Probability of finding missing cards I'm thinking about such problem.
Let it be 108 different cards to collect. They come up in a 4-cards packages and we can assume that all 4 cards in the package are different, but we can have some of the cards to be the same in different packages. We can also assume all cards are equally likely to be in a package.
Now, suppose you have collected 104 cards and you are lacking only four.
What is the smallest number of packages you have to buy to have a probability of collecting all remaining four cards greater than 50%?
 A: This can be modelled as a Markov chain. There will be five states in the
chain: $S_0,\ldots,S_4$ where $S_k$ represents the state in which
you require $k$ more cards. So you start in $S_4$ and want to get to $S_0$.
Let $0\le l\le k$. At each stated there is a transition probability to go
from $S_k$ to $S_l$. It is the probability of drawing $k-l$
of your wanted $k$ cards in the next packet. This follows a hypergeometric
distribution and is
$$\frac{\binom{k}{k-l}\binom{108-k}{4-k+l}}{\binom{108}{4}}.$$
Build a transition matrix $A$ from these and consider $A^n$.
One of the entries of $A^n$ is the probability that you finish your
collection in $\le n$ goes. So just find the smallest $n$ such that
this is $>0.5$.
A: Here is a solution via the Principle of Inclusion and Exclusion (PIE).  Let's say the cards are numbered from 1 to 108 and the missing cards are numbers 1 through 4.  We buy $r$ packages, and we would like to find the probability that all of the missing cards are included.  
There are $N = \binom{108}{4}^r$ possible arrangements of cards in the packages, all of which we assume are equally likely.  We want to count the number of arrangements in which all the missing cards have been found.  To that end, say an arrangement has "Property $i$" if it does not include card number $i$, for $i=1,2,3,4$.  If an arrangement has none of the properties then we have a complete set of cards.  Let $S_i$ be the number of arrangements having $i$ of the properties.  Then
$$\begin{align}
S_1 &= \binom{4}{1} \binom{107}{4}^r \\
S_2 &= \binom{4}{2} \binom{106}{4}^r \\
S_3 &= \binom{4}{3} \binom{105}{4}^r \\
S_4 &= \binom{4}{4} \binom{104}{4}^r 
\end{align}$$
and the number of arrangements having none of the properties, by PIE, is 
$$N_0 = N - S_1 + S_2 - S_3 + S_4$$
The probability that we have a complete set after buying $r$ packs is 
$$p_r =\frac{N_0}{N}$$
Computation shows that $p_{48} = 0.487962$ and $p_{49} = 0.502332$, so we need to buy 49 packs to have a 0.5 chance of having a complete set.
