Originally Posted by

**Swammerdami**
... when 20,000,000 >> 20,000 >> **53 >> 1**, where ">>" denotes "MUCH greater than" and pk is probability of exactly k hits, ... I won't attempt to prove the conjecture pk ~= (1/k!) *** p0** * (1-p0)^k

[Note corrections in red]

And as soon as I stepped away from keyboard, proof became trivial!

Let's substitute s = 20 million; w = 20 thousand. We'll leave 53 alone.

Given s >> w >> 53 >> 1,k we seek to prove that pk = C(53,k) C(s-53,w-k)) / C(s,w)

is approximated with pk ~= p0 (1 - p0) ^ k / k!

Change the C(.) to factorials: pk ~= 53! (s-53)! w! (s-w)! / (w-k)! (s-53-w+k)! s! k! (53-k)!

Change a! / (a-b)! to the approximation a^b whenever a >> b; and rearrange a bit to get k! pk ~= 53^k w^k (s-w)^53 / s^53 (s-w)^k

Solve for p0 to get p0 ~= (s-w)^53 / s^53 and recall that, when s >> w >> 53, p0 ~= 1 - 53w/s or 1 - p0 ~= 53w/s

Observing that (s/(s-w))^k ~= 1, substitutions now produce k! pk ~= p0 (1-p0)^k

Q.E.D.