When is the hypergeometric function ever needed if it can be computed so much more efficiently?

Question

So I recently tried to compute some probabilities, and through my journey learned all about the hypergeometric distribution. I finally learned how to calculate the probability of having at least one red and at least one black ball in a draw of $4$ balls out of an urn filled with $4$ red, $4$ black and $52$ white balls.

$$\sum_{r=1}^4 \sum_{b=1}^4 \frac{\displaystyle{4 \choose r}{4 \choose b}{52 \choose 7-r-b}}{\displaystyle{60 \choose 7}}$$

But then someone pointed out to me that this probability can be computed much more efficiently $($obviously after I finally figured out how to implement the above in my preferred programming language$)$, like this:

$$1-2~\frac{\displaystyle\binom{56}{7}}{\displaystyle\binom{60}{7}}+\frac{\displaystyle\binom{52}{7}}{\displaystyle\binom{60}{7}}$$

So, I am wondering, why or when would I ever prefer to compute the actual probability $($hypergeometric function$)$ instead of trying to compute the probability of the event$($s$)$ not happening, and simply subtract from $1$.

Answer 1

You are extremely narrow-focused ! Look, you have basically discovered the following identity:

$$\sum_{n=1}^\infty \sum_{k=1}^\infty{a \choose n}{b \choose k}{c \choose s-n-k}={a+b+c \choose s}+{c \choose s}-{a+c \choose s}-{b+c \choose s},$$

where $s=a+b-1$. Now, obviously, this is not “computationally efficient” for whatever

“real-life application” you can currently think of $($because, you know, that's what math is

really all about: computational efficiency, and real-life applications $\ldots)$ And indeed, giving

positive integer values to a, b, c is very “combinatoricky”, and very “probabilistic”, and very...

dull, boring, not to mention completely uninteresting... Now, if you really want to impress

the girls, you have to write one of those weird $\pi$ series, that leaves anyone and everyone

simply awe-struck... $~$ “And just how exactly are we going to do that ?”, you might ask. Well,

by giving fractional values to our three parameters. Try, for instance, $a=b=\dfrac12$ , and $c\in\mathbb N$.

You will end up with some pretty cool infinite series expressions for numbers of the form

$p+\dfrac q\pi$ , where $p,q\in\mathbb Q.~$ For instance, using the fact that $\displaystyle{1/2\choose n}=-\frac{\displaystyle{2n\choose n}}{(-4)^n(2n-1)}~,~$ we

arrive at $~\displaystyle\sum_{n=0}^\infty\bigg[\frac1{4^n(2n-1)}{2n\choose n}\bigg]^2=\frac4\pi$ . Computationally efficient ? Heck, no ! Impressive ?

Hell, yes ! ;-$)$