Expected number of draws without replacement from a deck of cards before drawing at least one of each card of interest Given a deck of cards of size $n$, containing $a$ of card a, $b$ of card b, $c$ of card c, $d$ of card d, and $n - (a+b+c+d)$ other cards, on average how many draws from this deck without replacement will it take to have at least 1 of each card a, b, c, and d?
This is easy to calculate for each card separately eg. just card a:
$$\frac{n+1}{a+1}$$
But I can't figure out how to calculate the number of draws to get all of them.
 A: For $m \in \Bbb{Z_{\geq 4}}$, let $p(m)$ denote the probability that at least one of each of the $4$ types is present, in the first $m$ cards.  Obviously, for $m \geq n$, you have that $p(m) = 1.$
Let $t(m)$ denote $p(m) - p(m-1)$.
Then, $t(m)$ denotes the probability that it takes exactly $(m)$ drawn cards to get at least $1$ of all $4$ types.
Then, the desired computation will be
$$\sum_{m=4}^\infty m \times t(m).$$
Therefore, the problem has been reduced to computing $p(m)$.

In my experience, there are $3$ distinct strategies that may be used to compute $p(m)$:

*

*Inclusion Exclusion

*The direct approach

*Recursion.

Also, it may be possible to use generating functions, which I am totally ignorant of.
To me, Inclusion-Exclusion seems to take the least creative thought, so that is what I will use.  Perhaps not coincidentally, it might be viewed as the least elegant of the approaches.
The approach that will be taken will be to compute
$$p(m) = \frac{N\text{(umerator)}}{D\text{(enominator)}} \tag1$$
where $~\displaystyle D = \binom{n}{m} ~: ~m \leq n.$
So now, the problem has been reduced to using Inclusion Exclusion to compute $N$ as a function of $m$.

Given any set $E$ with a finite number of elements, let $|E|$ denote the number of elements in the set $E$.
Let $G$ denote the set $\{1,2,\cdots,n\}$.
Assume that the elements in $G$ are assigned the following properties:

*

*$\{1,\cdots,a\}$ are assigned the property of being of type $A.$

*$\{a+1,\cdots,a+b\}$ are assigned the property of being of type $B.$

*$\{a+b+1,\cdots,a+b+c\}$ are assigned the property of being of type $C.$

*$\{a+b+c+1,\cdots, a+b+c+d\}$ are assigned the property of being of type $D.$

*$\{a+b+c+d+1,\cdots, n\}$ are assigned the property of being of type NULL.

Let $S$ denote the set of all subsets of $G$ that contain exactly $m$ elements.
Let $S_1$ denote the subset of $S$ that do not contain any elements of type $A$.
Let $S_2$ denote the subset of $S$ that do not contain any elements of type $B$.
Let $S_3$ denote the subset of $S$ that do not contain any elements of type $C$.
Let $S_4$ denote the subset of $S$ that do not contain any elements of type $D$.
Then, the problem has been reduced to computing :
$N = |S| - |S_1 \cup S_2 \cup S_3 \cup S_4|.$
Let $T_0$ denote $|S| = ~\displaystyle \binom{n}{m}.$
For $r \in \{1,2,3,4\}$, let
$$T_r = \sum_{1 \leq i_1 < i_2 < \cdots i_r \leq 4} |S_{i_1} \cap \cdots \cap S_{i_r}|.$$
That is, $T_r$ represents the sum of $~\displaystyle \binom{4}{r}$ terms.
Then, in accordance with Inclusion-Exclusion,
$$N = \sum_{r=0}^4 (-1)^r T_r.$$
Therefore, the problem has now been reduced to computing $T_1, \cdots, T_4$, as a function of $m$.

I am going to assume that the expression $~\displaystyle \binom{n}{k}$ will be used for $n \in \Bbb{Z^+}$, and $k \in \Bbb{Z_{\geq 0}}.$
Further, it is assumed that if $k > n$, then $~\displaystyle \binom{n}{k} = 0.$

$\underline{\text{Computation of} ~T_1}$ 
$~\displaystyle |S_1| = \binom{n-a}{m}.$ 
$~\displaystyle |S_2| = \binom{n-b}{m}.$ 
$~\displaystyle |S_3| = \binom{n-c}{m}.$ 
$~\displaystyle |S_4| = \binom{n-d}{m}.$
$~\displaystyle T_1 = ~$ the sum of the $4$ terms above. 

$\underline{\text{Computation of} ~T_2}$ 
$~\displaystyle |S_1 \cap S_2| = \binom{n-a-b}{m}.$ 
$~\displaystyle |S_1 \cap S_3| = \binom{n-a-c}{m}.$ 
$~\displaystyle |S_1 \cap S_4| = \binom{n-a-d}{m}.$ 
$~\displaystyle |S_2 \cap S_3| = \binom{n-b-c}{m}.$ 
$~\displaystyle |S_2 \cap S_4| = \binom{n-b-d}{m}.$ 
$~\displaystyle |S_3 \cap S_4| = \binom{n-c-d}{m}.$ 
$~\displaystyle T_2 = ~$ the sum of the $6$ terms above. 

$\underline{\text{Computation of} ~T_3}$ 
$~\displaystyle |S_1 \cap S_2 \cap S_3| = \binom{n-a-b-c}{m}.$ 
$~\displaystyle |S_1 \cap S_2 \cap S_4| = \binom{n-a-b-d}{m}.$ 
$~\displaystyle |S_1 \cap S_3 \cap S_4| = \binom{n-a-c-d}{m}.$ 
$~\displaystyle |S_2 \cap S_3 \cap S_4| = \binom{n-b-c-d}{m}.$ 
$~\displaystyle T_3 = ~$ the sum of the $4$ terms above. 

$\underline{\text{Computation of} ~T_4}$ 
$~\displaystyle T_4 = \binom{n-a-b-c-d}{m}.$ 
