216: Romantic Drama Equation

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Romantic Drama Equation

Title text: Real-life prospective-pairing curves over things like age can get depressing.

Explanation

This explanation may be incomplete or incorrect: The formulars are missing If you can address this issue, please edit the page! Thanks.

The equations in the comic and the graph show how many different love pairs can be made if you know the number of females and males in a group. The text explains that it was inspired by TV Romantic Drama (in this case, the gay drama Queer as Folk), 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.

It should be noted that the chart assumes that the ENTIRE cast, male AND female, will ALL be of the same sexuality (homo OR hetero).

The graph makes a note that it only holds true for large casts. Case in point, with a cast of only four people: a two-to-two female-to-male ratio will have four straight pairings to two gay pairings, while a three-to-one female-to-male ratio will have three straight pairings and three gay pairings.

The title-text mentions that Randall made a chart of his own prospective dating pool as he gets older, and was depressed by the results.

The formulas may be derived as follows:

Each straight couple needs to include one of the x males and one of the (n-x) females so there are x(n-x) possible ways of combining one of each. E.g., if there are n=5 people, of whom x=2 are male, then there will be 3 possible pairings involving the first male, and three possible pairings involving the second yielding 2(5-2)=6 possible pairings.

Each gay couple needs to include either two males or two females. To choose two males, we can start with any of the x males and choose any of the (x-1) remaining males. However, that counts each possible pairing twice. E.g., Adam&Steve got counted when we chose Adam first and Steve second, and again when we chose Steve first and Adam second. To avoid double counting the possible couples, we therefore need to divide that total by 2. So there are x(x-1)/2 possible male-male pairings. Similar reasoning involving the (n-x) females tells us that there are (n-x)(n-x-1)/2 possible female-female pairings. Multiplying these out and combining the male and lesbian couples together, we get the total number of possible gay couples is [x^2 - x + n^2 - nx - n - xn + x^2 + x]/2. That simplifies to [n^2 - n + 2 x^2 - 2 xn]/2. The left two terms can be combined together as n(n-1) and the right two terms can be combined together as 2x(x-n). Since the sum of these terms was divided by 2, we get that the total number of possible same-sex pairs is n(n-1)/2 + x(x-n), which is what the cartoon says.

Transcript

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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Discussion

This can't be right, even at 50/50, the number of gay pairings far outnumbers the number of straight pairings.80.235.105.134 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) 122.200.61.203 (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/2 141.101.98.229 (talk) (please sign your comments with ~~~~)

The intuitive explanation for this is that if there are equally many men and women (i.e., x = n/2), each individual can pair with (n/2 - 1) others of the same gender, but with n/2 of the opposite gender. So each individual has 1 more pairing with the opposite gender than with the same gender. Taken across the population, that leads to a difference of n/2. 172.71.126.145 10:13, 11 March 2024 (UTC)

There is a typo in his formula for gay casts. The + should be a -. 199.27.128.159 (talk) (please sign your comments with ~~~~) No, he's right. Notice the x-n term. x<n, so x-n is negative.108.162.215.61 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. 162.158.23.191 (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 172.68.78.100 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. 172.68.26.251 04:15, 21 March 2017 (UTC)

I think a better way to explain the gay pairing equation would be to look at it like this: n(n-1)/2 are all possible pairings, since you take each person(n), pair them with everyone except themselves(multiply by n-1) and then divide by 2 to eliminate pairing the same people twice. Then substract x(n-x) which is the equation for all straight pairings. - x(n-x)=x(x-n), hence the second part of this equation.(Hope I'm editing correctly)

It actually comes out that if each individual's gender is chosen randomly, the expected value for straight casts and for gay casts is the same (not too hard to prove via induction).