1. Calculating enormous numbers

I need to find P. The answer must be between 0 and 1.
Microsoft Excel got stuck with these numbers.

2. Originally Posted by totalnewbie
I need to find P. The answer must be between 0 and 1.
Microsoft Excel got stuck with these numbers.
$P = \frac{ \frac{p^L}{L!} }{ \sum_{K = 0}^L \frac{p^K}{K!} }$

Let's simplify this thing somewhat:
$P = \frac{ \frac{p^L}{L!} }{ \sum_{K = 0}^L \frac{p^K}{K!} } \cdot \frac{ \frac{L!}{p^L} }{ \frac{L!}{p^L} }$

$P = \left ( \sum_{K = 0}^L \frac{p^K}{K!} \cdot \frac{L!}{p^L} \right )^{-1}$

$P = \left ( \sum_{K = 0}^L \frac{(L - K)!}{p^{L - K}} \right )^{-1}$

Now, let n = L - K:
$P = \left ( \sum_{K = 0}^L \frac{(L - K)!}{p^{L - K}} \right )^{-1} = \left ( \sum_{n = L}^0 \frac{n!}{p^n} \right )^{-1}$

$P = \left ( \sum_{n = 0}^L \frac{n!}{p^n} \right )^{-1}$

This still looks pretty bad. But note that $\{ a_n \} = \frac{n!}{p^n}$ 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
$P \approx 0.998812$.

-Dan

3. Here is my attempt at persuading excel to solve the question. I got ~.76056

bignum.xls

4. We can't say that L!/K! = (L-K)! because L!/1 is not equal to (L-1)! while (0!=1!=1).

5. Originally Posted by totalnewbie
We can't say that L!/K! = (L-K)! because L!/1 is not equal to (L-1)! while (0!=1!=1).
I am completely baffled where such a relation came from.

6. Originally Posted by colby2152
I am completely baffled where such a relation came from.
topsquark said that let n = L - K.