I need to find P. The answer must be between 0 and 1.
Microsoft Excel got stuck with these numbers.
$\displaystyle P = \frac{ \frac{p^L}{L!} }{ \sum_{K = 0}^L \frac{p^K}{K!} }$
Let's simplify this thing somewhat:
$\displaystyle P = \frac{ \frac{p^L}{L!} }{ \sum_{K = 0}^L \frac{p^K}{K!} } \cdot \frac{ \frac{L!}{p^L} }{ \frac{L!}{p^L} }$
$\displaystyle P = \left ( \sum_{K = 0}^L \frac{p^K}{K!} \cdot \frac{L!}{p^L} \right )^{-1}$
$\displaystyle P = \left ( \sum_{K = 0}^L \frac{(L - K)!}{p^{L - K}} \right )^{-1}$
Now, let n = L - K:
$\displaystyle P = \left ( \sum_{K = 0}^L \frac{(L - K)!}{p^{L - K}} \right )^{-1} = \left ( \sum_{n = L}^0 \frac{n!}{p^n} \right )^{-1}$
$\displaystyle P = \left ( \sum_{n = 0}^L \frac{n!}{p^n} \right )^{-1}$
This still looks pretty bad. But note that $\displaystyle \{ a_n \} = \frac{n!}{p^n}$ is a convergent sequence (that converges rather rapidly to 0.) So all we need to do to get an estimate of this number is take the first several terms. Taking the first 50 terms I get that
$\displaystyle P \approx 0.998812$.
-Dan
Here is my attempt at persuading excel to solve the question. I got ~.76056
bignum.xls