Coupon Collector Score Suppose you have a non-uniform coupon collector problem. But, rather than quotas, each coupon gives you points. There are $n$ different groups of coupons, and the probability of receiving a coupon of each group is $p_i$. There are $N_i$ distinct coupons in group $i$ with each coupon within a group equiprobable. The number of points you receive for getting a new coupon in group $i$ is:
$$\dfrac{N_i}{p_i(N_i-k_i)}$$
where $k_i$ is the number of coupons you have already gathered from group $i$. So, obviously, the points for the first coupon of each group is just $\dfrac{1}{p_i}$. But, once you get more coupons from each group, each coupon is worth more.
Question: What is the expected number of points you will have after collecting $m$ distinct coupons?
I attempted to try to brute force the expected value of the score, but I was not having much luck. And when I tried to calculate it by monte carlo simulation, I wound up with a fairly large variance.
I can easily calculate the expected value for points after drawing one coupon. That is just the number of groups. And from there, I think for 2 coupons among $n$ groups, I think the expected number of points will be:
$$\sum_{i = 1}^n \sum_{j = 1}^n \dfrac{p_ip_j\left(\begin{cases}1 & i\neq j \\ \tfrac{N_i-1}{N_i} & i = j\end{cases}\right) \cdot \left(\dfrac{1}{p_i} + \dfrac{1}{p_j}\begin{cases}1 & i \neq j \\ \tfrac{N_i}{N_i-1} & i = j\end{cases}\right)}{\sum_{k=1}^n p_k\left(\begin{cases}1 & k \neq i \\ \tfrac{N_i-1}{N_i} & k = i\end{cases}\right)}$$
This just looks like a nightmare to calculate even for two coupons. Any advice on how to make this a bit easier to calculate? I am looking at similar coupon collector problems, such as the one here, but the addition of points for coupons is really throwing a wrench into my attempt at creating a generating function. I was hoping to expand my attempt at calculating the expected points from two coupons into a generating function, if I could see a pattern. I did not.
 A: I initially misread the post; this answer computes the expected number of points after $m$ draws, not after getting $m$ distinct draws.
Call a group dead if all of its coupons have been chosen previously, and alive otherwise.

Lemma If there are currently $a$ alive groups, the expected number of points from your next coupon is $a$.
Proof: Suppose you have already $k_i$ coupons from the $i^{th}$ group, for each $i\in \{1,\dots,n\}$. Let $X$ be the number of points from your next coupon. For each $i\in \{1,\dots,n\}$, let $G_i$ be the event that the next coupon is from group $i$.
$$
E[X]=\sum_{i=1}^n P(G_i)E[X\mid G_i]=\sum_{i=1}^n p_i\cdot 
\left(\begin{cases}
\displaystyle\frac{N_i-k_i}{N_i}\times \frac{N_i}{p_i(N_i-k_i)} & k_i < N_i
\\
0 & k_i=N_i
\end{cases}\right)
=\text{#}\{i\mid k_i<N_i\}
$$

Now, let $X_t$ be the number of points you get on draw number $t$, for $t\in \{1,2,3,\dots\}$. Then
\begin{align}
E[X_t]
  &= E[\text{# alive groups before turn $t$}]
\\&= \sum_{i=1}^n P(\text{group # $i$ alive before turn $t$})
\\&= \sum_{i=1}^n P\left(\bigcup_{j=1}^{N_i}\{\text{$j^\text{th}$ coupon in group $i$ not yet chosen}\}\right)
\\&= \sum_{i=1}^n \sum_{j=1}^{N_i} (-1)^{j-1}\binom{N_i}{j} \left(1-p_i\frac{j}{N_i}\right)^{t-1}
\end{align}
The last equation is a routine application of the principle of inclusion-exclusion. Therefore, if you draw $m$ coupons total, the expected number of points is
$$
\begin{align}
\sum_{t=1}^mE\left[ X_t\right]
&= \sum_{i=1}^n \sum_{j=1}^{N_i} (-1)^{j-1}\binom{N_i}{j} \sum_{t=1}^m\left(1-p_i\frac{j}{N_i}\right)^{t-1}
\\&= \sum_{i=1}^n \sum_{j=1}^{N_i} (-1)^{j-1}\binom{N_i}{j} \frac{1-(1-p_ij/N_i)^m}{p_ij/N_i}
\\&= \boxed{\sum_{i=1}^n \frac{N_i}{p_i}\sum_{j=1}^{N_i} (-1)^{j-1}\binom{N_i}{j}\frac1j\cdot  \left(1-\left(1-p_i\frac{j}{N_i}\right)^m\right)}
\end{align}
$$
