explain xkcd - User contributions [en]

Talk:Blue Eyes

2025-12-27T09:59:43Z

2001:8003:DC65:7E01:4460:4138:3D1A:2B74:

Is it really incomplete on the grounds that Joel hasn't be identified? Explanations of comics 57-59 leave no more explanation of "Scott" than that he appears to be Randall's friend. The fact that we don't have a last name for him doesn't make either [[Scott]] or those comic explanations incomplete. Similarly, not have a full identifier for "Joel" in this one doesn't, in my opinion, warrant an incomplete tag. I'm removing the tag. If anyone object, revert it. [[User:Djbrasier|Djbrasier]] ([[User talk:Djbrasier|talk]]) 19:22, 22 May 2015 (UTC)

The proof for this puzzle is incomplete, if not wrong. The theorem is too weak, it should be: "Theorem: N blue eyed people with Nth order knowledge of all N people being logicians, N people having blue eyes, and any blue eyed person will leave as soon as possible after deducing they have blue eyes, will be able to leave on the Nth day." This may seem pedantic, but it really gets to the heart of the problem, which is trying to illustrate the use of orders of knowledge. In the theorem as stated, just N blue eyed people will leave on the Nth day, the proof for the inductive steps does not hold. You need to further assume that the person is able to deduce the hypothesis (which should be proven). In other words, you say X-1 people would leave on the (X-1)th day by hypothesis, so the Xth person knows he can leave on the Xth day. But you did not prove that the Xth person can actually deduce this, namely that he has all the information necessary to do so. In the correctly stated hypothesis, you then need to show that N + 1 people with (N+1)th order knowledge of all those things can deduce that the N people would leave if it was just them, and further that N+1 people have (N+1)th order knowledge of all these things. This is very important, and holds true (Since N+1th order knowledge is equivalent to knowing the N people have the Nth order knowledge necessary to fulfill the hypothesis, and by symmetry if the N logicians can figure it out the (N+1)th can too. Also, they have (N+1)th order knowledge of people leaving as soon as they can and everyone being a logician since in the proper statement of the puzzle it should be noted this is common knowledge, and the guru makes the knowledge of someone having blue eyes common knowledge.). Then you have a full proof, since you have now included that they can actually deduce the inductive step. Again, this may seem pedantic, but is really necessary both to be correct and as it illustrates the key of the puzzle, namely the guru gives 100th order knowledge of someone having blue eyes (this is the main problem people have, realizing the concrete piece of information the guru gives). [[User:Jlangy|Jlangy]] ([[User talk:Jlangy|talk]]) 00:29, 9 July 2015

I think that the use of the word 'Guru' here confuses Zen and Indian mysticism. The puzzle depends on intentionality, a necessary condition which is not made clear by the word 'Guru'

What I don't follow here is that there's no clarification that the Guru is talking about someone different each time. Just because she says "I see someone with blue eyes" N times doesn't mean that there are N people with blue eyes; she could be talking about the same person every time, or each of two people half the time, etc. Can anyone clarify this?
Thanks - [[Special:Contributions/108.162.218.47|108.162.218.47]] 13:20, 28 October 2015 (UTC)

^^^ The Guru speaks only once on the first day and then is silent the rest of the time. --[[Special:Contributions/172.69.22.125|172.69.22.125]] 00:18, 10 December 2023 (UTC)

(EDIT: Observe the process of comprehension in action...or don't? I've been thinking about my own brain, with itself, long enough for one day, I'm tired.)
So, maybe I am indeed just "dumb", as the wiki insists. Clearly, I do not have a perfect understanding of formal logic. But frankly, my read of this puzzle is that "formal logic" just enables you to jump to ridiculous conclusions.
Let's theorize a simpler version of this puzzle. There are now only two people besides the Guru on the island, both with blue eyes. We'll call them Bill and Ted (totally bogus, I know). No matter how logical Bill and Ted might be, when Bill hears the Guru say "I see a person with blue eyes" to himself and Ted, and Bill has seen Ted's blue eyes himself, why would Bill assume anything about his own eye color? It would seem to Bill that Guru was just talking about Ted's eyes, and Ted would believe the reverse. Even knowing* that Ted would leave that night if Ted deduced he had blue eyes too, I still don't see why Bill would jump to the conclusion that the Guru was talking about him - he remains in the dark, as does Ted, and neither of them can be any more certain of anything than they previously were. Adding 98 more blue-eyed people, let alone doubling the island's population with irrelevant brown-eyers, hardly reduces the confusion.
* This was the point at which I began to think I had understood it, but then I became unsure again. Like I said in the "edit", my brain is tired.
--So, that settles it, I do not understand how the puzzle can be true, and I'm not convinced that it actually is. Knowing Randall is, in general, smarter than me...I still do not have the ability to completely accept that he's always right, or that I'm always wrong to ignorantly question his rightness. I have long maintained that certain well-respected "systems of knowledge", of which formal logic is a textbook example, have been respected too well for too long for not-good-enough reasons. To me, they seem to be founded on an assumption which is itself founded on nothing. I'm not trying to insult Randall or anyone else, I'm just utterly failing to comprehend. I will appreciate if anyone else attempts to educate me on the subject, but I may prove an intractable student, since I am unable to extend much faith or trust (or even, on a day where my mood is worse than today, the moderate degree of politeness as I've already managed) to a teacher. [[Special:Contributions/173.245.54.52|173.245.54.52]] 19:18, 30 October 2015 (UTC)

In your simplified version of the puzzle, Bill sees Ted has blue eyes. 
Here's Bill reasoning: 
- Either my eyes are blue or not. 
- If my eyes are not blue, then Ted knows that his eyes are blue, because the Guru said at least one of us has blue eyes, and he'll leave the island tonight. 
- Let's wait. If Ted doesn't leave tonight, that means he doesn't know his eyes are blue, and therefore my hypothesis is false. 
When Bill sees Ted doesn't leave that night, he can deduce that he has blue eyes. 
Ted can do the same reasoning. 
After that first night, both will know they both have blue eyes. 
--[[Special:Contributions/108.162.228.5|108.162.228.5]] 14:09, 14 December 2015 (UTC)

;Superrationality

The solution relies on the fact that "at least 1 blue" is new information which triggers a cascade.

Wouldn't the entire population of the island be able to conclude that everyone else on the island knows there is at least 1 blue eyed individual already?

For example, every person on the island will see at least 99 blues and 99 browns. From this, they can assume that everyone else on the island can see at least 98 blues and 98 browns. Of course, the actual numbers will differ, but 98 is the lower limit for all perspectives.

A blue will see 99 blues and 100 browns, so he will assume that all other blues can see at least 98 and all browns can see at least 99 blues. Similar logic for a brown or any observer.

[[User:Flewk|Flewk]] ([[User talk:Flewk|talk]]) 09:26, 26 December 2015 (UTC)

:No, not quite. Although Blue #100 can see 99 other blues, he doesn't know that he's a blue-eye. So although he knows that each of those 99 can see 98 other blue-eyes, *none of them know their own eye colour either*. So in Blue #100's perspective, each of those 99 would have to assume that they *may* only be able to see 98, as you said - however, therefore, those 98 (from the perspective of any of the 99, itself being imagined from the perspective of #100) for the same reasons can *themselves* only assume that each of *those* 98 can see at most 97... etc. There is no lower limit until the Guru provides one. [[Special:Contributions/165.225.213.87|165.225.213.87]] 16:27, 2 September 2025 (UTC)

The solution here is different to Randall's solution, and I think is actually incorrect for two reasons that add confusion and prevented me from understanding the solution until I'd thought about Randall's solution and realised these are actually different.
* It seems to falsely presume that the Guru is speaking to them each day, when this is explicitly not the case in the puzzle.
* I also believe it is incorrect to state that the brown-eyed people can be disregarded. The solution is actually dependent on a *combination* of hypothesis testing and on theory of mind; not just one or the other. It matters that everyone is also thinking about what the brown-eyed people around them must be thinking, otherwise you can't explain why mistakes will not happen with brown-eyed people getting on the ferry when they're not supposed to, and screwing up everybody else's logic.
- If you're on the island and you have blue eyes, there are two hypotheses: either there are 99 people with blue eyes or 100. If there are 99, then everyone one of those 99 people is thinking "either there are 98 people with blue eyes, or there are 99" (and therefore you do not have blue eyes). Blue-eyed people also know that if there are 99 of them, then the brown-eyed people are thinking, "Either there are 99 blue eyed people, or 100." If there are 100, then the brown eyed people are thinking, "Either there are 100, or 101". To summarise, blue eyed people are deciding between 99 or 100, and presuming that other blue eyed people are either suspecting there could be 98/99, or 99/100, while presuming that brown-eyed people are either suspecting there are 100/101, or 99/100.
- If there are 99, then blue-eyes are thinking 98/99, and brown-eyes are thinking 99/100. Blue eyes will plan to leave if the 98th day passes and nobody has left, brown-eyes will plan to leave if the 99th day passes and nobody has left.
- If there are 100, then blue-eyes are thinking 99/100, and brown-eyes are thinking 100/101. Blue eyes will plan to leave if the 99th day passes and nobody has left, brown-eyes will plan to leave if the 100th day passes and nobody has left.
- So you know that if you have brown eyes, you'll watch all the blue-eyes leave on the 99th day. And you know that if you have blue eyes, you'll watch all the brown-eyed people hold back in case their day is the 101st. If you're allowed to leave, there will be no situation where brown-eyed people mistakenly leave on the 100th day, thus confusing things. If you're not allowed to leave, there'll be no reason for you to mistakenly make an attempt to leave on the 99th day.
- Thinking about this fact - what the brown-eyed people are thinking - also reveals why the Guru's comment matters, and adds information, even though it should seem to most people as if no information is being added (because they can all already see that blue-eyed people exist). I think this is a key part of why the problem is so tricky. [[Special:Contributions/108.162.249.155|108.162.249.155]] 07:42, 10 March 2016 (UTC)

The new information the guru gives is nothing more than a common marker (the day of the announcement) to use as a starting point for counting days. Before the announcement, being unable to communicate with each other, they were unable to coordinate a means of figuring out their own eye color.
[[Special:Contributions/172.68.59.105|172.68.59.105]] 21:52, 22 September 2016 (UTC)

That's wrong. In that case the browned eyed people would do the same, but they can't.

What I'd like to know: If there were 100 blue-eyes, 200 brown-eyes, 300 grey-eyes and 400 red-eyes, and the Guru says "I don't see anyone with a unique eye color", would that permit everyone to leave (except the Guru herself) using the same logic? Meaning the blue-eyes leave again on day 100, the brown-eyes on 200, the grey-eyes on 300, and the red-eyes on 301.

I think it would actually be days 99, 199, 299, and 300, because the 'what if there were only two blue-eyes' case would be solved on day 1 - i.e. both would see only one blue-eye and deduce that they are also a blue-eye, and both would leave - so everything gets moved up by one day.[[Special:Contributions/141.101.76.16|141.101.76.16]] 13:52, 4 January 2018 (UTC)

This was bugging me today - specifically the guru doesn't seem to actually give any information, because with at least 3 blue-eyed people, everyone on the island knows that the guru sees people with blue eyes, and also everyone knows that everyone knows the guru sees people with blue eyes. So for a while I thought the brown-eyed people must have as much information as the blue-eyed people, and either they both could leave or neither could leave.

:Consider this, if there were only 2 people with blue eyes, everyone would know that she sees someone with blue eyes beforehand, but everyone wouldn't know that everyone knows that, as the 2 people with blue eyes would not know if the other person with blue eyes can see anyone with blue eyes, so the 2 people with blue eyes would deduce they have blue eyes when the other person doesn't leave the first day, as themselves having blue eyes would be the only explanation for that person not leaving the first day. The dispute here is if you can extend that chain of reasoning past there being only two people. After all, with 3 people, as you said, everyone knows that everyone knows that she can see someone with blue eyes already, but when you consider the people who have blue eyes, everyone doesn't know that everyone knows that everyone knows that, even though each individual would personally know that everyone knows that everyone knows that, as the people with blue eyes know that if they don't have blue eyes, then the 2 people with blue eyes they see would only see one person with blue eyes and know that the Guru can see someone with blue eyes, but wouldn't know that the other person with blue eyes knows that. But would everyone follow such a chain of logic and make assumptions based one people not leaving based on days with significantly lower numbers passing that they personally know that no one would expect a possibility of anyone leaving on? This whole question is hinged on people following perfect logic that is based on other people following the same perfect logic that would predict this. If perfect logic necessitated people leaving in such a manner, then everyone would know the rest of the people would follow this rule and the solution would hold, but if it didn't, then it would be just as consistent if no one left. This is basically circular reasoning about using logic to predict the actions of people who are acting according to the same reasoning as yourself. Both this answer being correct and it wouldn't wouldn't violate pure logic based on a system of reasonable seeming logical principles and the terms of the question.--[[Special:Contributions/172.70.127.94|172.70.127.94]] 04:35, 12 October 2022 (UTC)

After a bunch of reading and testing possibilities I think I've actually figured it out now and why only blue-eyed people leave, but I haven't seen an actual good explanation for it yet, so here's my explanation: The information the guru gives is that, NO MATTER HOW MANY BLUE-EYED PEOPLE THERE ARE, they can figure out they have blue eyes. It is important that it is possible for blue-eyed people to be able to solve it for EVERY number, even if everyone knows there is more people than that number (basically because, everyone doesn't know that everyone knows there is more than that number and there's a gap in their logic without knowing that). If there is any number for which blue-eyed people cannot figure it out, then any solution (namely, what I thought before testing possibilities) would require that there is a number N of blue-eyed people that cannot leave, but a number N+1 of blue-eyed people that can leave. This is self-contradicting though. If N blue-eyed people can't figure it out, than N+1 people (regardless of eye color) can't get meaningful information from the action of those N blue-eyed people. And since they can't get meaningful information from N people's actions, N+1 people can't tell if there are N people and that individual is not blue eyed, or N+1 people and they are blue-eyed. It would be logically incorrect for them to assume an eye color at that point, which means they don't know if they can leave, and then N+2 people are similarly unable to get meaningful information from N+1 people's actions, and so on. Because a single blue-eyed person cannot figure it out, more blue-eyed people (regardless of number) cannot make any assumptions without additional information. Then the guru effectively states a single blue-eyed person could leave immediately (which means 2 could leave the next night confidently, and thus 3 the next, and so on in an UNBROKEN chain). Kejardon - [[Special:Contributions/162.158.214.82|162.158.214.82]] 11:15, 9 January 2020 (UTC) (I doublechecked and edited/corrected my post kind of at the same time, so your reply doesn't make sense anymore, sorry Lupo. Feel free to delete these two sentences if you change/delete your reply)
:You bring up 2 points. First about the common marker. This is true, but it contains more information than "start counting from today," because every blueeyed person has 2 scenarios: With 99 blue eyed people and with 100. The numbers are not important, and it would also be needed with 1,2,3,etc. blue eyed people. The point about the brown eyed people: The brown eyed people have no way to conclude (remember, they are "perfect logicans" and their task is to figure out, not to make an estimated guess) that their own eyes are in fact brown, and not red, or green just like the Gurus. If the guru just said: "Start counting", then noone woul leave at any given night. --[[User:Lupo|Lupo]] ([[User talk:Lupo|talk]]) 11:05, 9 January 2020 (UTC)

I figured this out in less than a minute... there were so many warnings about the solution being convoluted that I thought I couldn't possibly have it right. It's not really that confusing and I've seen waaaaaaay harder logic problems.

This image is listed under "My Hobby" for some reason, despite not even being a comic, let alone a "My Hobby" comic. [[Special:Contributions/162.158.111.151|162.158.111.151]] 00:58, 9 September 2019 (UTC)How sign edit

It's been years and I stumbled back across this post, and I think I finally understand what has been bothering me.
The examples always go up to 3 blue-eyed people, and then we just assume that it follows a the pattern, but a pattern has to be confirmed to all states to be a proof. This is why Fermat's last theorem remained unproven for years even though we had solved the N =1,2,3 cases. To expand on the biggest criticism, what new information does the guru give? 
-Someone learns something : Only true from the perspective of someone who sees no-one with blue eyes. 
-Someone could learn something : Only true from the perspective of someone who sees exactly one person with blue eyes. 
-No-One can learn anything : True from the perspective from someone who sees multiple people with blue eyes. 
This is where the problem occurs, because this only resolves with 3 blue-eyed people otherwise we don't learn anything, but how we acknowledge that we don't learn anything matters. 
-Everybody must know that no-one can learn : True from the perspective of people who see people who must see multiple people with blue eyes. 
-Everyone must be aware of the previous fact : I guess it does go on... 
In the end, it's all about the meta-data. 
I know, you know, they just don't have any proof... 
Crow --[[Special:Contributions/108.162.245.65|108.162.245.65]] 00:12, 30 September 2021 (UTC)

Apostle's Solution: The first person to look into the water and see their own reflection. For a logic puzzle, people seem to forget about the situations actual logic. {{unsigned ip|162.158.74.213}}
:What part of: "There are no mirrors or reflecting surfaces" do you not understand? --[[User:Lupo|Lupo]] ([[User talk:Lupo|talk]]) 09:54, 18 May 2020 (UTC)
:What can you even do to an island to remove all reflecting surfaces? Absolutely everybody would leave the the night after the weather is good enough to see their reflection in the sea.
::Then the weather is never clear enough to see their reflections. It's equally valid to question why the ferryman would only free people who knew their eye color, or whether some people would want to stay on the island instead of leaving...and far ''more'' valid to point out that perfectly logical beings like the problem describes ''don't exist''. If you're not willing to suspend your disbelief enough to accept the premise of a logic puzzle, that's fine, but it's no more ridiculous than suspending your disbelief enough to accept that hobbits and wizards exist in Middle Earth. [[User:GreatWyrmGold|GreatWyrmGold]] ([[User talk:GreatWyrmGold|talk]]) 03:29, 17 July 2021 (UTC)

If there are N islanders with blue eyes, every person sees at least N-1 blue eyed persons. Person A with blue eyes seeing N-1 blue eyed persons has to consider the possibility that there might be a person B who only sees N-2 blue eyed people and acts accordingly. However, everyone in that scenario will agree that it is absolutely impossible for anyone to see N-3 or less blue eyed people and that case cannot possibly be considered by anyone. Therefore the induction step is invalid and nobody will ever leave the island for N>3. [[User:Boopers|Boopers]] ([[User talk:Boopers|talk]]) 18:07, 15 October 2021 (UTC)

:So, Boopers, you would buy Situations 0 through 2 above (under "Question 2" of "Randall's follow-up questions"), and you would even buy Situation 3, namely "If I see 3 blue-eyed people, they might each be in situation 2 watching to see whether anyone in situation 1 leaves the second night (I know nobody will leave that night, but they wouldn't know this), in which case they would leave together on night 3; or else they might be in situation 3 just like me, in which case we'll all leave together on night 4", but having bought that, you won't buy Situation 4? [[Special:Contributions/165.225.213.98|165.225.213.98]] 14:03, 27 August 2025 (UTC)

If a person B only sees N-2 blue eyed people, then surely they have to consider the possibility of a person C who only sees N-3 people. Sure, person B doesn't actually exist in this scenario, but you'd have to consider that person A would have to consider what a hypothetical person B would have to consider. It's turtles all the way down. [[Special:Contributions/172.70.110.171|172.70.110.171]] 00:17, 19 October 2021 (UTC)

No. It doesn't turtle down and nobody inside the scenario is free to consider any hypothetical persons. Instead, everything that one of the persons may consider about another person on the island has to be consistent with his observations of the persons who are actually there. And that breaks down at person C in my example. Yes, person B from the example does not exist, but the important thing that was being missed is that person C can be proven by everyone involved to be impossible which is not the case for B.
Lets just consider the scenario with 4 blue eyed persons, the smallest number where the argument breaks down. Now lets choose two completely arbitrary persons A and B (And I really mean completely arbitrary - doesn't even require for A and B to be different persons). It can be shown that A knows that B sees at least 2 blue eyed persons. Now since A and B were chosen arbitrarily, that means that every person on the island knows that every person on the island sees at least 2 blue eyed persons. This observation is of great importance, because it contradicts an invalid assumption implicitly made in the induction step. The induction step relies on the incorrect assumption that at the end of day N knowledge is gained that the number of blue eyed persons is larger than N. That assumption works out well for 2 or 3 blue eyed persons, but completely breaks down at 4 blue eyed persons right at day 1, since we have shown that everyone is already aware that there is more than 1 blue eyed person before the end of day 1.
This invalid assumption is extremely well hidden in the presented proofs. In the "Intuitive Proof" section, it is hidden inside the wrong assertion that the simplified problem supposedly is equivalent to the original one when it is absolutely not. And in the "Formal Proof" section, the islanders are just being assigned a reasoning without there being a proof to why they would reason this way. Of course following this flawed reasoning will lead to the desired result, but if they were the perfectly logical thinking people like the problem described, they wouldn't just miss the fact that there are cases where no knowledge can be gained on day 1 and therefore on any of the following days, and they would incorporate that into their reasoning. [[User:Boopers|Boopers]] ([[User talk:Boopers|talk]]) 18:31, 13 November 2021 (UTC)

Boopers, let me see if I can convince you it works for n=4 by making it a story. There are 4 blue-eyed people on the island; Arnold, Boopers (you), Carl, and Denise. You don't know your eye color, but have always thought of yourself as a brown-eyed person, and if you had to guess, you would guess you have brown eyes; after all, almost everyone does. When you hear the guru speak, you think that since there are only 3 blue-eyed people, A, C, and D. They are dear friends of yours, and you are very sad that after 3 days, you will never see them again. So you have a busy 3 days coming up, since there are things you've put off doing that you must do quickly, before A, C, and D leave your life forever.

A, while a good friend, has a habit of borrowing things and not returning them, and you don't want your stuff going with A on the boat. So you spend day 1 getting A to return your lawnmower, snowblower, and the dozen books he has borrowed over the years and not returned. You get all your stuff back. A good day 1.

While C is a friend, you haven't always treated him well. So day 2 you spend with C, doing whatever you can to ensure he has a good day on his next-to-last day on the island, and apologize to him for the ways you have wronged him in the past. He accepts your apology, and you are pleased that you haven't left things unresolved with your friend who will leave the island forever in 2 days. A great day 2.

On day 3; you confess to Denise (who is married to Arnold) your undying love for her. You've never told her this, out of respect for her marriage, but you've always suspected that your feelings were returned. Now that you will never see Denise or her husband again after noon tomorrow, so there will be no repercussions. To your delight, you discover that your feelings are indeed returned, and you and Denise spend what you think will be your last night together making love until the early morning hours. A perfect day 3. And since A and D are leaving on the boat the next day, and you, with your brown eyes, are staying on the island, probably forever, there will be no repercussions.

The next day, when you know A, C, and D will be leaving at noon, you rise early, and just before noon, you head to the pier, wanting to see Denise once more before she leaves on the noon boat. A, C, and D are of course there too. But to your great surprise, when the "all aboard" is sounding, A, C, and D do not move to get on the boat.

How can this be? What has happened? Again and again you go over the logic that says "if there are only 3 blue-eyed people on the island, they will all leave on day 3" and find it to be ironclad. And you know that your friends A, C, and D are perfect logicians, and can reason the same way you do, and you certainly know that they would never violate the rule that if you know your eye color, you must leave. And yet they remain! It seems impossible! How can this be? Unless...

Does a possibility occur to you to explain their inexplicably remaining on the island? You are certain of your logic that says "*If* there are exactly 3 blue-eyed people on the island, they will leave on day 3". So it must be that there are more than 3 blue-eyed people on the island. But you re-check the eye color of everyone else you can see, and they are all brown. Do you realize what must have happened?

Do you now know your eyes are blue? Do you get on the boat on day 4? [[User:Yp17|Yp17]] ([[User talk:Yp17|talk]]) 19:14, 9 February 2022 (UTC)

I think that helps highlight why the induction as presented doesn't work. Indeed, I believe everyone leaves on the 4th night, and that the Guru provides no information at all -- only a random token which could not have existed before since communication was impossible.

Once the Guru speaks, the solution becomes possible for only the case of blue-eyed people, because only then can every person on the island be sure they are counting the same color. But the content of what she says is meaningless, as everyone already knows what she says. She could have simply said, "Blue eyes", and the same result could be accomplished (I in fact nominate that for a harder form of the puzzle).

Following this, the expected reasoning happens, but there is no reason for it to stop at the time presented. Waiting one day to see if there is one blue-eyed person that will leave is meaningless and completely irrational -- every person on the island can see enough blue-eyed people to know, with certainty, that every person knows there is more than one blue-eyed person. Therefore, waiting that first night is wasted, since the outcome is known with absolute certainty. Waiting must logically begin only on the first day where the outcome is not certain.

To find that day, the relevant metric is the lowest number of blue-eyed people that can be known to exist to every person on the island. That number is 97 -- consider a person A, blue-eyed, calculating the knowledge of another person B, also blue-eyed. (If either are brown-eyed they can mutually estimate more blue-eyed people, so this is the relevant case.) Since A is uncertain of their eye color, they must for the worst case assume it to be brown; therefore, they estimate 99 blue-eyed people, of which B is one, whom therefore sees 98 blue-eyed people in this scenario. B cannot be sure of their own eye color, so they will see 98 blue-eyed people, and when modelling a further person (C, say) following the same logic could conclude they themselves are not blue-eyed and therefore C will only be able to guarantee 97 blue-eyed people (A's model is, A is brown, and B's model is B is brown, so C's model excludes A and B and is uncertain about C.)

Importantly, this does not induct! There is no rational way in the given 100/100/1 distribution for someone to not be certain of 97 people, even though there is a temptation to chase a clear recursive pattern. The reasoning behind it does not hold -- A does not need to worry about B->C->D's model, because no one has that much uncertainty -- it is incorrect to model D, because you can observe everyone's computation at most two steps removed from directly. When establishing a pairwise estimate of the distribution of eye colors, there are only two points of uncertainty -- the observer, and the person being modelled. A knows, with certainty, that there is no situation where someone will model the behavior of another and find a lower bound of less than 97, because even if A is brown-eyed, they know every other person on the island will see at least 98 blue-eyed people, regardless of all other factors.

The salient modelling question is, therefore, only how much "worse" than 98 it can be. One may be tempted to assign another -1 to the new observer who then must assign another -1 to their target, but that is illusory -- in our previous example, the uncertainty with which C considers D is the same uncertainty that A already accounted for, i.e. the uncertainty of the observer. No matter who is modelling who, there is only one observer, one modeled-observer, and one modeled-observer-target. Further extrapolation isn't necessary. Therefore, since A knows they are looking at someone with blue eyes, that person B can conclude that of the 98 people they assuredly see, they must subtract only one more for the worst case -- the estimate of C, who knows not their own eye color. B doesn't need to account for their own eye color, because A already did that in their own uncertainty, and they know that B is blue-eyed -- no matter who they select of the 99 blue eyed people they see, that person will see (at least) 98 blue eyed people, and gives no greater uncertainty than the case where A is brown eyed and they are considering someone else blue-eyed, lowering the bound to 97. It in no circumstances is estimated lower than this.

That means that every knows, with certainty, there are 97 blue eyed people at a minimum, and therefore the first interesting day is the 97th on the "original timeline" -- but that clearly isn't the case, because everyone can already count more than 97 blue-eyed people. Indeed, everyone can count 99 blue-eyed people; it is only in question for each observer whether they are a 100th blue-eyed person, and the chain of reasoning exists to expose that. The worst-case 97 holds only if his own eyes are brown, and another blue-eyed person considers a third blue-eyed person. Since there is no way to communicate, there is no way to improve upon that uncertainty, so that must be the point of first decision:

The first interesting night requiring a decision is establishing whether 97 people have blue eyes, or more than 97 people. This could be known with ceratinty by the original reasoning, which persists until the 97th night. However, everyone knows, with certainty (due to the above) that every night prior to the 97th night will result in no action; therefore the only logical course of action is to begin with the first point of new information. They "start" with the 97th day, and leave on the 100th, for a total of 4 days. [[User:Dokushin|Dokushin]] ([[User talk:Dokushin|talk]]) 07:55, 26 February 2023 (UTC)

:You are 99% of the way there... to explaining why the solution does not work at all. You are absolutely correct that there is no reason for anyone to assume that anyone else would ever assume less than 97 blue-eyed people, and therefore the induction does not carry through. But this means they have absolutely no grounds whatsoever to start making any assumptions about who could possibly figure anything more out and therefore who & when could possibly be getting onto the boat. In the presented scenario, no one ever leaves. It could only be different if the guru spoke about seeing 98 blue-eyed persons. [[Special:Contributions/172.70.85.125|172.70.85.125]] 11:11, 30 September 2023 (UTC)

::No, neither of you are "99% of the way there". Dokushin's claim that the induction doesn't work is supported by nothing more than arbitrarily terminating the inductive chain after 3 steps, a move which is itself backed up by a whole lot of "I can terminate the chain after 3 steps, because, well, because I can terminate the chain after 3 steps". No, logic doesn't work that way. He says "every[one] knows, with certainty, there are 97 blue eyed people at a minimum". No, everyone knows, with certainty, that there are *99* other blue-eyed people. They can each see that. But none of them know their *own* eye colour, and seeing everyone *else's* eyes doesn't help them in the slightest. Not at first. They each have to work that out for themselves, symmetrically and in parallel, one step at a time, one day at a time, one empty boat departure at a time, starting from only what the Guru told them - i.e. 1. Not 97. That's a totally arbitrary number. [[Special:Contributions/165.225.213.98|165.225.213.98]] 13:59, 9 September 2025 (UTC)

;Unstated assumption (about motives)
While trying to solve this, I started questioning my assumptions about the motives of the islanders. It turns out my assumptions were correct, but I think they deserve to be explicitly stated:
# Everyone wants to leave the island ASAP.
# An islander only "figures out" their eye color through logical deduction (no guessing or playing the odds)
# The Guru's statement is meant to help others leave the island. (but as stated above, it may not be the optimal means of helping others leave -- "nobody has a unique eye color")
:Regarding the 1st point: Yes, that needs to be stated.--[[User:Lupo|Lupo]] ([[User talk:Lupo|talk]]) 11:37, 9 January 2020 (UTC)
:Regarding the 2nd point: It should be mentioned, but could be implied by the words "figure out" and by the fact that everyone is a "Perfect logican". --[[User:Lupo|Lupo]] ([[User talk:Lupo|talk]]) 11:37, 9 January 2020 (UTC)
:Regarding the 3rd point: If it was not meant to help people leave, it would still do the same job. Also the statement from the Guru "nobody has a unique eye color" would be wrong and misleading! In that case everyone would wrongly assume: "I must have green eyes, as otherwise the Guru would have an unique eye color." - The alternative statement "I see noone with a unique eye color" would work. --[[User:Lupo|Lupo]] ([[User talk:Lupo|talk]]) 11:37, 9 January 2020 (UTC)

Storming lighteyes (not sorry) [[User:SilverMagpie|SilverMagpie]] ([[User talk:SilverMagpie|talk]]) 17:50, 10 January 2020 (UTC)

;Guru gives synchronizing signal to begin recursive cascade, without which it can’t start.
We might wonder why it is noted that the islanders have been on the island “all these endless years”, but the recursive cascade had never happened. The reason is that There must be a time zero, T0, from which the wait period is defined. Without that discrete starting point, there is no reference point to begin waiting to see who leaves. The Guru provides a starting event, visible (or audible) to all, where they can say “this is T0”, and our deductions can now proceed. So it synchronizes everyone to begin their waits.

I finally got this one, and while this may be repeating others' explanation of "what info does the Guru provide", I'll try another way in the simple way that was MY "aha!"/satori moment in hopes it'll help someone else too:
(A) as others have pointed out, the case of 100 reduces down to 1 (if there were 1 blue-eye, that statement would do it in 1 day, so it cascades for cases of 2, 3 etc.);
(B) once the Guru "triggers" that cascade, you just count the other people with blue eyes, and wait that number of nights; when it hits zero, leave. (It works for 1 (leave that night), for 2, (leave next night), 3 (leave the 2nd night)....)
(C) SO: the Guru saying it the statement is basically "I am hereby "triggering the "Blue-Eye Cascade". It's NOT so much whether ANYONE CAN SEE A BLUE-EYED PERSON - obviously everyone on the island can!

* '''It's the Guru BEING A LOGICIAN, AND TRIGGERING A LOGICAL CASCADE FOR THEM. S/he could've ALSO said BROWN - and THAT would've had the OTHER effect - because that'd have been giving THAT "hint"/"clue" (!!!!!!!)'''
[[User:Abner Doon|Abner Doon]] ([[User talk:Abner Doon|talk]]) 16:12, 16 April 2023 (UTC)

what if there is 1 blue-eyed person on the island outside of houses while the guru is saying that, and everyone else is in houses with their windows and doors shut? [[User:Plushiefan4111|plushie fan]] ([[User talk:Plushiefan4111|talk]]) 15:47, 22 November 2023 (UTC)

^^^ Trying to "break" the problem in this way leads to no greater understanding of the stated logic problem or its possible solution(s). --[[Special:Contributions/162.158.167.19|162.158.167.19]] 00:43, 10 December 2023 (UTC)

You could point at water or dirt then point at your eyes and tilt your head questioningly as if to say "Which one? Brown or Blue?"
Simple. [[User:Psychoticpotato|Psychoticpotato]] ([[User talk:Psychoticpotato|talk]]) 21:44, 4 April 2024 (UTC)
:"No, you don't ''look'' like you're crying..." [[Special:Contributions/172.70.162.19|172.70.162.19]] 23:28, 4 April 2024 (UTC)

Isn't the problem a little more complex, because the way Randall phrased it, Anyone who *knows* their eye color can leave. regardless of whether the eye color is brown or blue {{unsigned ip|108.162.238.76|22:10, 11 June 2024}}
:The information of there being "(at least) one person with <colour> eyes" eventually triggers every person who can see 99 pairs of eyes of <colour> and 100 pairs of eyes of <unstated colour> to leave on day 100, in a way that a person of <unstated colour> eyes cannot be. Nor even, with the benefit of another day but no further information (not even the fact of the departures), could they know that they are not personally <yet another colour>ed (e.g. another green-eyed person, like the guru...).
:Until the declaration of there being a blue-eyed person present, there's no base from which the blue-eyed can (eventually) logically deduce they should leave. It would take a similar declaration about brown-eyedness to (eventually) lead to all browns leaving, but that's not stated.
:As far as I can work out, there's no statement(s) that the Guru could make which would lead to the Guru leaving, without a defined subset of possible eye colours to rule out. They could accept that the logically-derived departures of all blue-eyed ''and'' brown-eyed residents meant that they were in neither the brown nor the blue cohorts, themself, but that only helps if green is known to be the ''sole possible'' third option. Four or more (or undefinedly many) leaves them still uncertain. [[Special:Contributions/172.69.195.63|172.69.195.63]] 09:02, 12 June 2024 (UTC)
:Shorter version: seeing 100 blue-eyed people and 99 brown-eyed ones (and 1 green-eyed one), you can ''expect'' to see the blue eyes leave on the 100th day. But that doesn't help identify your own eyes. (If the 100 do ''not'' leave, you would have learnt that you are the 101st blue-eyed person, someone that each of the other hundred blues observes as their own potential 'other hundred', and will join the departure the day after after establishing that you are one of the 101.) But every person can imagine having ''any'' type of eye, if the logical proof of being blue does not end up resolving to apply to them. [[Special:Contributions/172.69.195.184|172.69.195.184]] 12:38, 12 June 2024 (UTC)

I think that the invalid nature of the solution lies in the assumption that all islanders will adopt the same strategy to determine the number of blue-eyed islanders. It may appear that this is addressed in the premise of all islanders being perfect logicians, but this is not the case when there are numerous equally logical strategies. {{unsigned|Ningnong|22:38, 3 September 2024}}
:That there are numerous equally logical strategies (which ones were you thinking of?) needn't be a logistical problem if all logistically perfect islanders are aware of each of them and then enact the ''first'' strategy that provides its solution. (i.e. if there was some other reason to act, prior to the stated time to do so, it wouldn't matter that there was still the comic's logistical 'answer'; the prior conclusiveness would have been obeyed already...)
:In order to have a conflict, you'd need a rival logic that would apply in a different order to different people. The only real difference between people is which eye colour they have (you see 100 pairs of eyes of the other (non-guru) colour from you, 99 of your own – though of course you don't know this until the logic tells you that this must be the case) but that's already the threshold value that kicks in the 'perfect logic' reaction from those that need to (if it had not kicked in, then you'd have waited another day to resolve the next possible hypothesis). With some reason to have acted the day ''before'', everybody in your position would have acted the day before.
:There cannot be two (valid) logics that suggest different conclusions to different people, as given. The closest you can get is instead having a 101:100 split (or other similar issue of uneven non-guru numbers), in which everybody who could be "the 101st" is seeing 100+100 and could consider being 101st of ''either'' colour. (Prior to that, someone who sees 101+99 can wonder if they are "the 100th", but the alternative is go be "the 102nd" of the alternative, whose 'proof day' has yet to come). But the balanced numbers, ironically, ensures an imbalance in the potential 'proof-days' that means that two different interpretations cannot become 'true' on the same day. [[Special:Contributions/172.70.91.58|172.70.91.58]] 01:17, 4 September 2024 (UTC)
::For example, they could proceed as if the Guru said it two days ago. Or a week ago. Or 90 days ago. The 'solution' hinges on the assumption that they all start their deductive process at the same point. In fact, if they picked a different strategy (still requiring a hive mind) they could actually figure it out much more quickly. [[User:Ningnong|Ningnong]] ([[User talk:Ningnong|talk]]) 22:20, 4 September 2024 (UTC)

"If P then Q" is true is P is false.

Theorem 1: If exactly one person has blue eyes, no one will ever leave the island. It is not the case that exactly one person has blue eyes, so this theorem is true.

Alternatively,

Theorem 1: If exactly 97 people have blue eyes, they will leave the island as soon as the guru makes his statement. Everyone knows that not exactly 97 people have blue eyes, so this theorem is also true, so 96 or so days can be cut from the original inductive proof. {{unsigned ip|172.68.3.7411:30, 1 October 2024}}
:No, lost me there. Looking at the second, if there are 97 people on the islend with blue eyes (assume 103 with brown eyes, plus green-guru), then all blue-eyed can see 96 blues (could be 97, if they are one) and 103 brown eyes (could be 104, if they are one) and 1 green eye (could be 2, if they are one). And no other colours (though they could ''be'' the one!). But I'm not sure how that helps shortcut.
:Some people will see numbers of eyes in others and know that ''at least'' so many days need to be 'wasted' before it becomes clear whether one must assume that one is the extra person with blue eyes. This raises the possibility that the perfectly logical inhabitants can skip over the 'waste days', perhaps. But if you're a blue-eye you'd know you'd need to waste a number of days that is less than must be known to be wasted by a non-blue-eye, and you don't actually know whether you're in either camp, so is it possible that everyone can make the assessment of how many days everyone else thinks should not bother to waste, and get straight to the post-waste (or maybe within a day or two of it, depending upon your personal ideas)? Well, not really, because blue-eyed and non-blue-eyed would have different "If I am not, then we can skip N, if I am, then we could skip N-1" values. Without communicating what value of N each person considers (breaking the rules, and would immediately reveal if you are the same/different from the comminicator), there seems to be no logical amount of 'assumed days to not bother wasting' to shortcut everyone to jump straight to (or nearly straight to) whether the exodus happens.
:As to the first form, If exactly one person has blue eyes, they get told there is someone with blue eyes, they can't see anyone with blue eyes, therefore they know their eyes are blue, therefore they leave.
:I also can't really parse <code>"If P then Q" is true is P is false.</code> in the first place. As I read it, given a test/result of "If P then Q", and the result of that is true (by which you mean, Q is true), then P must be true and then... well, not sure where you go with the rest. But that's already potentially wrong, because Q could be true without P (unless you specify "Iff P then Q"). But perhaps you need to explain that more. [[Special:Contributions/172.69.79.138|172.69.79.138]] 12:42, 1 October 2024 (UTC)
::The statement "If P then Q" is commonly discussed when discussing logic. If P is true, then Q must be true for the entire statement to be true. If P is false, it does not matter if Q is true or false (e.g. "if 2+2=5, then the sky is green" is a true statement because the first part is false). Applying this to the Blue Eyes problem, the theorem "If exactly one person has blue eyes, then no one will ever leave the island" is true because it is not the case that exactly one person has blue eyes. All the theorems that start with "if there are exactly N blue-eyed people on the island" with N < 98 are true because everyone knows there are at least 98 people with blue eyes. It does not matter what the second part of the theorem is. With all those nonsense theorems, the induction falls apart.
::That said, I am not as smart as Randall so I am very likely missing something. Perhaps a different wording of the theorems would clear things up for me. But as the solution is stated, you could jump right to Theorem 97 and replace the 97th night with the first or 1000th or -9999th and the Theorem would still be true.
::I made a typo earlier. "If P then Q" is true '''if''' P is false. There are lots of explanations out there that are better than mine.([[Special:Contributions/172.68.1.158|172.68.1.158]] 00:52, 2 October 2024 (UTC))
:::The if/is typo did not help. Also arguable whether "This statement is false" can be true if you later specify that the statement is not an applicable statement.
:::Everyone knows there are at least 98 people, but everyone does not know that "everyone knows there are at least 98 people".
:::#Everyone (blue) knows that there are at least 99 people, to contrast with everyone (not-blue) who knows that there are at least 100 people.
:::#Everyone (blue) can imagine that the 99 blues they see are thinking the same as them (if they are blue), or thinking that it's 98 blues (if they aren't).
:::#Everyone (not-blue) can imagine that the 100 blues they see are thinking "at least 99", in the one case (of unknowable self-realisation) or "at least 100" in the other.
:::#*In first-order analysis, everyone can imagine there are anywhere from 99 to 101 blue-eyed people... 99: blue-eyed does not imagine they are; 100: blue imagines they are blue, ''or'' not-blue imagines they are not-blue; 101: not-blue imagines they are blue
:::#Infering the thoughts of the (99 or 100) people of either eye-colour must apply the unknown quality of the inferer's eyes to the way that the inferee's own inferences must be shaped.
:::#*This 'second-order' analysis gives the 'infered inferences' a range of 98 to 102: a blue imagining they are not blue would anticipate another blue (imagining they are not) establishing a theoretical 98 'true-blues' (not the original inferer, nor the inferer's theoretical inferer). A non-blue who imagines wrongly could imagine another non-blue imagining wrong (their +1 misassessment added to the other's +1) to give an upper-limit of 102.
:::#*But note that the 98 'lower limit' (at two orders of logical carry-through) is only applicable to blues (who do not yet know they are blues, and know that all other blues do not yet think they are blues). Non-blues have 2nd-order 'lower limit' that is greater than 98 (i.e. 99: Non-blue imagining correctly, imagining a non-blue imagining wrongly)
:::#Beyond this, it calls into question the original truth or falsity of the assumptions made by the 'master inferer' in their recursive inferences, since the 'truth' of their own eyes becomes again part of the bounding logic-chain of "they might be imagining one less (or more!) such person than I am imagining".
:::#Everybody can probably agree (wordlessly) that there are more than N blue-eyeds, such that N steps ''could'' be skipped to get straight to the point of every blue-eyed person that can be seen (by a non-blue) leaving at the first opportunity. But how to agree upon that N, when some Ns are N+1?
:::Worse than that, the step in the solution to prompt someone to leave is that they observed that all those others who should leave did not, bringing the realisation that they should (anyone who actually should not leave, in the possibility that they theoretically must eventually, would have been prompted to the day ''after'' everyone a tually left, but only if they actually had not). If everybody has their own idea of N and N+1, and they wait to see if the Ns leave, then nobody leaves for 'their' N and then the next day ''everybody'' assumes their N+1 applies and leaves.
:::If only everybody knew what N everybody else was thinking about..? But they aren't allowed to communicate. Except in just the ''one'' way: By leaving (or not) each day. All non-departing days, from the guru's initiating statement onwards, is a logically agreeable increment of the value that eventually becomes (the correct) N.
:::When the day count gets high enough that theoretically everybody that a person sees as blue ''should'' have left, this prompts the blueness-realisation in that person (and all fellow blues, who had themselves down as the unknown quantity instead) and creates a departure event upon the next day. (And thus forestalls all non-blues from erroneously thinking themselves "the last self-unknowing blue" and trying to depart the day after.)
:::Simple? [[Special:Contributions/172.70.160.230|172.70.160.230]] 08:30, 2 October 2024 (UTC)

Why can't the guru just say "I see 100 blue eyed people and 100 brown eyed people" on day one?[[Special:Contributions/104.23.190.137|104.23.190.137]] 14:56, 6 March 2025 (UTC)
:Then there wouldn't be a puzzle. (Or, if you prefer, "only the guru knows, probably hence why they're the guru".) You might as well ask why there's the rules on what people must and must not do (not openly speak/sign-language to others about their own status, and leave immediately upon having the required knowledge, plus ''never'' try to look into any decent reflective surface). Given the setting (presume some sort of island of myth), probably there's some sort of enchantment/geas/divine-command laid down upon the community, whether intended to be so restrictive (yet have the logical loophole that allows the guru their single utterance of no ''immediate'' value) or an unfortunate mistake by whatever power it was that blanketed them all within this scenario. [[Special:Contributions/172.70.162.57|172.70.162.57]] 16:08, 6 March 2025 (UTC)

Why is this in [[:Category:My_Hobby|My Hobby]]? Like, what's the hobby? [[User:Giraffequeries|Giraffequeries]] ([[User talk:Giraffequeries|talk]]) 00:48, 22 August 2025 (UTC)
:The fact that the [[37: Hyphen]] "My hobby" comic is included in this page? [[Special:Contributions/84.43.20.118|84.43.20.118]] 01:02, 22 August 2025 (UTC)

Blue Eyes

2025-12-27T09:58:34Z

2001:8003:DC65:7E01:4460:4138:3D1A:2B74: /* Explanation */

{{comic
| date =
| title = Blue Eyes
| lappend = blue_eyes.html
| extra = yes
| before = <center><table width="90%"><tr><td><center><big>If you like formal logic, graph theory, sappy romance, bitter sarcasm, puns, or landscape art, check out my webcomic, [https://www.xkcd.com/ xkcd]</big>.

[[File:frame.jpg|link=https://www.xkcd.com/]]

Blue Eyes:

The Hardest Logic Puzzle in the World</center>

<big>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."

I've done my best to make the wording as precise and unambiguious as possible (after working through the explanation with many people), but if you're confused about anything, please let me know. A word of warning: The answer is not simple. This is an exercise in serious logic, not a lateral thinking riddle. There is not a quick-and-easy answer, and really understanding it takes some effort.</big>

<center>[[File:hyphen.jpg|link=https://www.xkcd.com/]]</center>

<center><big>I didn't come up with the idea of this puzzle, but I've written and rewritten it over the the years to try to make a definitive version. The guy who told it to me originally was some dude on the street in Boston named Joel.</big></center>
</td></tr></table></center>
}}
{{TOC}}
==Explanation==
{{incomplete|
*The date is incorrect, the comic was available long before October 11, 2006. The earliest date we have is [https://web.archive.org/web/20041024201125/http://68.57.186.221:8080/ October 24th, 2004] (see fourth link on the page), and the earliest version of the comic is from [https://web.archive.org/web/20041109034300/http://68.57.186.221:8080/blue_eyes.html November 4th, 2004].

*Additionally, both the puzzle [https://xkcd.com/solution.html and the solution] (here's an [https://web.archive.org/web/20061102070433/https://xkcd.com/solution.html earlier version of the solution])were modified and updated several times since its release. Update the explanation and the [[#Trivia]] with this info.}}
{{misc page}}

[[xkcd]]'s [http://xkcd.com/blue_eyes.html Blue Eyes] puzzle is a logic puzzle posted around the same time as comic [[169: Words that End in GRY]]. [[Randall]] calls it "The Hardest Logic Puzzle in the World" on its page, but whether it really is the hardest is up to speculation.

The page contains two comics. On the top is [[82: Frame]], and at the bottom is [[37: Hyphen]]. These particular comics may have been chosen intentionally, as 82 involves a mind screw (and formal logic can be pretty mind-screwy to the uninitiated) and 37 involves linguistic ambiguity, which Randall has explicitly gone out of his way to avoid (interestingly, [[169]] involves the infuriating ambiguity caused by misquoting riddles). That said, Randall could have simply picked those comics out of a hat to plug for his comic (which he also does explicitly), and the date of release could also be completely random.

Randall cites "some dude on the streets in Boston named Joel" as his source for the comic idea, although he's rewritten it.

I think that the use of the word 'Guru' here confuses Zen and Indian mysticism. The puzzle depends on intentionality, a necessary condition which is not made clear by the word 'Guru'

==Puzzle==
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 always see everyone else 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?

==Solution==
Randall's solution can be found at [http://xkcd.com/solution.html xkcd.com/solution.html]. Here are some observations that help simplify the problem:

No one without blue eyes will ever leave the island, because they are given no information that can allow them to determine which non-blue eye color they have. The presence of the non-blue-eyed people is not relevant at all. We can ignore them. All that matters is when the blue-eyed people learn that they actually are blue-eyed.

There are two ways in which blue-eyed people might leave the island. A lone blue-eyed person might leave on the first night because they can see that no one else has blue eyes, so the Guru must have been talking about them. Or an accompanied blue-eyed person can leave on a later night, after noticing that other blue-eyed people have behaved in a way that indicates they have noticed that their eyes are also blue.

The problem is symmetrical for all blue-eyed people, so this means they will either all leave at once or all stay forever.

===Theorem===
If there are N blue-eyed people, they will all leave on the Nth night.

===Dual logic===
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.

===Intuitive 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, they 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.

It's obvious that our theorem holds for the "simpler" puzzle, and this "simpler" puzzle is perfectly equivalent to the original puzzle, so our theorem must hold for the original puzzle too.

Another way of looking at it is to use selective attention. Although each blue-eyed person can see each other blue-eyed person on the island, they don't need to. The only thing they need to know in order to determine whether to leave on night N is whether they can see an Nth person with blue eyes. On night 1, they only need to see 1 other blue-eyed person to not leave; on night 2, they can see 2 other blue-eyed people, so they don't leave; and so on until night 100, when they can't see a 100th blue-eyed person, and then leave.

===Formal proof===
To prove this more formally, we can use mathematical induction. To do that, we'll need to show that our theorem holds for the base case of N=1, and we'll need to show that, for any given X, *if* we assume that the theorem holds for any value of N less than X, then it will also hold for N=X. If we can show both these things, then we'll know the theorem is true for N=1 (the base case), for N=2 (using the inductive step once), for N=3 (using the inductive step a second time) and so on, for whatever value of N you want.

Base case: N=1. If there is just one blue-eyed person, they will see that no one else has blue eyes, know that the Guru was talking about them, and leave on the first night.

Inductive step: Here we assume that the theorem holds for any value of N less than some arbitrary X (integer greater than 1), and we need to show that it would then hold for N=X too. If there are X blue-eyed people, then each will reason as follows: "I can see that X-1 other people have blue eyes, so either just those X-1 people have blue eyes, or X people do (them plus me). If there are just X-1 people with blue eyes, then by our assumption, they'll all leave on night number X-1. If they don't all leave on night number X-1, then that means that there is an Xth blue-eyed person in addition to the X-1 that I can see, namely me. So, if they all stay past night number X-1, then I'll know I have blue eyes, so I'll leave on night number X. Of course, they'll also be in exactly the same circumstance as me, so they'll leave on night number X too."

This suffices to prove our theorem. The base case tells us the theorem holds for N=1. That together with the inductive step tells us that it therefore holds for N=2, and that together with the inductive step again tells us that it holds for N=3, and so on... In particular, it holds for the case the original puzzle asked about, N=100, so we get the conclusion that the 100 blue-eyed people will leave on the 100th night.

==Randall's follow-up questions==
After giving his solution, Randall posed three questions for further thought about the puzzle. They are answered below, in a different order than his.

===Question 2===
:''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?''

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... All the levels are relevant. (It's like [https://www.youtube.com/watch?v=U_eZmEiyTo0 the Princess Bride scene] where Vizzini is trying to think about what Wesley would choose in part based upon Wesley thinking about what Vizzini would choose in part based upon... "So clearly I cannot choose the one in front of me!") Each layer of thinking about what someone else might be thinking about can decrement by 1 the number of blue-eyed people visible to the lattermost imagined person, so it turns out that even the base case with N=1 blue-eyed person is relevant. As the days go by, some of the more far-fetched "he might be thinking that I might be thinking that he might be thinking that I might be thinking that..." hypotheses get ruled out. But it's only after night N-1 that the blue-eyed people rule out all the possibilities in which they have brown eyes, whereas the brown-eyed people only learn on night number N that they don't have blue eyes.

It might help to think of all the different situations people might be in. (Remember brown-eyed people always are situated where they can see one more blue-eyed face than blue-eyed people can.)

'''Situation 0.''' If I see 0 blue-eyed people, I can leave right after the announcement on night 1.
'''Situation 1.''' If I see 1 blue-eyed person, then they might be in situation 0 and about to leave on night 1; or else they might be in situation 1 just like me, in which case we'll both leave together on night 2.
'''Situation 2.''' If I see 2 blue-eyed people, they might each be in situation 1 watching to see whether anyone in situation 0 leaves the first night (I know nobody will leave that night, but they wouldn't know this), in which case they would leave together on night 2; or else they might be in situation 2 just like me, in which case we'll all leave together on night 3.

⋮
'''Situation N.''' If I see N blue-eyed people, they might be in situation N-1 watching to see whether any people in situation N-2 leave on night N-1 (I know nobody will leave that night, but they wouldn't know this), in which case they would leave together on night N; or else they might be in situation N just like me, in which case we'll all leave together on night N+1.
⋮

Even though I start out in situation 99, I need to worry that the blue-eyed people might be in situation 98, so I need to wait long enough for people in situation 98 to figure out what's going on, and then see whether they act like they are indeed in situation 98. But if they're in situation 98, then they're worrying about whether all the blue-eyed people might be in situation 97, so they're going to need to wait long enough for people in situation 97 to figure out what's going on. Of course, that requires waiting long enough for people in situation 96 to figure out what's going on, and so on, down all the way to situation 0. All the levels are relevant, and it takes a separate day to eliminate each level, which is why the entire process takes N days.

===Question 3===
:''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?''

Consider an analogy. I've heard that miners used to take canaries down into mines because canaries pass out more quickly in poor air than miners do. Suppose you know the canary will do fine for 98 or so seconds, and then pass out if the air is bad. As you watch the canary for those 98 seconds, there's a sense in which you're just verifying something you already know (it'll do fine), but it seems more accurate to say that your best detector for the quality of the air takes 98 seconds to give you a reading, and you're waiting 98 seconds to see what that reading is.

When the blue-eyed people wait 98 or so days to leave, that's because their best available detector of their own eye-color takes 98 or so days to give a reading. (This detector involves watching what the other blue-eyed people do, and of course they themselves are waiting on a detector that takes 97 or so days to yield its result...) There's a sense in which they're "simply verifying something that they already know", but it seems more accurate to say that they're waiting for their best available detector of their own eye-color to deliver its reading.

===Question 1===
:''What is the quantified piece of information that the Guru provides that each person did not already have?''

Before the Guru speaks, the hypothetical chain of A imagining B imaging C imagining D... imagining Z seeing N blue-eyed people cannot terminate uniquely. Z seeing no blue-eyed people can consider whether they are blue-eyed. This means it is not {{w|Common knowledge (logic)|common knowledge}} that there are blue eyes. Once the Guru makes their pronouncement it is common knowledge, and every chain of reasoning must terminate at 1 blue-eyed person and Z above would have to conclude that they had blue eyes. From then on, every midnight the common knowledge that there are N blue-eyed people increments by 1 as everyone sees nobody leaving on the ferry.

Stated another way, there's only one stable set of beliefs for the blue-eyed people that would allow them to have so many exist on the island indefinitely. That is if each blue-eyed person believed not only that they have brown eyes, but also that every other blue-eyed person believed, incorrectly, that they had brown eyes. Logic reduces this to "all unknowing blue-eyes believes that all known blues-eyes believe that they have brown eyes". The Guru eliminates that particular possibility.

Another straightforward way to understand why the Guru's information is important is thus. Each blue-eyed person knows two sets of information: what the actual situation is on the island (both now and in the past), and what would happen in a hypothetical situation. Each blue-eyed person then needs only to compare the actual situation to a known hypothetical one, and if it matches up, then they take the corresponding action. Consider this: If there were only one blue-eyed person, and the Guru never made the announcement, they would not leave on day 1 because they would not know that N is greater than or equal to 1. Now let's add a second blue-eyed person. Blue-eyes 2 would not be able to inductively determine whether or not to leave on night 2, because blue-eyes 2's knowledge of whether or not to leave on night 2 is dependent on what blue-eyes 1 does on night 1 if and only if blue-eyes 1 knows what to do on night 1. If blue-eyes 1 doesn't know that N is greater than or equal to 1, then blue-eyes 1 doesn't know what to do on night 1. Thus, their lack of leaving gives blue-eyes 2 no new information, since it was an uninformed action and blue-eyes 2's inductive reasoning was dependent on blue-eyes 1 knowing what to do, and so the inductive process never takes off for the hypothetical situation. This means a hypothetical situation for N people cannot be induced. As such, blue-eyes 100 does not have certain knowledge of the hypothetical situation that would occur on nights 99 and 100, and so even though they know N = either 99 or 100, they can't take action on either of those nights, because they have no certain hypothetical situation to compare reality to, and as such cannot have certainty about the actions they should take.

==Transcript==
:[The 'comic page' is mostly text, with two actual images, neither of which relate to the text.]
:[The first image is that of the comic [[82: Frame]]. It is positioned above the title.]
:[The second image is that of the comic [[37: Hyphen]]. It is positioned directly above the last sentence.]

==Trivia==
The web page which contains the puzzle has no {{w|CSS|style sheet}}. The font size of the heading and subheading is increased with deprecated HTML tags, rather than with heading tags. The way the page is displayed therefore depends on the browser's settings. Despite this fact, due to a similarity of default settings between computers, most computers will display the page with a white background, black text, and the {{w|Times New Roman}} font. However, it has two line breaks after every paragraph instead of HTML paragraph breaks, meaning that paragraph spacing will not vary between browsers, relative to the font size.

{{comic discussion}}

[[Category:My Hobby]]
[[Category:Multiple Cueballs]]
[[Category:No title text]]
[[Category:No date]]
[[Category:Comics with lowercase text]]

Talk:2526: TSP vs TBSP

2025-12-27T09:57:45Z

2001:8003:DC65:7E01:4460:4138:3D1A:2B74: edit error

The title text refers to lyrics in the Alanis Morissette song Ironic -MonteCarloe [[Special:Contributions/172.70.126.211|172.70.126.211]] 16:59, 8 October 2021 (UTC)

I wonder why Randall used 1024^4 instead of 2^40. -MonteCarlo [[Special:Contributions/162.158.74.150|162.158.74.150]] 17:10, 8 October 2021 (UTC)
:To highlight the similarity between 1000 and 1024, I suppose.--[[User:Pere prlpz|Pere prlpz]] ([[User talk:Pere prlpz|talk]]) 18:03, 8 October 2021 (UTC)
:And a byte is 1024^0 bytes, kilobyte=1024^1, megabyte=1024^3, etc. Talking entirely in terms of 2^10N loses the sheer simplicity and easy approximation.
:(Though not as much as 'short' -illions. 5.45 trillion? That's 5, then the decimal padded/shortened to exactly three figures (450), ''then'' three sets of three zeros. Stupid system.) [[Special:Contributions/172.69.54.136|172.69.54.136]] 18:20, 8 October 2021 (UTC)

A table with common or interesting volumes measured in teraspoons would be a good addition.--[[User:Pere prlpz|Pere prlpz]] ([[User talk:Pere prlpz|talk]]) 18:03, 8 October 2021 (UTC)

he did kilonife kiloknife because the k in knife stands for kilo [[Special:Contributions/172.70.34.91|172.70.34.91]] 18:26, 8 October 2021 (UTC)Bumpf

I thought this was a play on storage (esp. hard drives) advertising 1000-based capacities (1TB=1000GB) whereas computers tend to measure in 2^ sizes (1TB=1024GB) to make them seem bigger on the packaging. IIRC some cloud platforms distinguish between GB and GiB of Ram, for instance. [[User:Mike-jed|Mike-jed]] ([[User talk:Mike-jed|talk]]) 19:38, 8 October 2021 (UTC)

XKCD again with its timing… Just 3 days ago I used a TBSP when I should have used a TSP. Two comics earlier, #2524, was about a long-period comet coming close to Earth and on the same date (October 4th) as a long-period comet coming close to Earth in the movie "Your name". [[User:Fabian42|Fabian42]] ([[User talk:Fabian42|talk]]) 21:47, 8 October 2021 (UTC)

1 mL = 1 cm * 1 cm * 1 cm. 1 m = 100 cm. 1 m^3 = 100 cm * 100 cm *100 cm = 10^6 cm^3. 1 tera-whatever = 10^12. 10 ^ 12 / 10 ^ 6 = 10 ^ 6. I respectfully submit that the estimate of how many cubic meters there are in a tea-teraspoon is off by three orders of magnitude, and should be approx. 5 * 10 ^6 cubic meters, not 5 * 10 ^ 3 cubic meters. However, it is late afternoon locally, and my math skills typically erode severely at this time of day, I could well be embarrassingly wrong. Someone meticulous and more awake should check and cast a deciding vote. -E. Schori

Alanis Morissette describes a situation where the ratio between available spoons and required "nifes" is 10:1 (10 kspoon vs 1 knife). Why doesn't the title text say "one teraspoon when all you need are 10^8 kilonife"? Is there a reason I don't get? Or does it simply sound better to have a single kilonife? Or are the numbers from the title text arbitrary? This conundrum is totally bugging me out! -mape

Most importantly: What is a nife? In German "Nife" is the name for Earth's inner core (Ni+Fe = Nife). But in English the word doesn't seem to exist. I get the joke of a knife being 1k nifes (or nives?) but the joke would be even better if "nife" existed in the first place. [[User:Elektrizikekswerk|Elektrizikekswerk]] ([[User talk:Elektrizikekswerk|talk]]) 11:20, 11 October 2021 (UTC)
: Addendum: If "nife" does indeed not exist wouldn't that mean that a (kilo)nife is simply nothing? (Yes, I know that I'm overthinking this, but I find it curious) [[User:Elektrizikekswerk|Elektrizikekswerk]] ([[User talk:Elektrizikekswerk|talk]]) 11:23, 11 October 2021 (UTC)

It seems nearly impossible to add a citation to the statement that "All these units have fairly limited uses in cooking". Perhaps it might be better to state something along the lines of "All these units are very large and not commonly used in cooking", and then cite the NIST.gov Metric Cooking Resources page (https://www.nist.gov/pml/weights-and-measures/metric-cooking-resources) which lists TSP & TBSP but does not list Teraspoons and Tebispoons? [[User:MrYellow04|MrYellow04]] ([[User talk:MrYellow04|talk]]) 17:33, 11 October 2021 (UTC)

I retract my previous statement. After reviewing the FAQ for this wiki, I've learned that the "citation needed" template is meant to be used only when it conveys a deeper sense of humor, not necessarily as an actual call for a citation. I apologize for my previous statement; I am new to this wiki and did not understand all of the etiquette and common practices. I have not edited the wiki page, and the original statement "All these units have fairly limited uses in cooking" that I had previously discussed possibly changing has not been altered. [[User:MrYellow04|MrYellow04]] ([[User talk:MrYellow04|talk]]) 17:42, 11 October 2021 (UTC)
:Aye, it's bit of an in-joke that arose from Randall's use of the phrase for mostly-inarguable axioms in What-Ifs, rather than his Wikipedia Protestor comic. But if it had to be linked to something, that page is as logical as any other.
:Some people think it's overused. Other people dislike the people who think it's overused and then are removing what these others think is actually quite a funny joke. Anybody who comes fresh to the Wiki (more used to other Wikis), like yourself, can be a little discombobulated by all the use of this particular shibboleth.
:(Me, I think there's nothing wrong with an in-joke, in moderation. It's a bonus for the uninitiated as they gradually lose the 'un-' through exposure and realisation. Welcome to the club!)[[Special:Contributions/162.158.92.131|162.158.92.131]] 22:05, 11 October 2021 (UTC)

Talk:2526: TSP vs TBSP

2025-12-27T09:56:51Z

2001:8003:DC65:7E01:4460:4138:3D1A:2B74:

Talk:3085: About 20 Pounds

2025-12-27T09:08:00Z

2001:8003:DC65:7E01:4460:4138:3D1A:2B74:

Wow - first here! I can't help thinking 'about 20 pounds' could be exactly 10 kg! 0r even one Newton?! [[User:RIIW - Ponder it|RIIW - Ponder it]] ([[User talk:RIIW - Ponder it|talk]]) 05:50, 6 May 2025 (UTC)
::No you weren't. I got here about two hours ago. (sorry if i'm being mean or condencending) {{unsigned ip|172.69.176.6|08:39, 8 May 2025|or disingenuously late...}}
:"One Newton" and "10 kg" are totally different things. "10 kg" would cause 1 Newton of gravitational force if you were in a world with about 1% of Earth's gravity, though. --[[Special:Contributions/172.69.109.86|172.69.109.86]] 09:53, 6 May 2025 (UTC)
::Oops! In my rush I should have checked and put 100 Newtons. I was relying on 10kg being about 22 pounds, or rather the other way around, and then a particle having mass not weight and Science using Metric units. Apologies. [[User:RIIW - Ponder it|RIIW - Ponder it]] ([[User talk:RIIW - Ponder it|talk]]) 11:41, 6 May 2025 (UTC)
::(Moved your reply up a bit. You seemed to respond to "20 pounds are...", below, ''and'' split their timestamp signature from their message. And forgot to sign properly, at first, so I got edit-conflicted ''twice'' whilst trying to post myself and correct your initial error. Please take a bit more care, everybody. [[Special:Contributions/172.70.163.53|172.70.163.53]] 11:52, 6 May 2025 (UTC))
:20 pounds are approximately 9.072 kg, so not exactly 10 kg (in fact, it rounds to 9). [[Special:Contributions/172.70.134.55|172.70.134.55]] 10:02, 6 May 2025 (UTC)
::That's the wrong way to think about it. "Exactly 10kg" is "exactly 22.0462lbs", but that (to the nearest single significant figure) is legitimately "about 20lbs". See any given step in [[2585: Rounding]], especially where that 'disagrees greatly' with an adjacent step.
::As with any Oracle (that's worth its omphalos), it may be giving an ''entirely true'' answer which nevertheless is deliberately phrased as ambiguous and misinterpretable, the possible supernatural complement to the 'exact words' genie contract. As with the [[2741: Wish Interpretation]] genie, the Oracle ''may'' slip into less "unhelpfully helpful" mode immediately after, though for different reasons. However, "burritos are ''pretty'' good" also suggests that there's some other thing that is ''more'' good, so — again — it's giving a sufficient response to what they (now) should do, but not a perfect one.
::As I write, the explanation (probably needs a general rewrite) doesn't mention anything about the burritos except as title text, or I would have ensured the famed exact-words/vague-detail was noted in that bit. (Shorter than here.) [[Special:Contributions/141.101.98.82|141.101.98.82]] 11:46, 6 May 2025 (UTC)
:Nobody said anything about "exactly 20 pounds". 20 pounds is about 10 KG and about 100 Newtons when considered as a force rather than a mass. The comic says "about 20 pounds". [[Special:Contributions/172.70.230.63|172.70.230.63]] 22:36, 6 May 2025 (UTC)
Though I don't think it at all merits being described as a reference, I am minded of the {{w|The Usenet Oracle}} (at least when I knew of it). Though, if it ''was'' to be a deleliberate shout-out, I'd expect a few more actual in-jokes. [[Special:Contributions/172.70.86.130|172.70.86.130]] 06:10, 6 May 2025 (UTC)

I bet Randall is in some kind of force-interaction-related, What-if-induced rabbit hole right now (or has been at the time of writing). Wondering what the next comic will be about. [[Special:Contributions/172.71.144.175|172.71.144.175]] 08:39, 6 May 2025 (UTC)

"Nature of ... 20 pounds" is a reference to the koan "A monk asked Tozan, 'What is the nature of Buddha?' He replied, 'Three pounds of flax.'" Someone can add this to the explanation. [[Special:Contributions/172.70.111.115|172.70.111.115]] 08:57, 6 May 2025 (UTC)
:There is a similar story in the Principia Discordia. When asked what is the meaning behind POEE, a Discordian cabal, Malaclypse the younger answered "five tons of flax." [[User:FlavianusEP|FlavianusEP]] ([[User talk:FlavianusEP|talk]]) 16:26, 6 May 2025 (UTC)

"something that doesn't interact with electromagnetism cannot be 'seen', as photons will pass through it completely unaffected": is this supposed to be true ? I thought photons interacted with gravity, and even the phrase before states that gravity is believed to affect everything. [[Special:Contributions/172.68.151.93|172.68.151.93]] 09:17, 6 May 2025 (UTC)
: Indeed, photons do interact with gravity. What I had in mind when writing that were ''direct'' interactions -- of course everything interacts with everything else via a second-order interaction, <X> -> gravity -> <X> for any particle/field <X>. I can clarify that if nobody gets to it before I get around to it. [[User:Linkhyrule5|Linkhyrule5]] ([[User talk:Linkhyrule5|talk]]) 21:45, 6 May 2025 (UTC)
:We can ''infer'' Dark Matter (and, for that... *ahem* ...matter, also Dark Energy) from what the photons in the universe are telling us that does not look anything like what 'light(-interacting) matter' ''should'' be doing. As with some searches for black holes (most particularly, when the theory is that the unseen mass of the universe is a lot of small black holes drifting in the void, not acreting enough to create secondary visible effects), whether or not light is being gravitationally lensed by things (that we cannot directly see) is part of the way that we're narrowing down what-and-where DM is.
:And, I think, currently it seems to be considered that it's residing in a webwork of DM tendrils, at extragalactic (indeed, cosmological) scales, such that where the tendril cross is where they draw 'normal' matter together enough to be any given galaxy. But that's in an "explains all(/many) known facts" way, and might yet be incorrect. e.g. if there's side-dimensions (equally undetectable, at least visually) that change the inverse-square dropoff of gravity at large enough scales to govern galactic rotation rates by just enough to fit observations, or we have some other misunderstanding/scientific blind spot that further study may correct.
:Or, in short, think Brownian Motion. We can't see a handful of air molecules (not by normal, even microscope-enhanced, human vision), they might as well be invisible. But, by what we see of more visible particles, suggests that they exist as something. Conversly, the æther, a proposed medium for light, was thought to exist in a similar all-pervasive manner (insofar as trivial human experience, though less physically 'interactive' than wind), but deeper checks (as to whether its effects on light were as they should have been) dismissed it as a possible concept.
:Depending upon interpretation of the comic (I originally read it as "all dark-matter particles are ~20lbs in mass WIMPs/nano-MACHOs/whatever", but it seems that others take it as "''all of'' dark-matter particles is a single ~20lbs mass particle"; and that's make the oracle-invokers' attitudes more logical, if not the universe), there actually being Dark Matter, but it being just 20lbs of 'something' ''somewhere'' in the whole universe, makes it a needle in a galactic-supercluster-sized haystack.
:Detecting ''that'' would be difficult in the extreme. Even if it's somehow within a few hundred metres of the experimenters. There are ways to {{w|Cavendish experiment|observe the movements of small masses at small distances}}, but when you don't even have a clue ''if'' it exists (or is moving/has moved, and how), it's fairly hopeless. Gravitational lensing of light would be impractical at such distances/masses. LIGO may be very clever, insofar as merging high-mass objects at long distances, but not really for this. Event Horizon Telescope's ability to see a black hole('s accretion disk) via Very Long Baseline Interferometry is also totally useless here.
:I think I'd ''also'' settle for the burritos, given that certainty that I wasn't going to find what I'm looking for via any obvious route. (Assuming I couldn't ask the Oracle to ''show me'' the Dark Matter, rather than just answer questions about it. And noting that, if not for the indicated progression of the conversation, I might have assumed the oracular voice were really from the pentagram (more usual for demonology, not oracularities!) and that the dark blob ''was'' the 20lbs of Dark Matter. Which, of course, it ''does not deny'', so maybe my headcan[n]on ''is'' that the summoned Oracle ''is'' the DM, being deliberately evasive, and successfully so. That would satisfy it being both that which Ponytail seeks, ''and'' the entity of which Ponytail summons in order to seek it! Cueball, however, is currently just seeking food, which (one assumes) the DM-slash-Oracle is not.) [[Special:Contributions/172.68.229.25|172.68.229.25]] 12:48, 6 May 2025 (UTC)

My physics skills are rusty but 20 pounds is much more than the Planck mass. Doesn't this imply that Randall's dark matter particles would be black holes? [[Special:Contributions/172.68.243.107|172.68.243.107]] 10:05, 6 May 2025 (UTC)
:Yes, you are right that 9 kg is about 417,000,000 times more than the Planck mass (21.76 μg), but no, that doesn't imply that 9 kg dark matter particles would be black holes, for that particle can be larger than 417,000,000 Planck lengths (1 Planck length is c. 1.616255×10–35 m, so above 7 rm, this particle would not collapse into a black hole). [[Special:Contributions/172.68.245.81|172.68.245.81]] 10:23, 6 May 2025 (UTC)
::No, unfortunately that doesn't work - if it's not a fundamental particle then it interacts with forces that aren't gravity, and if it is then its Compton wavelength (which is the relevant size for a fundamental particle) is about 1/417000000 Planck lengths. The Planck mass is the maximum possible mass for a fundamental particle (running coupling constants might change the limit from the low-energy observed Planck mass, but not by nine orders of magnitude up). [[Special:Contributions/172.68.55.40|172.68.55.40]] 03:23, 8 May 2025 (UTC)
Since it's Star Wars day and the 20 lbs. reference would be causing a massively large amount of mass, would it be safe to say that they "sense a great disturbance in the force?" [[Special:Contributions/67.84.20.42|67.84.20.42]] 10:20, 6 May 2025 (UTC)

Back in 2005, when the kg was an actual object's mass, there was an article about what a five pound (~2.268 kg) electron is, but it was deleted, for it is a "trivial result of special relativity". [[Special:Contributions/172.68.245.81|172.68.245.81]] 10:23, 6 May 2025 (UTC)

Since pounds are a measure of weight, and weight is a measure of the gravitational attraction between an object and its "planet", what is the reference planet that is being used to define the weight of the Dark Matter particle?
 Should we assume that Earth's surface is being used as the reference, even though we have no measurements that suggest DM particles are around us, and no reason to assume that the particles would even notice that Earth has a "surface"?
 If Randall wanted to use mass, then he should have used the imperial unit of slug, but I suppose saying that a DM particle is 0.62162 slugs might not give the readers quite the same impression as using 20 pounds. [[User:Galeindfal|Galeindfal]] ([[User talk:Galeindfal|talk]]) 13:38, 6 May 2025 (UTC)
:I might be missing some humour here, but the pound is actually a measure of mass, just like the gram, so it doesn't vary from a planet to another. You might have fallen prey to the second paragraph of the {{w|pound-force|wikipedia article about the pound-force}}, which states: 'Pound-force should not be confused with pound-mass (lb), often simply called "pound"' [[Special:Contributions/172.71.127.160|172.71.127.160]] 14:35, 6 May 2025 (UTC)
:Indeed, when I was at school in England in the sixties, I was taught the pound was a unit of mass, and the Imperial unit of force was the poundal, the force needed to accelerate one pound mass at one foot per second per second.--[[Special:Contributions/172.69.195.4|172.69.195.4]] 12:00, 8 May 2025 (UTC)
::You'd also have been taught that pounds were divided into 240 pennies, I imagine. ;) [[Special:Contributions/172.70.163.150|172.70.163.150]] 13:43, 8 May 2025 (UTC)

Is this, by chance, the Internet Oracle? [[Special:Contributions/104.23.187.126|104.23.187.126]] 13:49, 6 May 2025 (UTC)
:I don't see anything like the pentagram with candles at its web site. The comic seems more like they're summoning a daemon. [[User:Barmar|Barmar]] ([[User talk:Barmar|talk]]) 14:10, 6 May 2025 (UTC)

Any idea where Randall came up with "20 pounds"? Why not 19 or 21 (blackjack!)? Why not use Newtons (too figgy?)? Only thing I can think of is that, in America at least, many people think they are "about 20 pounds overweight." I think that's too much of a stretch (pants???) to be the answer here. [[Special:Contributions/172.68.27.170|172.68.27.170]] 14:07, 6 May 2025 (UTC)
:I think it's just humorous, adding to the imprecision / casualness of "about 20". Imperial measurements feel "less scientific" than metric. [[Special:Contributions/162.158.146.124|162.158.146.124]] 16:26, 6 May 2025 (UTC)
:The amount of people confusing mass and weight/force in this thread is pretty disappointing for an xkcd forum. You can't convert pounds into Newtons. [[Special:Contributions/162.158.172.143|162.158.172.143]] 16:38, 6 May 2025 (UTC)
:The poster meant Kilograms rather than Newtons I assume. Under the assumption that the oracle is satire for large language models and AI chats, the "20 pounds" shows the kind of tone and data the AI has been tuned to provide. It's a way they behave when outside the core domain they are well-trained for, producing a certain brand of mistakes, awkwardnesses, and uncanny valleys, that may be quite humorous when first encountered. [[Special:Contributions/172.70.111.77|172.70.111.77]] 22:33, 6 May 2025 (UTC)
I don't think it's accurate to say (as the explanation does right now) that 20 lbs is too little to detect through gravitational interaction. Throwing some numbers together: a 20lbs-sphere of Osmium, the heaviest stable element, is about 4.5cm in radius. If a 20lbs point mass flies by just above the surface of that sphere, it would generate a gravitational force of about 2.5 micronewtons (hooray for Gauss's theorem). That's the weight of a few grains of salt - small, but definitely detectable. If they're all really really fast, or there's always lots of them around at any given time or something, that might wash out any measurements (someone more knowledgeable about dark matter can probably comment what the expected velocity and flux density of 20lbs-dark-matter-particles would be where we are). But in principle, rather measurable! [[Special:Contributions/162.158.172.142|162.158.172.142]] 17:00, 6 May 2025 (UTC)

"A particle that interacts with nothing except gravity, could only be detected by a gravitational telescope." -- Detected by a whatnow? Is that a thing which exists? Google had nothing for "gravitational telescope" when I searched for it. 
Additionally, are there any theoretical physicists out there who can weigh in on how plausible the "20 pound particle that doesn't interact with anything else" theory is? [[User:MeZimm|MeZimm]] ([[User talk:MeZimm|talk]]) 19:33, 6 May 2025 (UTC)
:Well, we haven't seen any such thing. But... that is of course the point. So, by logical extension, it ''must'' be true! :o [[Special:Contributions/172.70.58.130|172.70.58.130]] 21:05, 6 May 2025 (UTC)

There’s a few places online that I’ve seen people ''frustrated'' about dark matter because it’s been so hard to detect, even hoping that it turns out to be experimental error or otherwise not real (for example, [http://www.collectspace.com/ubb/Forum3/HTML/005665.html this collectSPACE thread]) just because it has been so “annoying” to readers of science news to not have a solution for so long. I think this comic could be kind of a joke along those lines (“what if it turns out dark matter ''doesn’t'' matter at all and we’ve been wasting our time?”) but played out in a ridiculous way because dismissing any scientific research, however frustrating, as “a waste of time” would be very un-XKCD.{{unsigned ip|104.23.90.225|21:32+22:01, 6 May 2025}}

Sounds an awful lot like a primordial black hole (a genuine dark matter hypothesis), doesn't it? Anyone who knows what the Hawking temperature of a black hole of that mass would be? (A black hole with a Hawking temperature less than 2.7 K would absorb more energy than it gives off in Hawking radiation and would therefore be stable in the current age of the universe.)

EDIT: answering my own question, apparently about 10²² K. [[Special:Contributions/162.158.91.110|162.158.91.110]] 23:20, 6 May 2025 (UTC)

Shouldn’t there be a comma after ‘well’ (when it is said by the oracle in the last panel), because there’s a comma after ‘dear oracle’ in the first panel? You can’t say that because there’s a break you don’t need a comma because there’s one in dear oracle [[User:Broseph|Broseph]] ([[User talk:Broseph|talk]]) 07:04, 7 May 2025 (UTC)
:Well. Perhaps it may need ''some'' punctuation. But... It doesn't necessarily have to represent a mid-sentence pause.
:Not that I like "you pause for breath, you add a comma" writing, also often used to justify the Oxford cases. Yes, a "clause-pause" coincides with a clausing comma. But if you're representing various other pauses or switches of tone there are mdashes, ellipses, parenthises of various types, the full-stop-slash-period and, in the comic example, the clear newline (and speechline?) acting as a rhetoric 'hold'. (The "Dear oracle," runs on ''into'' the next bit, so maybe that comma indicates symantic ''joining'' that was never required or implied with the effective "Well <mystical equivalent of taking a breath>" merely as a filled-pause.)
:Also note that it's not unknown for Randall to be inconsistent with whether or not to fully punctuate the 'ends' of speech-blocks (and has occasionally added them in post-publication edits; not entirely sure he's ever ''solely'' edited for that, but definitely has done when correcting some bigger mix-up/whatever). House-style seems to be fairly loose on that principle. It's effectively "reported speech", most of the time, so you can perhaps even put it down to the character's voice sort of... Trailing off, or even just halted altogether. [[Special:Contributions/172.68.229.67|172.68.229.67]] 09:46, 7 May 2025 (UTC)

"However, even if a particle does interact via a given force, an interaction is possible only if energy is conserved."
I am puzzled by this statement, which started the paragraph about 20 lbs being ludicrous for a fundamental particle. I think conservation of mass-energy is one of our most fundamental understandings of the universe. The rest of the paragraph does not develop further on the idea of conservation violation. If you know what was intended by this statement and think it is important, please clarify it and put it back in.
I liked the use of ludicrous mass. Brought to mind Space Balls and {{w|ludicrous speed}}. (I haven't thought of a way to make an appropriate tangential reference in the article.) [[Special:Contributions/162.158.41.36|162.158.41.36]] 19:34, 7 May 2025 (UTC)

I pride myself on being a physics nerd, but also somewhat a writing nerd. I read the article, and the stuff after the mention of a gravitational telescope, and thinking of the non-physics nerds who might be reading it, and... well, this is the closest I've come to wishing there was an explainexplainxkcd.com. // I feel sheepish for not having any revisions to propose, and grateful to whomever wrote this version, but damn. Blame it on the subject material, I suppose. [[Special:Contributions/172.70.135.206|172.70.135.206]] 20:04, 7 May 2025 (UTC)

I thought this was also just a reference to AIs like ChatGPT hallucinating facts and saying them with confidence. [[Special:Contributions/172.68.0.190|172.68.0.190]] 01:15, 8 May 2025 (UTC)
:Traditionally, any actual Oracles were famed for being mostly true but (in the long run) only actually useful for a fortunate few (like "if you attack now, a great kingdom will fall!", but we aren't saying ''whose''; or "trust in wooden walls!", which you then have to work out whether it might mean your navy at sea, rather than a stockade on land). Pretty much the ''opposite'' of our current expectations of AI (can be useful, but not necessarily truthful).
:And I read this comic as having the ungarnished truth (in-universe, at least), but that doesn't really help anyone to know (except the local burrito vendor). Typical 'mystical contract' sort of thing. [[Special:Contributions/172.70.163.150|172.70.163.150]] 13:56, 8 May 2025 (UTC)

to communicate with something from the 'dark realm' 
:It's a black hole.

Talk:2056: Horror Movies

2025-12-27T09:00:38Z

2001:8003:DC65:7E01:4460:4138:3D1A:2B74:

Horror is about my least favorite genre of film. Westerns rank around the same. I find both pretty boring, though there are some suspense films, when done right, that I do like. [[User:N0lqu|-boB]] ([[User talk:N0lqu|talk]]) 17:55, 8 October 2018 (UTC)
:I agree about horror but the Western {{w|A Fistful of Dollars}} and some more by {{w|Sergio Leone}} are on my personal top movie list. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 18:02, 8 October 2018 (UTC)
::Actually, I can enjoy almost any genre if its a humorous spoof of the genre. I love Blazing Saddles, for example. And there are rare examples in genres that can appeal to outsiders. I may have to give Sergio Leone's movies a try. [[User:N0lqu|-boB]] ([[User talk:N0lqu|talk]]) 18:13, 8 October 2018 (UTC)
:::Oh, from the {{w|Dollars Trilogy}} you should start with the third, running three hours and still the best. And for comedy horror this is a widely unknown great one: {{w|Little Shop of Horrors (film)}}. Check the cast. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 19:00, 8 October 2018 (UTC)
Early XKCD issues frequently referenced Randall's fear of velociraptors. There hasn't been a reference to this fear in many years though. Did Randall forget this trope? Is he no longer scared of dinosaurs? [[Special:Contributions/108.162.216.112|108.162.216.112]] 19:55, 8 October 2018 (UTC)
: It could be he hasn't encountered a velociraptor recently and has therefore lost his fear. [[User:N0lqu|-boB]] ([[User talk:N0lqu|talk]]) 21:04, 8 October 2018 (UTC)
: It could also be he has encountered a velociraptor recently but she was friendly and assured him humans are not among her prey list and she just eats evil T-rex professors who screw around with the speed of light. [[Special:Contributions/141.101.76.58|141.101.76.58]] 10:47, 9 October 2018 (UTC)
:I've always seen it as a LOVE for velociraptors, not a fear, that he's often brought them up. As Jurassic Park has told everybody they're killing machines, it makes the most sense to mention them in regards to attacks. :) [[User:NiceGuy1|NiceGuy1]] ([[User talk:NiceGuy1|talk]]) 05:24, 12 October 2018 (UTC)

I understand the idea: I don't care about horrors, but I like fantasy and sci-fi movies and was never stopped by someone classifying some fantasy/sci-fi movie as horror ... like in case of Dracula, Frankenstein, The Exorcist ... or Underworld, Constantine, Alien ... wait, is Alien classified as horror or not? -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 22:34, 8 October 2018 (UTC)
: Alien is quite surely a horror movie, I would say. Aliens (=the second movie) sets a different tone and includes elements of an action movie. Sebastian --[[Special:Contributions/172.68.110.34|172.68.110.34]] 06:47, 9 October 2018 (UTC)
: You'd like Sunshine then. Personally, even that had too much horror for me. --[[Special:Contributions/162.158.38.196|162.158.38.196]] 11:17, 9 October 2018 (UTC)

If we take it as reverse analogy, the title text hints he didn't liked/watched the part where Jeff tried to convince the executives to build the park in the original movie, as it was observed as a Horror part for him. --[[User:Nachuo|Nachuo]] ([[User talk:Nachuo|talk]]) 06:41, 9 October 2018 (UTC)

I don't think the explanation should say that Randall is the one talking in the title text. I have to assume that that is still a conversation between Cueball and White Hat. I doubt Randall is talking to himself there... [[Special:Contributions/162.158.78.130|162.158.78.130]] 14:26, 9 October 2018 (UTC)
:You are right, it's a conversation between Cueball and White Hat. But as it's explained before Cueball represents Randall, not uncommon at xkcd. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:59, 9 October 2018 (UTC)

Haha, I'm totally with Randall on this one. I just don't get horror movies. ¯\_(ツ)_/¯ --[[User:Sensorfire|Sensorfire]] ([[User talk:Sensorfire|talk]]) 17:40, 9 October 2018 (UTC)
: There's actually some serious research which had been done. In most case the pleasant sensation some people experience, comes from "watching terrible things happen to *other* people". i.e.: it's due to their brains suddenly realizing that *they* did not get any horrible thing. It hapenned to someone else and they managed to "survive" it. [[Special:Contributions/162.158.111.133|162.158.111.133]] 17:26, 10 October 2018 (UTC)
::It's been known since the 1950s that the physical reaction is the same to horror, love, and other excitement, and that the mind doesn't automatically make the distinction. It's been speculated that watching a horror movie with someone else evokes attachment.

Yeah, I agree too, don’t much like horror movies, but have been intending to watch Jurassic Park for the science and dinosaurs, despite the horror theme. I think the part of the book that stuck with me the most was the idea of “what could they possibly be doing with THREE CRAY 1 SUPERCOMPUTERS?!” 😜 [[User:PotatoGod|PotatoGod]] ([[User talk:PotatoGod|talk]]) 18:41, 9 October 2018 (UTC)

Why am I getting deja vu? Is there a similar xkcd comic or a similar joke on another comic?
[[Special:Contributions/172.69.134.9|172.69.134.9]] 01:37, 10 October 2018 (UTC)

A recent edit changed it from Cueball being sarcastic in his supposed enthusiasm for horror films, to one where he is a genuine fan. I believe this is incorrect, and he really is being sarcastic. But I didn't want to change it back without some discussion. Who thinks Cueball is a fan, who thinks he's just being sarcastic, and who thinks we should try to include both theories in the explanation? [[User:N0lqu|-boB]] ([[User talk:N0lqu|talk]]) 13:58, 10 October 2018 (UTC)
:I'm coming around to the idea that he's a real fan. Since I'm not, I just assumed he was being sarcastic. :-) I've added that possibility and tweaked it a little. [[User:N0lqu|-boB]] ([[User talk:N0lqu|talk]]) 14:28, 10 October 2018 (UTC)
:I believe your first impression was correct - to me, the caption reinforces the idea that the response in the comic panel is a sarcastic answer to White Hat. [[User:Ianrbibtitlht|Ianrbibtitlht]] ([[User talk:Ianrbibtitlht|talk]]) 14:58, 10 October 2018 (UTC)
:I'm 100% certain that Cueball is being sarcastic. [[User:Tarcas|Tarcas]] ([[User talk:Tarcas|talk]]) 15:44, 10 October 2018 (UTC)
:He's being sarcastic, no question. Him being sarcastic fits in with the caption. AND the title text, sarcasm means the conversation is simply continuing in the title text. Plus Cueball is generally the "on-screen" representation of Cueball, him being sarcastic would be Randall being sarcastic, which, again, fits in with Randall's sentiment in the caption and title text. :) You've fallen prey to the usual ExplainXKCD fallacy of introducing doubt and uncertainty where there is none, LOL! And I SO agree with Randall on this comic, and I have the added thing that horror movies don't work on me anyway (even if I WAS interested in getting scared), outside of cheap jump scares. [[User:NiceGuy1|NiceGuy1]] ([[User talk:NiceGuy1|talk]]) 05:24, 12 October 2018 (UTC)
:In this case Cueball doesn't represent Randall but a typical horror fan; his statement illustrates Randall's perception of the fundamental absurdity of horror fandom. He's a nice guy who can't imagine taking pleasure in other people's distress or the sight of gore.[[Special:Contributions/172.69.54.75|172.69.54.75]] 22:10, 28 October 2018 (UTC)

The idea of breeding an island full of killers is actually a good description of the horror movie The Brood.[[User:JeroenRL|JeroenRL]] ([[User talk:JeroenRL|talk]]) 14:10, 10 October 2018 (UTC)

I don't really agree with the section that says Randall admits he's criticising from a position of ignorance. He's shown that he's not ignorant of horror movies, he merely doesn't understand why they're popular. [[Special:Contributions/172.69.55.166|172.69.55.166]] 15:22, 10 October 2018 (UTC)

"Death, uh... finds a way." -Ian Malcolm, Serial Killer Park --Electro-- [[Special:Contributions/162.158.186.192|162.158.186.192]] 05:28, 17 October 2018 (UTC)

On the strange popularity of horror movies: a morbid fascination with violent events makes sense in the dangerous ancestral environment. People who listened with rapt attention to stories of terrible things happening to other people would have been more likely to avoid or survive them. The guy who said "I don't care to hear about the lion attack in such gruesome detail. It was a terrible thing that happened and it makes me feel afraid," was less likely to escape the next attack, and less likely to have become your ancestor. Today's entertainment is yesterday's education. [[User:Bob Stein - VisiBone|Bob Stein - VisiBone]] ([[User talk:Bob Stein - VisiBone|talk]]) 14:57, 28 October 2018 (UTC)