I have a variation of the Coupon Collector's Problem which can be state as follows.
"Given a set $N$ of $n$ different kinds of coupons and a subset $K\subset N$ of $k$ kinds of coupons, with $k<n$, what is the expected number of draws you need to obtain at least one copy of each coupon of the set $K$. Each draw takes a single coupon from $N$ with same probability for each kind."
My previous approach was to enumerate the elements of $K$, lets say $l_1,l_2,...,l_k$ and associate each element $l_i$ with an integer $t_i$ which is the number of the trial in which we get $l_i$.
I'm "defining" a success to be "get a element of $K$ which we didn't get yet".
Then I tried to compute the probability of getting all the coupons from $K$ in $T$ trials base on the multiplicative rule. For simplicity, I considered $t_i<t_j$ if $i<j$ by now: we need $t_1-1$ fails, which have probability $1-\frac{k}{n}$ each and then a success with probability $\frac{k}{n}$. After that, the probability of a success decrease to $\frac{k-1}{n}$ since we already an element of $K$ and we are going to have more $t_2 - t_1 - 1$ fails with probability $1-\frac{k-1}{n}$.
With this line of thinking I got the probability of getting $l_i$ in trial $t_i$ is $\prod_{i=1}^k (\frac{k-i+1}{n})(1-\frac{k-i+1}{n})^{t_i - t_{i-1}-1}$ given that $i<j$ imples $t_i < t_j$ and may define $t_0 = 0$ and $t_k = T$ to keep it consistent.
So to consider all the possible orders for getting $l_1,...,l_k$ we can multiply by the number of ways to order this list, it is $k!$. My next idea was to take in count the fact that if we fix $t_i$, for all $i \neq j $, and let $t_j$ vary, we will have exclusives situations and we could compute the probability of getting all the coupons in $K$ in $T$ trials by making summations over $t_1,t_2,...,t_k$. But things get really complicated very soon.
I don't even know if that is a reasonable approach to the problem or if there is one easier.
Any help would be appreciable.
Thank You
I don't think this is substantially different from the original coupon collector problem.
The expected number of draws to collect the first coupon is $1/(k/n) = n/k$. The expected number of additional draws to collect the next coupon is $1/((k-1)/n)=n/(k-1)$. etc.
Therefore they find the $k$ coupons in, on average, $nH_k$ draws.