Probability that a given Poisson variable samples greater than its mean $\lambda$, provided $\lambda > D$ Given a random variable $X \sim \text{Poisson}(\lambda)$ such that $\lambda > D$, with $\lambda, D \in \mathbb{N}$, what is the probability that a sample obtained from $X$ is greater than $\lambda$?
In other words, what is the value of $\mathbb{P}(X > \lambda)$?
I think that calculating the probability of $X$ being greater than the said value $D$ analysing this for every possible mean greater than $\lambda$ is a possible path to follow, but I'm not sure of how difficult to calculate this would be.

Poisson's CDF states that, for $X \sim \text{Poisson}(\lambda)$ and $k \in \mathbb{N}$, $$ \mathbb{P}(X \leqslant k) = \mathsf{e}^{-\lambda} \sum\limits_{i = 0}^{k} \frac{\lambda^i}{i!} $$
So, the value I'm asking should be the summation of the above probability's complement from $D+1$ to infinity, or even $$ 1 - \sum\limits_{p = 0}^{D} \left( 1 - \mathsf{e}^{-p} \sum\limits_{i = 0}^{D} \frac{p^i}{i!} \right) $$
Is this the correct value? I already know that $\lambda > D$, so maybe there should be some conditional probability involved, but I'm not sure.
If this is correct, what I'm asking is if there isn't a more concise way to calculate this, with less summations or none at all. This is because I'll need to calculate this value extensively in a computer program and computational time is very, very precious.
 A: I am unsure where you use $D$. For a Poisson-distributed random variable:
$$
P(X > x) = 1 - P (X \leq x) = 1 - \sum_{i=0}^x \frac{e^{-\lambda}\lambda^i}{i!}
$$
It does not matter what $x$ is as long as $x$ is a nonnegative integer. If $x = \lambda$ the answer is still:
$$
1 - \sum_{i=0}^\lambda \frac{e^{-\lambda}\lambda^i}{i!}
$$
Unless I have misunderstood you.
Update - 2018-10-08
It climbs to $0.5$ very slowly. Here is some R code that will calculate entries. The first function calculates a single entry, the second will calculate a sequence for plotting.
PGTMean <- function(lambda){
  q <- seq_len(lambda + 1) - 1
  1 - sum(exp(-lambda + q * log(lambda) - lgamma(q + 1)))
}
PGTMSeq <- function(n = 100){
  L <- seq_len(n)
  vapply(L, PGTMean, double(1))
}

For the sequence $10^0, 10^1, \cdots, 10^8$ the values of the Poisson being greater than the mean are:
0.2642411 0.4169602 0.4734378 0.4915906 0.4973404 0.4991590 0.4997340 0.4999159 0.4999734

A plot of the first 10,000 entries is below.

