double dixie cup problem - upper bound There is a set of $n$ cards to collect, in each pack there is exactly one card. The sets are distributed uniformly among the packs (i.e. the probability for obtaining a particular card is $\frac{1}{n}$). If you pull a card that is already in your collection, you give it to a friend. Now, how can one find the best upperbound (in $O$- notation) of the expected number of cards you have to buy so you and your friend both have all the $n$ cards?
I read something about the double dixi cup problem and thought that this might be equivalent to the aboth, but still I don't know how to find an upper bound.
 A: Welcome to MSE!
You're exactly right -- this problem is exactly the "double dixie cup problem". To that end, you can see a solution in Newman's "The Double Dixie Cup Problem"
(pdf here, for instance). The argument is a bit more delicate than the classical case, so I'll omit it. For more details you can read the above paper, or the (excellent) survey by Ferrante and Saltalamacchia (pdf here).
As a tl;dr, the expected number of coupons you need to buy
is
$$O(n\log n + n\log \log n)$$

Edit:
Here's a sketch, as requested.
The idea is to let $p_i$ be the probability that you haven't collected all $2n$ coupons after you look at the $i$th one. Then the expected time to collect all $2n$ is $\sum_{i=0}^\infty p_i$ (do you see why?). A clever argument shows
$$
p_i = 
\left . \frac{ \{ (x_1 + \ldots + x_n)^i \}_2 }{n^i} \right |_{x_1 = 1, \ldots, x_n = 1}
$$
where by $\{ \cdot \}_2$, I mean to keep only the terms in this polynomial for which every variable $x_i$ shows up at most $1$ time. So $x_1 x_3$ and $x_1 x_2 x_3$ would be allowed, but $x_1^2 x_4$ would not be.
So we're interested in
$$
\sum_i p_i = \sum_i
\left . \frac{ \{ (x_1 + \ldots + x_n)^i \}_2 }{n^i} \right |_{x_1 = 1, \ldots, x_n = 1}
$$
But it's not hard to see that
$$
\left \{ e^{(x_1 + \ldots + x_n)} \right \}_2 = \sum_i \frac{ \{ (x_1 + \ldots + x_n)^i \}_2 }{i!}
$$
So all that's left is to relate our sum of interest to this new sum, then compute this new sum. This is, again, a bit delicate, but we can show that our sum (and thus the expected number of coupons we'll need to buy) reduces to an integral
$$
n \int_0^\infty 1 - \left ( 1 - \frac{1 + t}{e^t} \right )^n dt
$$
One then computes the asymptotics of this integral as $n \to \infty$ to get the claim.

I hope this helps ^_^
