Probability calculation with large numbers I do have a probability measure:
$P = 1 - \dfrac{k!\, \binom{2^{32}} {k}}{(2^{32})^k}$,
where $k$ is an positive integer. 
Yet, I do have trouble evaluating it in terms of a numerical plot, as the values are too large. Even simplifications by expanding the binomial coefficient didn't help. I'd be interested in a plot of P over k and of a numerical result for 
$P(k) \overset{!}{=} 0.5$.
 A: $$
1 - \frac{k!\left(\matrix{2^{32}\\k}\right)}{\left(2^{32}\right)^k}
$$
lets expand we find
$$
1 - \frac{k!\frac{2^{32}!}{k!\left(2^{32}-k\right)!}}{\left(2^{32}\right)^k} = 1 - \frac{2^{32}!}{\left(2^{32}\right)^k\left(2^{32}-k\right)!}
$$
notice that
$$
\frac{2^{32}!}{\left(2^{32}-k\right)!} = 2^{32}\left(2^{32}-1\right)\cdots\left(2^{32}-(k-1)\right)
$$
matching the terms with the numerator
$$
P = 1 - \frac{2^{32}\left(2^{32}-1\right)\cdots\left(2^{32}-(k-1)\right)}{\left(2^{32}\right)^k} = 1 - 1\cdot\left(1-\frac{1}{2^{32}}\right)\cdots\left(1 -\frac{(k-1)}{2^{32}}\right)
$$
thus you get
$$
P = 1 - \prod_{i=1}^{k-1}\left(1-\frac{i}{2^{32}}\right)
$$
I am still working on the last part, namely 
$$
\prod_{i=1}^{k-1}\left(1-\frac{i}{2^{32}}\right) = ?
$$
Going by your preliminary work it should be $0.5$ but I need to see it from the above. 
A: Using $1-a=\exp(\ln(1-a))\approx\exp(-a)$ for small values of $a$ and
$$
1+2+…+(k-1)+k=\frac{k(k+1)}2\approx \frac{k^2}2
$$
one gets
$$
P(k)=1-\prod_{j=1}^{k-1}\left(1-\frac j{2^{32}}\right)\approx 1-\exp\left(-\frac{k^2}{2^{33}}\right)
$$
Solving this for $P(k)=0.5$ results in the approximation
$$
k\approx \sqrt{-\ln(1-0.5)·2^{33}}=77 162.74…
$$
