ASSUME: This is strictly one dude. X girls.
Initially i figured it was a simple function
Mathematically speaking:
We will make Y(X) a determinitive function to describe the type of sexual actvity
Let X=the number of girls
We will also assume that the male involved as well as each female is invariable. That is, that fatties still only count as one person. You can’t be 400lbs and count yourself as two people here (same goes for the girls). IE: you can’t bang a fatty and say that you’ve had a 3 way. This also precludes a 450lb version of skunkape from halfheartedly masturbating and claiming he actually had a “twosome” of sorts.
Thus: let M = the person solving the problem = 1
Now, that said i would simply think that
Y(X) = M + X
since M=1
Y(X) = 1 + X
Now here is where the personal opinion comes in. It is my opinion that anytime there are more than 5 people having sex at once, it’s an orgy, such that:
Y(X) >= 5 = Orgy
and
Y(X) < 5 = X*some
Rearranging:
Y(X) = Orgy for X>=4
and
Y(X) = X*some for X<4
However, after some creative input from Kcuv, it was decided that another variable was required. His understanding is that an orgy is actually defined by multiple people having sex in the same room, wheres an X*some could stretch to X=infinity. Using this assumption:
Let F = The number of people who M is linked to via a sexual act at any one time. To maximize bragging rights, we will let Fmax = the maximum number of people linked during the act in question.
The following equation was derived:
Y(X) = (M + X)F^-1
Rearranging:
Y(X) = (M + X)/(F+1)
Using this logic
Y(X) = 1 would result in the only admissible X some. All else would tend towards orgy as F approached 0.