I have a simple question but I can't seem to find the answer anywhere. Say that I have a set $\mathbb Z$ and a subset of that $\mathbb X$.
I want to draw elements from $\mathbb Z$ until there is at least a 50% chance that I have drawn all the elements of $\mathbb X$.
Clearly if $|\mathbb X| = 1$ then I just need to draw $\frac{|\mathbb Z|}{2}$.
But how many will I need to draw if $|\mathbb X| > 1$?
EDIT:
GBQT made a good comment, I'm talking about drawing with replacement. Drawing element $e$ from $\mathbb Z$ does not preclude $e$ from being drawn again on subsequent draws. But $\mathbb X$ is finite, as in it has each element of $\mathbb X$ is unique.
The number of draws is the unknown here, it will be represented by the variable $x$. If $x < |\mathbb X|$ then the chance of drawing all elements of $\mathbb X$ is 0%. Otherwise the chance of drawing all elements of $\mathbb X$ is: All combinations with repetition that can be made from $\mathbb Z$ of size $x - |\mathbb X|$, divided by all combinations with repetition that can be made from $\mathbb Z$ of size $x$.
Combination with Repetition can be calculated by the Binomial Coefficient: $\binom {n + k - 1}{k}$ Where $n$ is the size of the set and $k$ is the size of the combinations.
So the equation this question asks for is actually: $$\frac {\binom {|\mathbb Z| + x - |\mathbb X| - 1}{x - |\mathbb X|}}{\binom {|\mathbb Z| + x - 1}{x}} = 0.5$$
That already looks hairy, and it doesn't bet nicer as it's expanded out: $$\frac {(|\mathbb Z| + x - |\mathbb X| - 1)!x!}{(|\mathbb Z| + x - 1)!(x - |\mathbb X|)!} = 0.5$$
Really the only way that this can be simplified is removing the constants: $$\frac {(|\mathbb Z| + x - |\mathbb X|)!x!}{(|\mathbb Z| + x)!(x - |\mathbb X|)!} \frac {|\mathbb Z| + x}{|\mathbb Z| + x - |\mathbb X|} = 0.5$$
Alternatively it could be visualized as: $$\frac{\underset {i = x - |\mathbb X| + 1}{\overset {x}{\LARGE \mathrm \Pi}}i}{\underset{i = |\mathbb Z| + x - |\mathbb X|}{\overset {|\mathbb Z| + x - 1}{\LARGE \mathrm \Pi}}i} = 0.5$$