X-sided dice probability I have an $x$-sided dice and I roll it $y$ times.
What is the probability that I will see at least $z$ of the $x$ sides?
e.g. If I rolled a $6$-sided dice $8$ times, what is the probability I see at least $4$ of the sides?
But I want to do with much bigger numbers
Specifically, I want to know  what is probability I see at least $z$ of the $160,000$ sides if I rolled a $160,000$-sided dice $25,000$ times.
Thanks in advance!
 A: The chance of a side missing out is $$\left({x-1\over x} \right)^y$$
The chance of two sides missing out is $$\left({x-2\over x}\right)^y$$
The usual approximation by an exponential causes a large error.
I will use the binomial distribution.
On average, you will see $$\mu\approx x(1-e^{-y/x})$$ different sides.  The expected value of the square is
$$x(1-\left({x-1\over x}\right)^y)+ \\
(x^2-x)(1-2\left({x-1\over x}\right)^y+\left({x-2\over x}\right)^y) $$
So the variance is
$$\sigma^2= x(\left({x-1\over x}\right)^y-\left({x-2\over x}\right)^y \\
+x^2(\left({x-2\over x}\right)^y-\left({x-1\over x}\right)^{2y})$$
This is about 1500 for your numbers, compared with 19000 when I neglected the x^2 term.  As I mention in comments, this is close to $$\sigma^2\approx xe^{-y/z}-(x+y)e^{-2y/x}$$
For large $x$ and $y$, it will be very close to a normal distribution with that mean and variance.  Sixty-eight percent of the time $z$ will be between $\mu-\sigma$ and $\mu+\sigma$ (note $\sigma$ is the square-root of $\sigma^2$ defined above), and ninety-five percent of the time it will be between $\mu-2\sigma$ and $\mu+2\sigma$.
A: Possible practical approach (without necessarily needing to approximate):
Define $P(a,b)$ to be the probability of seeing at least $z$ distinct sides after $a$ additional rolls having seen already $b$ distinct sides. Then we have a simple recurrence
$$P(a,b)=\frac{b}{x}P(a-1,b)+(1-\frac{b}{x})P(a-1,b+1)$$
Base cases: $P(a,b)=0$ if $a+b<z$ and $P(a,b)=1$ if $b \geq z$.
You want $P(y,0)$.
This is around three hundred million subproblems for $y=25000$.
For example, for $z = 23145$, we get about $0.504$ (using double-precision floating point with no concern for numerical issues, took maybe a minute to compute by computer).
($23145$ chosen as an "interesting" value of $z$ using the coupon collector problem to calculate how many distinct coupons to collect from $160000$ to give expected number of draws of about $25000$, ie $160000 (H_{160000}-H_{160000-23145}) \approx 25000$ where $H_n$ is the $n$th harmonic number; can also get $23145$ from Empy2's answer.)
