explain xkcd - User contributions [en]

Talk:246: Labyrinth Puzzle

2021-05-02T23:42:18Z

172.68.37.38:

'''SOLUTION 2 DOES NOT WORK''' ONE QUESTION, NOT TWO, AND YOU DONT KNOW WHO THE LIAR IS. one answer of "this is the door" doesnt help beacuse you wont know the answer of the other guy. in which solution one will always get you the same door, and you pick the other one. solution 2 just gives you two different doors. no matter who you ask you will randomly get a door. if the guy says the truth you wont know that hes the truthfull one or not. AGAIN thruthful guy points to a door, the liar would have( if you asked him instead, OR could even ask both) pointed to the other door. again, asking that question does not give you two answers, because you. dont. know. who. the. liar. is.

'''SOLUTION 2 DOES WORK'''
Solution 2 does work. It IS one question. The question consists of a premise and a concluding statement "If I were to ask you ..." being the premise and "What would your answer be?" being the concluding statement of the question. You are not asking two questions. You are asking ONE question about another question. It may be a logical dodge but it is the SAME logical dodge as the first solution, except that it is asking them about themselves rather than about their brother which takes the form "If I was to ask your brother..." "what would his answer be" In many ways it's neater and more logical and does NOT require the brothers to know how the other would answer.

Solution 1: In mathematical terms takes the form (-1) × (+1) or (+1) × (-1) the result of both being =-1 however for it to work we have to assume that the brother being asked knows what his brother would say. However, nowhere in the question is this usually stated. So we're actualy dealing with either
a × (+1) or a × (-1) with a being an unknown variable which is either +1 or +1 or 0 (zero denoting that the brother doesn't actually know). This is a truer representation of the kind of answer you could expect to get from Solution 1.

Solution 2: As you are asking them about themselves then you are multiplying by themselves in mathematical terms so either (-1)×(-1) or (+1)×(+1). Multiplying either a positive or negative by itself will result in a positive. The brother doesn't need to know what he would answer, he just needs to know what answer HE would give IF he was asked the question.

[[User:SpiroExDeus|SpiroExDeus]] ([[User talk:SpiroExDeus|talk]]) 14:24, 19 November 2020 (UTC)

Just ask which color is the sky.. {{unsigned|‎175.110.37.200}}
:Oh, although the strip doesn't explicitly say so; in those riddles you can normally only ask one question. --[[User:St.nerol|St.nerol]] ([[User talk:St.nerol|talk]]) 23:00, 27 January 2013 (UTC)
::There's another (more traditional) three-guard variation where one guard always tells the truth, one guard always tells a lie and the third alternates between pure truth and pure lie (and you don't know which flip they're currently flopped upon). But you ''still'' only get to ask one question of one guard. Have fun with that one. My personal solution certainly has a degree of convolution, but I've heard other workable answers. [[Special:Contributions/178.98.31.27|178.98.31.27]] 02:24, 21 June 2013 (UTC)
::: I hope no-one considers this a spoiler to say but there is a trivial solution to the 3 guards problem, whether third guard is a spear handler or one that flips between truth and false hood, just try "Did you know that the pub in the village behing the freedom door is serving free beer all day, as long as their stocks last?" Then follow the guards through whichever door they use. Alternatively substitute beer for another commodity the guards may desire. [[Special:Contributions/141.101.98.175|141.101.98.175]] 00:59, 7 November 2016 (UTC)
:::@‎175.110.37.200, you would know which one lies but you would not know which door leads out. [[User:Tharkon|Tharkon]] ([[User talk:Tharkon|talk]]) 23:13, 10 October 2013 (UTC)
::::Eh, well, even if you had a perfect question to ask in this case, a lot of good would that do you: it'd only reveal the truth behind the setup, that ''none'' of the doors lead out. :p -- [[Special:Contributions/173.245.51.210|173.245.51.210]] 08:20, 8 November 2013 (UTC)
:::::Well yes it says that in the title-text. But good pick-up. [[Special:Contributions/108.162.219.58|108.162.219.58]] 02:31, 6 February 2014 (UTC)

One question, of one guard. I really like the original form of this riddle. It's a bit of a trick, though. It is crucial that the guards "know" each other's rules, but this is not even implied. And if it was stated in the question, that would probably be a good enough clue to get you to the answer. Of course, once you know the answer it seems trivial, but I wonder what percentage of people actually worked it out for themselves? Another good one is Monty Hall, even though that is pure, straightforward probability. [[Special:Contributions/108.162.219.223|108.162.219.223]] 18:11, 17 January 2014 (UTC)

:With two guards, they wouldn't need to know each others role. If they know their own role - which they do - each can infer the role of the other. [[Special:Contributions/162.158.34.137|162.158.34.137]] 13:01, 21 April 2016 (UTC)

:I think somebody needs a hug! [[Special:Contributions/108.162.219.223|108.162.219.223]] 18:11, 17 January 2014 (UTC)

The whole problem with this entire riddle is that if they are both liars you are screwed! Nothing in the riddle establishes a fact that they aren't liars. Now if there was a known truth teller in the riddle that explains the nature of the guards or the narrator does it, then the above solution works. {{unsigned ip|108.162.216.28

As you aren't given a limit to the number of questions, you can just ask each guard if they're the stabby guard. If two say yes, the third one is the truthful guard and you can ask him which way the exit is. If two say no, the third one is the lying guard and you can ask him where the exit isn't. No tricky questions so the stabby guard shouldn't stab you.[[Special:Contributions/162.158.255.195|162.158.255.195]] 18:14, 14 August 2015 (UTC)

I have a solution, but you need to ask multiple questions: 
''If the Stab Guard tells the truth:'' 
Ask each guard, firstly, "Are you the Stab Guard?" 
Truth Guard will answer "No." 
Stab Guard will answer "Yes." 
Liar Guard knows the answer is no, but, because he lies, will answer "Yes." 
The one who said no is the Truth Guard, so you can ask him which door leads to freedom. 
''If the Stab Guard lies:'' 
Point to the guard on the left, and ask each guard, "Does that guard lie?" 
If that guard is Truth Guard, then Truth Guard will answer "No," while Stab Guard and Liar Guard answer "Yes." 
If that guard is a liar, then Truth Guard will answer "Yes," while Stab Guard and Liar Guard answer "No." 
Whichever guard gives a unique answer is Truth Guard, so you can ask him which door leads to freedom. [[User:NickOfFørvania|NickOfFørvania]] ([[User talk:NickOfFørvania|talk]]) 23:37, 3 November 2015 (UTC)

:It's been solved on puzzling.stackexchange.com (given a specific definition of a tricky question): http://puzzling.stackexchange.com/questions/43092/xkcd-inspired-logic-puzzle [[Special:Contributions/141.101.98.130|141.101.98.130]] 12:14, 24 September 2016 (UTC)

I remember a book where the main character kicked one guard in the face and asked if it hurt. {{unsigned ip|162.158.252.137}}

There's another unspoken rule: That the lie is either a yes or a no. If you asked the liar something, he could lie and say, "I don't know," which would leave you with nothing. Also, as Stabby MacStabberson does not appear to have any restrictions on what he tells you (that is, he has the choice between truth or falsehood,) there's no sure way out even if he wasn't tasked with stabbing you.[[Special:Contributions/162.158.255.69|162.158.255.69]] 05:32, 30 April 2016 (UTC)

Who said the Stab Guard has a true sense of complex? He could just stab you anytime. [[User:Dontknow|Dontknow]] ([[User talk:Dontknow|talk]]) 00:37, 16 March 2017 (UTC)

What's wrong with "Are you gonna stab me?"? They'll either answer or not and stab you or not, that's around 2 bits, which seems like it should maybe help decide in a space of 3. I ask Lie, he says yes, yet doesn't stab me. I ask True, he says no, and doesn't stab me. I ask Stabby, and he says no and doesn't stab me because it's a really simple question. [[Special:Contributions/108.162.216.48|108.162.216.48]] 16:44, 29 December 2020 (UTC) with [[Special:Contributions/108.162.216.48|108.162.216.48]] 16:44, 29 December 2020 (UTC)

Here's a solution, but it's quite stupid, and might not work, depending on if the guards appreciate their mothers:
Just ask a random guard, point to another, and say this: "That guard told me your mom was wearing their shirt last night!"
Then stand back, and let the chaos ensue.[[Special:Contributions/108.162.216.126|108.162.216.126]] 16:14, 25 November 2019 (UTC)

"What would you say is the way out?" It doesn't seem tricky, but it's actually a meta-question. The guard knows what they would say is the way out. If the guard tells the truth, they would say the correct door. The guard tells the truth about telling the truth and says the correct door. If the guard lies, they know they would lie about what the correct door is. The guard lies about lying and says the correct door. [[Special:Contributions/162.158.79.47|162.158.79.47]] 21:19, 3 February 2020 (UTC)

Assuming there are three doors (as shown in the picture), at least one door leads out, and at least one door does not, the single non tricky question of "Which doors lead out?" will always yield a useful answer. For two safe doors, if two door are pointed at, they are safe. if one door is pointed at, it's deadly. If only one door is safe, it's the opposite. If there are only two doors, (as in the standard puzzle) there's nothing you can door, as any question that both the liar and the truth teller would answer identically and tells you which door to go through is a tricky question. [[Special:Contributions/172.68.37.38|172.68.37.38]] 23:42, 2 May 2021 (UTC)

718: The Flake Equation

2021-03-05T04:43:36Z

172.68.37.38: /* Explanation */

{{comic
| number = 718
| date = March 24, 2010
| title = The Flake Equation
| image = the flake equation.png
| titletext = Statistics suggest that there should be tons of alien encounter stories, and in practice there are tons of alien encounter stories. This is known as Fermi's Lack-of-a-Paradox.
}}

==Explanation==
The Flake equation is a parody of the {{w|Drake equation}}, which is an estimate of the number of detectable extraterrestrial civilizations in our galaxy. The Flake equation, however, provides an estimate about how many false or fake stories ''about'' aliens are likely to exist. It does so in similar manner as the Drake equation which find the a number of detectable civilizations by multiplying the number of stars by consecutive probabilities of a star having certain characteristics making detectable life possible. Note that the Flake equation does not even attempt to include the number of accounts which are intentional lies.

The word "Flake" is informally used to describe a crazy person, such as someone who would imagine an alien encounter.

The Flake equation finds the number of fake stories by multiplying number of people by consecutive probabilities of a person having certain tendencies to tell such stories and other people telling them on and spreading them on the internet. Just like in the Drake equation, exact numbers are unknown, but can be estimated, and the equation in the comic shows [[Randall|Randall's]] guesses about these values. See an [[#Explanations of values|explanations of values]] below.

The final results tells us that there should be about 100,000 stories about aliens that have reliable explanation. (The numbers given in the equation gives 126,000 stories). The data is obviously highly uncertain, and as with the Drake Equation, you can plug in your own numbers, but if you keep your guesses realistic, you will most likely get a very large number. This convinces the reader that the fact that there are many stories about aliens does not necessarily means that many people actually met aliens.

The title text refers to Fermi's Lack-of-a-Paradox as a parody of the {{w|Fermi paradox}}: The contradiction between the high estimated probable existence of extraterrestrial civilizations and the lack of establishing contact to such civilizations by humans.

Another comic parodying this equation is [[384: The Drake Equation]]. The credibility of paranormal reports in general is revisited in [[1235: Settled]], which posits that if such phenomena were real they should have been unambiguously captured on camera by now.

===Explanations of values===
{| class="wikitable"
!Symbol
!Assumed value
!Explanation
|-

|WP
|7,000,000,000
|World population at the time of the creation of the comic, taken as a starting value.
|-
|(CR + Mi)
|1/10,000 + 1/10,000
|Fraction of people who would falsely believe they had been visited by aliens. This is assumed to be people who are crazy, want to feel special, or misinterpret physical phenomena as alien sightings. It is assumed that a total of one in 10,000 people to at least one of the first two of these groups and another one in 10,000 belong to the final group, for a total of one in 5000 belonging to one of these three groups. Multiplied by the world population we get the number of people who falsely believe to have been visited by aliens
|-
|TK
|1/10
|The fraction of people who believe they have experienced an alien sighting that tell others about their experience. Multiplying with the previous values we get the number of people who falsely believe to have experienced an alien sighting, and tell others about it.
|-
|F0
|10
|Average number of people they tell about their "sightings". Multiplying with the previous values we get the amount of people this is the amount of people who hear about a false alien sighting from the "primary source".
|-
|F1
|10
|Average number of people that they decide to tell about the "firsthand" account. Multiplying with the previous values we get the amount of people who hear a second-hand account of a false story.
|-
|DT
|9/10
|The probability that the details will be slightly adjusted during the retelling process, making the account believable. The total is now the amount of believable-yet-false alien sighting stories.
|-
|AU
|1/100
|The proportion of people who have the willingness and ability to share this story with the with a broad audience. The total is now the amount of believable-yet-false alien sightings that are published to a wider audience.
|}

==Transcript==
:The Flake Equation:
:P = WP × (CR + MI) × TK × F0 × F1 × DT × AU ≈ 100,000
:Where:
::WP = World Population (7,000,000,000)
::CR = Fraction of people who imagine an alien encounter because they're crazy or want to feel special (1/10,000)
::MI = Fraction of people who misinterpret a physical or physiological experience as an alien sighting (1/10,000)
::TK = Probability that they'll tell someone (1/10)
::F0 = Average number of people they tell (10)
::F1 = Average number of people each friend tells this "firsthand" account (10)
::DT = Probability that any details not fitting the narrative will be revised or forgotten in retelling (9/10)
::AU = Fraction of people with the means and motivation to share the story with a wider audience (blogs, forums, reporters) (1/100)
:Even with conservative guesses for the values of the variables, this suggests there must be a ''huge'' number of credible-sounding alien sightings out there, available to anyone who wants to believe!

{{comic discussion}}
[[Category:Math]]
[[Category:Astronomy]]
[[Category:Science]]
[[Category:SETI]]
[[Category:Statistics]]
[[Category:Paranormal]]

2393: Presidential Middle Names

2020-12-03T03:26:44Z

172.68.37.38: /* Explanation */

{{comic
| number = 2393
| date = December 3, 2020
| title = Presidential Middle Names
| image = presidential_middle_names.png
| titletext = The bottom of the list remains unchanged. Poor Rutherford Birchard Hayes.
}}

==Explanation==Mocking both parties and the media by predicting the win of presidential candidate Biden.

==Transcript==
{{incomplete transcript|Do NOT delete this tag too soon.}}

{{comic discussion}}