Expected number of 5-sticker packs needed to complete a Panini soccer album To fill a football album of 200 stickers (images of players), how many envelopes do I need to buy if every envelope has 5 stickers and the album company says that all stickers has the same probability of appearing?
The answer is 233 but I can not reach it.
There is something I am missing (about the same probability of appearing: $0.5$?).
For me, I declared the variables $n=200$ and $k=5$.
Therefore, $$n!/k!*(n-k)! = 200!/5!*(195)!$$
My answer is $0.0254$ because something is missing, and I do not know what. Thanks! 
 A: The usual coupon collector problem, where you get the coupons one by one, would say the expected number to get a set is $200H_n \approx 200(\ln(200)+\gamma)\approx 1175$ coupons.  Naively you would expect to need $1175/5=225$ packs.  This will be correct if the stickers are randomly distributed in the packs, including the chance that there are duplicates in a pack.  I don't know how to calculate it, but suspect prohibiting duplicates will marginally reduce the expected number of packs.
A: This problem can be modelled through a Markov chain on $201$ states, labelled from $0$ to $200$, representing the number of stickers in the album. In we are in the state marked as $a$, we have fixed probabilities of transition into one of the states $a,a+1,a+2,a+3,a+4,a+5$, only depending on the value of $a$. For instance $a\to a$ happens with probability $\binom{a}{5}\cdot\binom{200}{5}^{-1}$, $a\mapsto a+1$ happens with probability $\binom{200-a}{1}\binom{a}{4}\cdot\binom{200}{5}^{-1}$ and so on. The wanted expected value can be computed from the transition matrix of such Markov chain (which is an upper triangular band matrix) with little numerical effort. Its closed form might be not so pleasant, but as already remarked by Ross Millikan, by the standard version of the coupon collector's problem it is reasonably close to $\frac{1}{5}\cdot 200\, H_{200}\approx 235$.
