other. Indeed. The brown-eyed people reason in exactly the same way as the blue-eyed people, since none of them knows the colour of their own eyes. But the people who do not actually have blue eyes have an advantage - they can see one more blue-eyed person than the blue-eyed people can. So, the day before the brown-eyed people would have concluded that they must have blue eyes, all the blue-eyed people leave. ---
**Author:** ![aldiboronti](https://avatars.discourse-cdn.com/v4/letter/a/9fc348/32.png) [@aldiboronti](https://boards.straightdope.com/u/aldiboronti)\ **Post date:** [December 9, 2011, 11:05pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/154 "2011-12-09T23:05:16Z")
> [@Galileo](#): > > How does the brown-eyed person know that his eyes are not blue, or green, or purple? Remember, they can’t communicate with each other. Well, that put the kibosh on that! Good point, I’d forgotten that other colors could figure. Are the browns then stuck on the island when all the blues have left? No logical out for them? ---
**Author:** ![Danger\_Man](https://avatars.discourse-cdn.com/v4/letter/d/6de8d8/32.png) [@Danger\_Man](https://boards.straightdope.com/u/Danger_Man)\ **Post date:** [December 9, 2011, 11:10pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/155 "2011-12-09T23:10:28Z")
I had lots of trouble with this, but suddenly I had a revelation and it all makes sense intuitively for me now. I know for sure there are 99 blue-eyed people. I know that person #2 knows for sure there are 98 blue-eyed people. I know that person #2 knows for sure that person #3 knows for sure there are 97 blue-eyed people. I know that person #2 knows for sure that person #3 knows for sure that person #4 knows for sure that there are 96 blue-eyed people. All the way to the bottom… I know that person #2 knows for sure that… …person #100 knows for sure there is at least one blue-eyed person. But this line of reasoning does not work at the bottom before the guru makes his announcement. As someone said before, it’s all in the higher level “he knows that he knows that he knows”, that won’t be complete until the guru makes his announcement. ---
**Author:** ![Galileo](https://avatars.discourse-cdn.com/v4/letter/g/85e7bf/32.png) [@Galileo](https://boards.straightdope.com/u/Galileo)\ **Post date:** [December 9, 2011, 11:16pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/156 "2011-12-09T23:16:32Z")
**Danger Man** , imagine that in the puzzle there are only 4 people. Two have blue eyes, and the other two have non-blue eyes. Then go through the logic. ---
**Author:** ![Danger\_Man](https://avatars.discourse-cdn.com/v4/letter/d/6de8d8/32.png) [@Danger\_Man](https://boards.straightdope.com/u/Danger_Man)\ **Post date:** [December 9, 2011, 11:22pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/157 "2011-12-09T23:22:50Z")
> [@Galileo](#): > > **Danger Man** , imagine that in the puzzle there are only 4 people. Two have blue eyes, and the other two have non-blue eyes. > > Then go through the logic. If I am the blue-eyed person: I know that the other blue-eyed person knows there is at least one blue-eyed person. The next day he knows that there were two and leaves. I know too and leave. If I am not the blue-eyed person. I can for certain know that one of the blue-eyed persons can know for certain that the other can know for certain there is at least one blue-eyed person. The next day I know for certain that the first guy knows for certain that the second guy knows for certain there are two blue-eyed persons and he leaves. As soon as the guru makes his announcement, my chain of reasoning starts working, and I can wait for my day and then leave. If everyone leaves before the day I was planning to leave, then I know I had brown eyes. But the guru completes there very last parts of everyone’s “I know that he knows that he knows…” reasoning. ---
**Author:** ![Danger\_Man](https://avatars.discourse-cdn.com/v4/letter/d/6de8d8/32.png) [@Danger\_Man](https://boards.straightdope.com/u/Danger_Man)\ **Post date:** [December 9, 2011, 11:38pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/158 "2011-12-09T23:38:51Z")
So when it comes to the 100th day and you decide to leave, it depends on you trusting the other 99 guys having made a logic deduction. In which all those 99 trust the 98 others having made the deduction. Those 98 trust the 97 having made the deduction… … those trust the 2 people having made the deduction about the last guy WHO IS TOTALLY DEPENT ON THE GURU TO MAKE HIS DEDUCTION. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [December 9, 2011, 11:45pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/159 "2011-12-09T23:45:47Z")
> [@Chronos](#): > > C: If there were exactly two blue-eyed people, then on the first night, each one might stay put, thinking that the other one might be the only one. There’s no “might” about not leaving - each is required to stay put, because neither is sure of his eye color. But the fact that no one departs the first night makes them both sure of their eyes are blue - so they both leave the second night. (BTW, the rules should probably be more explicit that an islander who learns his eye color is required to leave immediately.) ---
**Author:** ![Mosier](https://avatars.discourse-cdn.com/v4/letter/m/7ea924/32.png) [@Mosier](https://boards.straightdope.com/u/Mosier)\ **Post date:** [December 9, 2011, 11:46pm UTC](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531/160 "2011-12-09T23:46:31Z")
> [@Galileo](#): > > **Danger Man** , imagine that in the puzzle there are only 4 people. Two have blue eyes, and the other two have non-blue eyes. > > Then go through the logic. Each blue eyed person must consider the possibility that there is only 1 blue eyed person, who can not see anyone with blue eyes. On the first day, when the other blue eyed person does not leave, he must assume that the other blue eyed person sees at least 1 blue eyed person, making 2 total. Since he himself sees only 1 blue eyed person, he must conclude that he is the other blue eyed person. They both leave on day 2. [Previous page](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531.md?page=7) [Next page](https://boards.straightdope.com/t/blue-eyes-logic-puzzle/605531.md?page=9)