expected number of elements after fixed number of trials

Question

I have a practical problem but not math experience to solve it. Consider I have a bin with n distinct numbers (n=1047 in my case). I extract x elements without replacement (say x=10), take note of them and reinsert in the bin. I repeat this process a fixed number d of times (say d=50). The question is: at the end of the d extractions how many different elements I expect I have seen?

I expected it was something like a bernulli distribution, but in this case the probability of finding an element not seen before is not constant. Thanks.

Answer 1

For $i=1,\dots n$ let $X_i$ be Bernouilli distributed with $X_i=1$ iff number $i$ never shows up.

Then $\mathbb EX_i=\Pr(X_i=1)=(1-\frac{x}{n})^d$ and $X:=X_1+\cdots X_n$ equals the number of numbers that never show up.

Now apply linearity of expectation and symmetry to find $\mathbb EX$.

You are looking for: $\mathbb E(n-X)=n-\mathbb EX$.