Eliminating Repeat Numbers from a Hat Inspired by this question.
Inside a hat there are $n$ slips of paper with the integers $1$ through $n$ on them. The numbers on slips $\{1,2,\ldots,r\}$ are erased and replaced with the numbers $\{r+1,r+2,\ldots,2r\}$ respectively $\left(\text{assume }r\leq\frac{n}{2}\right)$. What is the expected value of papers that need to be picked out of the hat for all the numbers in the hat (without replacement) to be unique? (For each element of $\{r+1,r+2,\ldots,2r\}$, at least one paper containing that element has been chosen).
I have tried to use similar logic to the linked question, but I do not know how to incorporate repeats into the calculations.
 A: For notational convenience let's assume the numbers $1,\ldots r$ are the duplicated ones. Let $T$ be the number of draws required before everything remaining in the hat is unique. Then
$$
E(T)=\sum_{k=0}^n P(T>k)=\sum_{k=0}^n P\left(\bigcup_{i=1}^r B_{i,k}\right)\tag1
$$
where $B_{i,k}$ is the event that no slips labeled $i$ have been seen by the time $k$ draws have been made (and therefore both copies of $i$ are among the remaining $n-k$ slips in the hat). By inclusion-exclusion, we have
$$
P\left(\bigcup_{i=1}^rB_{i,k}\right)=\sum_{i=1}^r(-1)^{i-1}{r\choose i}P(B_{1,k}\cap B_{2,k}\cap\cdots\cap B_{i,k})
=\sum_{i=1}^r(-1)^{i-1}{r\choose i}{{n-k\choose 2i}\over{n\choose 2i}}.\tag2
$$
Plugging (2) into (1) looks like a mess, but we can simplify by interchanging the summations and using the "upper summation" identity
$$
\sum_{m=0}^n{m\choose a}={n+1\choose a+1}\tag3$$
to obtain
$$
E(T) = \sum_{i=1}^r(-1)^{i-1}{r\choose i}{{n+1\choose 2i+1}\over{n\choose 2i}}=\sum_{i=1}^r(-1)^{i-1}{r\choose i}{n+1\over2i+1}.\tag4
$$
This can be simplified further. Introduce an $i=0$ term in (4) and rearrange to obtain
$$
{E(T)\over n+1}=1-\sum_{i=0}^r(-1)^i{r\choose i}\frac1{2i+1}.\tag5
$$
Finally apply the identity
$$
\sum_{k=0}^n(-1)^k{n\choose k}\frac1{k+\frac12}={1\over\frac12{n+\frac12\choose n}}:={n!\over(\frac12)(\frac32)\cdots(n-\frac12)(n+\frac12)}\tag6
$$
(where ${n+\frac12\choose n}$ is a generalized binomial coefficient) to (5) and get
$$
{E(T)\over n+1}=1-\frac1{{r+\frac12\choose r}}.
$$ 
With $r=0$ this gives $E(T)=0$, as expected. With $r=1$ we get $E(T)=\frac13(n+1)$, and with $r=2$ we get $E(T)=\frac7{15}(n+1)$.
Note: Identity (6) can be derived by considering the binomial expansion of
$$
t^{-1/2}(1-t)^n = \sum_{k=0}^n(-1)^k{n\choose k}t^{k-1/2}
$$
and then integrating over $t$ from $0$ to $1$. It is the $x=1/2$ case of the more general identity
$$
\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k+x} = \frac{1}{x\binom{n+x}{n}}
$$
