Probability - Expected number of draws to get all 52 cards at least once drawing in groups of size n Imagine you have a deck of cards and want to be fairly sure that you draw each card once (with a perfectly fair, complete, and random deck on each draw, of course). You are drawing cards in groups of size n from the deck. What is the expected number of draws such that each card has been drawn at least once?
Similar to the coupon collector's problem, but not quite. How would one go about integrating the math for that algorithm with drawing multiple cards at the same time?
Edit: found some duplicate questions.
How to calculate the expected value of the coupon collector problem if we are collecting the coupons in groups of k?
Coupon Collector Problem with Batched Selections
 A: This is not a solution to the asked question, it is just an approximation
Let a deck with $M$ distinct cards, and each time you draw $n$ cards randomly, put inside again and draw again, etc... We will suppose that the probability to draw some card is the same for all cards, that is, $p=1/M$.
We will work this problem as a Markov chain: suppose you had drawn $k$ distinct cards (no matter in how many draws, ignore this) and you want to know the probability that drawing the next $n$ cards the state of distinct cards will change from $k$ to $k+j$, where $j\in\{0,\ldots,n\}$.
Then if we draw $n$ cards and all are repeated, we have that
$$\Pr[k\to k]=\frac{k}{M}\cdot\frac{k-1}{M-1}\cdots\frac{k-n+1}{M-n+1}=\frac{k^\underline n}{M^\underline{n}}$$
and in general
$$\Pr[k\to k+j]=\binom{n}{j}\frac{k^\underline{n-j}(M-k)^\underline{j}}{M^\underline n}$$
Then the expected change of $k$ from a draw is
$$\mathrm E[\text{change}]=\sum_{j=0}^n j\Pr[k\to k+j]=\frac1{M^\underline n}\sum_{j=0}^n j\binom{n}{j}k^\underline{n-j}(M-k)^\underline{j}\tag{1}$$
The last summation involves a well-known Chu-Vandermonde identity:
$$\sum_{k=0}^n \binom{n}{k}a^\underline k b^\underline{n-k}=(a+b)^\underline n$$
Then using some algebra in (1) we have that
$$\mathrm E[\text{change}]=\frac{n(M-k)}{M^\underline n}\sum_{j=0}^n \binom{n-1}{j-1}(k-1)^\underline{n-j}(M-k)^\underline{j-1}=\frac{n(M-k)}{M^\underline n}(M-1)^\underline{n-1}=n\frac{M-k}M=n\left(1-\frac{k}M\right)$$
The above means that from some draw the expected number of new cards is $n(M-k)/M$ (observe that this quantity is well-defined only when $0\le k\le M$), then (if Im not wrong, what is not sure) the expected number of different cards after $\ell$ draws is the recurrent sum
$$T_\ell:=\sum_{h=1}^\ell N_h\tag{2}$$
where $N_h:=n\left(1-\frac{\sum_{j=1}^{h-1}N_j}M\right) $ and $N_1=n$. I dont know if (2) have a closed form, anyway with different values of $\ell$ you can get an approximation for the minimal number of throws such that $T\ge M$.
EDIT:
It seems that (2) is closable, using some mathematical software I get the solution:
$$T_\ell=(6M\ell+\ell-\ell^3)\frac{n}{6M}$$
But this function for $M=52$ and $n=5$ never gets bigger than $\approx 34$ (this happen when $\ell=10$), so something is very wrong in my interpretation/calculation of the approximation. Probably the fastest way to approximate the expected number of draws is through some numerical modeling software as R.

In the wikipedia article about the Coupon collector's problem is stated that

Wolfgang Stadje has solved the case where the stickers are bought in packets, which contain no duplicates.[3] The results show that for practical applications, say packets of five stickers, the effect of packets is negligible.

Then this problem is practically the same than the original coupon problem.

From page 18 of this document there is an analysis for this case.
A: *

*M = total cards in the deck

*n = number of cards per draw

*k = expected unique cards drawn

*i = number of draws (of n cards)


I found the following approximations based on the accepted answer's setup. They get close enough to the values on page 21 of this document that I feel that they are mostly correct, and rather simple to use.
$$ k=M(−((1−\frac{n}{M})^i−1)) $$
The same equation solved for i to make things easy.
$$ i = \log_{(1-\frac{n}{M})}(1-\frac{k}{M}) $$
You can approximate $ i $ by setting $ k = M - .5 $ when solving the equation for $ i $.
