Talk:216: Romantic Drama Equation

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).