Explain xkcd: It's 'cause you're dumb.
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Revision as of 10:38, 3 December 2012
The equations in the comic and the graph show how many different love pairs can be made if you know number of females and males in a group. The text explains that it was inspired by TV Romantic Drama, but of course the formula is valid for any group of people. There are two graphs and equations - gay option is the case when we are looking for pairs with same gender, straight option in for heterosexual equations. The interesting/funny part about the results is that in most cases there are more possibilities when we consider the homosexual option. Also it is interesting to observe what is kind of obvious - in the heterosexual case the "best" case is if both genders are present equally and the possibilities drop very fast if there is substantial difference between genders.
- TV Romantic Drama Equation (Derived during a series of "Queer as Folk" episodes)
- [A table shows equations for possible romantic pairings in a TV show. The equation under "gay" is n(n-1) = 2+x(x-n); the equation under "straight" is x(n-x).]
- x: Number of male (or female) cast members.
- n: total number of cast members.
- [A graph plots pairings (for large casts) against cast makeup. Each of the above equations forms a curve. "Gay cast" starts high for an all male cast, dips down at 50/50 cast makeup, and then rises again for all female. "Straight cast" starts at zero for an all male cast, peaks at 50/50 cast makeup, and then drops to zero again for an all female cast. The two curves intersect at two points close to the middle.]
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This can't be right, even at 50/50, the number of gay pairings far outnumbers the number of straight pairings.18.104.22.168 20:10, 28 February 2013 (UTC) Moved from article page
- Not quite. Consider a cast of 4 with 2 male (A, B) and 2 female (C, D). Possible gay pairings - 2 (A-B and C-D). Possible straight pairings - 4 (A-C, A-D, B-C, B-D) 22.214.171.124 (talk) (please sign your comments with ~~~~)
- He says for large casts. For 2000 cast members, with 1000 of each gender, the gay couplings comes out at 999,000 and straight at 1,000,000. Presumably this is the small cross over the diagram alludes to. If you substitute x = n/2 into the equations, then you get (n^2-2n)/4 for the gay combinations and n^2/4 for the straight combinations, so for gender balanced cast size of n, the straight combinations outnumber the gay by n/2 126.96.36.199 (talk) (please sign your comments with ~~~~)
There is a typo in his formula for gay casts. The + should be a -. 188.8.131.52 (talk) (please sign your comments with ~~~~)
No, he's right. Notice the x-n term. x<n, so x-n is negative.184.108.40.206 03:14, 2 March 2014 (UTC)
This also the small implication that "Queer as Folk" was so dull that Randall produced this equation to occupy his mind during it. I often find my mind wandering while sat watching soaps with my other half. Drmouse (talk) 14:20, 3 January 2014 (UTC)
The first equation can also be understood more simply as the total number of possible pairings, minus the number of straight ones. 220.127.116.11 (talk) (please sign your comments with ~~~~)
- Good point! I wonder where exactly that small crossover region should be. n(n-1)/2 - x(n-x) = x(n-x) so n(n-1)/2 = 2x(n-x) so n(n-1) = 4x(n-x). Hm, he said for large casts, so I suppose Randall's making approximations based on the limit. As n -> infinity, n(n-1) -> n^2, and as x -> n/2, 4x(n-x) -> 2 n(n/2) which is also n^2. So it makes sense that the crossover region gets closer to just being one point at n/2. But can we calculate an exact trend? n(n-1)=4nx-x^2, so x^2-4nx+n(n-1)=0, so x=[4nx±sqrt(16-4(n)(n-1))]/2, so x(1-2n)=x-2nx=±sqrt[16-4(n)(n-1)]/2=±sqrt[4-n(n-1)], so x=±sqrt[4-n(n-1)]/(1-2n). Also, that can't possibly be right because it would give a negative answer but whatever, it's late, I think I did the approximated math right so that's good enough 18.104.22.168 05:53, 21 June 2017 (UTC)
I think all genders being constant isn't really an assumption of the graph. Obviously the graph only works for a single moment in time in a TV show, since the cast changes over time with the plot of the show (such as when people die in the show). The graph already needs to be re-drawn every time someone enters or leaves the cast. For the data we're tracking, a sex change operation is the same as, for example, a man leaving the show and a woman subsequently entering it. Sure, you could then also say that the cast being constant is an assumption of the graph, but that's not really accurate either. The graph simply doesn't observe the passage of time. You'd have to add a time axis for that, making the graph three-dimensional. 22.214.171.124 04:15, 21 March 2017 (UTC)