Less common probabilities and expected values related to the coupon collector problem 
A company brings out a new set of $n$ collectible cards. The cards are
made in equal number, so there is the same probability of any given
card being in a random set. They are purchased one-by-one.
Question 1: If I buy $x$ cards, on average how many unique cards $(k)$
will I have? $k$ does not need to be a whole number.
Question 2: How large does x need to be that k/n is over y%?
Question 3: If I buy $x$ cards, what is the probability that I have $j$
unique cards. $j$ is a whole number.

EDIT: Thank you to all who have replied! As a first time poster, it is amazing to get so many helpful responses so quickly. I have learnt that this known as the coupon collector's problem. There is a very straightforward answer to the question, "What is the expected value of x needed to complete the set on n cards"
\begin{align}
E[x]={}&x\sum_{i=1}^n\frac1i\;.
\end{align}
I am struggling to find equations as simple as this to questions 1 and 2, so will post a second edit once I have them. On question 3 "user 4" kindly gave what I think is the right answer:
\begin{align}
\frac{n!}{(n-j)! n^x}{x \brace j}
\end{align}
 A: Q1:
Let us ask first what is the expected value of the binary event $X_i$ which value is $1$ if a certain card was bought at least once and 0 if it was not bought. Clearly the expected value is just the probability to buy the card in at least one of $x$ trials:
$$
\mathsf E(X_i)=1-\left(\frac{n-1}n\right)^x.\tag1
$$
Now due to linearity of the expectation the expected number $k$ of unique cards bought at least once is:
$$
k=\sum_{i=1}^n\mathsf E(X_i)=n\left[1-\left(\frac{n-1}n\right)^x\right].\tag2
$$
Q2:
As I already mentioned in a comment the question is ambiguous. If you mean the ratio $\frac kn$ it is given by (1).
Q3:
The probability can be computed as follows:
$$
p_j(n,x)=\frac1{n^x}\binom nj j!{x \brace j}=\frac{n!}{(n-j)! n^x}{x \brace j},\tag3
$$
where ${x \brace j}$ is the Stirling number of the second kind
In the expression (3)
$n^x$ counts the number of ways to dictribute $x$ labeled cards into $n $ labeled boxes with empty boxes being allowed, $\binom nj$ counts the number of ways to choose $j$ out of $n $ labeled boxes, and $j!{x \brace j}$ counts the number of ways to dictribute $x$ labeled cards into $j $ labeled boxes with empty boxes being not allowed. The cards are labeled by the order of their arrival. The boxes are labeled by the types of unique cards.
Observe that we factually proved the relation:
$$
\sum_{j=1}^n jp_j(n,x)=n\left[1-\left(\frac{n-1}n\right)^x\right].
$$
