The probability of a die rolling a given number at least $x$ times over the course of $n$ rolls Say you have a $2000$-sided die, which you roll $2000$ times. I know the probability that you will get any given number (let's just say $1$) at least once in those $2000$ rolls is $1-.9995^{2000}$, i.e., $63.22\text{%}$. But how do you find the probability that you will roll a $1$ at least twice during a series of $2000$ rolls?
 A: If you are interested in the exact answer instead of an approximation, note first that the statement roll a one at least twice is the opposite of roll a one at most once.
It is much easier to calculate the latter probability than the former.
$P(\text{at least twice}) = 1 - P(\text{at most once}) = 1 - (P(\text{exactly once}) + P(\text{exactly zero times}))$
$P(\text{exactly once}) = 2000\cdot \frac{1}{2000}\cdot (\frac{1999}{2000})^{1999}$
$P(\text{exactly zero times}) = (\frac{1999}{2000})^{2000}$
So, $P(\text{at least twice}) = 1 - 2000\cdot \frac{1}{2000}\cdot (\frac{1999}{2000})^{1999} - (\frac{1999}{2000})^{2000}\approx0.2642$

This used the Binomial Theorem

For $n$ independent trials each with probability of success $p$, the probability of getting exactly $r$ successes is $\binom{n}{r} p^r (1-p)^{n-r}$

A: It is easiest to find first the probability $p$ of getting one or fewer $1$'s. Then the probability of getting a $1$ at least twice is $1-p$.
The probability we have zero $1$'s is 
$$\left(1-\frac{1}{2000}\right)^{2000}.\tag{1}$$ 
Now we find the probability of exactly one $1$. That $1$ can occur in any one of $2000$ places. The probability it occurs in a particular place is $\frac{1}{2000}$, and the probability it does not occur in the remaining $1999$ places is $\left(1-\frac{1}{2000}\right)^{1999}$, for a total of 
$$2000\cdot\frac{1}{2000}\cdot \left(1-\frac{1}{2000}\right)^{1999},\tag{2}$$
Add (1) and (2) to find $p$, and finally calculate $1-p$.
Another way: We use the Poisson approximation. The mean number of $1$'s is $\lambda=2000\cdot \frac{1}{2000}$. The number $X$ of $1$'s has approximately Poisson distribution parameter $\lambda=1$. So the probability that $X=0$ is approximately $e^{-1}\frac{1^0}{0!}$, and the probability that $X=1$ is approximately $e^{-1}\frac{1^1}{1!}$. So the probability that $X\ge 2$ is approximately $1-2e^{-1}$.
If you do both calculations, you will find that here the Poisson approximation gives an answer very close to the truth. 
A: $p = \dfrac{1}{2,000} = 0.0005, q = 1 - p = 1- 0.0005 = 0.9995, n = 2,000, k = 2, \mu = np = 2,000\cdot 0.0005 = 1, \sigma = \sqrt{npq} = \sqrt{0.9995} =0.9997$, We want to find: $P(x\geq 2)$. We can use normal approximation to the binomial distribution. Thus:
$P(x\geq 2) \approx P(x \geq 1.5) = P\left(z \geq \dfrac{1.5-1}{0.9997}\right) = P(z \geq 0.5) = 1- P(z < 0.5)=1-0.6915 = 0.3085$
