Binomial probability for large $n$, small $p$ 
*

*I need to compute the probability of getting more than $x$ "successes" in a large number of trials $\left(\,10^{11}\,\right)$ of an event with a small probability $\left(\,10^{-7}\,\right)$.

*Exact Binomial won't work, and the Poisson approximation does not seem appropriate. 


Thanks.
 A: As requested in comments:
You could use R: for example the probability of being strictly more than $9876$ could be about 
> pbinom(9876, size=10^11, prob=10^-7, lower.tail=FALSE)
[1] 0.8917494

This compares with the normal approximation with continuity correction of being above $9876.5$ giving the close
> 1 - pnorm((9876.5 - 10^11 * 10^-7)/sqrt(10^11 * 10^-7 * (1-10^-7)))
[1] 0.8915848

In either case the standard deviation is close to $100$ so here only values close to something like $10000 \pm 300$ will give interesting probabilities
A: I'm guessing that if the Poisson approximation does not seem appropriate, it's because the expected value and the variance (which, for the Poisson distribution, is the same as the expected value) are so big. In that case, approximating it by a normal distribution can serve.
If $X\sim\operatorname{Poisson}(\lambda)$ then the distribution of $\displaystyle \frac{X-\lambda}{\sqrt{\lambda}}$ approaches the standard normal distribution (i.e. the normal with expected value $0$ and standard deviation $1$) as $\lambda\to\infty$. In your case, you have $\lambda=10^4,$ which is plenty. If $X\sim\operatorname{Poisson}(10^4)$ then, for example,
$$
\Pr(9910 \le X \le 10050) = \Pr\left( \frac{9910-10000}{\sqrt{10000}} \le \frac{X-10000}{\sqrt{10000}} \le \frac{10050-10000}{\sqrt{10000}} \right)
$$
$$
\approx \Pr\left( \frac{9910-10000}{\sqrt{10000}} \le Z \le \frac{10050-10000}{\sqrt{10000}} \right) = \Pr(-0.9 < Z < 0.5)
$$
where $Z\sim N(0,1).$
