Estimate the number of trials needed to observe all the possible outcomes of an experiment I am stuck with the following problem:
Each package of Pokemon cards contains 1 of N possible legendary Pokemon. How many packs do you expect to buy to get all N? We assume all N legendary cards are equally likely with each purchase.
Here it is my trial of incomplete solution.
I model the problem as follows:
I have an unrn containing $N$ different objects labeled from 1 to $N$. 
I am performing draws with replacement from this urn.
I report the results of my experiment into an $N$ component vector, $\vec{k_n}=(k_1^n,\dots, k_N^n)$ where $k_i^n$ is the number of times that event $i$ has occurred out of $n$ trials. Clearly $\Sigma_{i=1}^{N}k_i^n=n$.
Therefore the probability of observing a given vector of outcomes $\vec{k}$ out of a number of $n$ drawings is
$$
p(\vec{k_n}=(\tilde{k_1}, ...,\tilde{k_N})|n\text{ trials})=\frac{n!}{\tilde{k_1}!\dots\tilde{k_N}!}p_1^{\tilde{k_1}}\dots p_N^\tilde{k_N}
$$
since all the legendary cards are equally probable we have that
$$
p_1=p_2=\dots=p_N=\frac{1}{N}
$$
and remembering that 
$$
\Sigma_{i=1}^{N}\tilde{k_i}=n
$$
We get
$$
p(\vec{k_n}=(\tilde{k_1}, ...,\tilde{k_N})|n\text{ trials})=\frac{n!}{N^n}\frac{1}{\tilde{k_1}!\dots\tilde{k_N}!}
$$
Arrived at this point I have no a precise idea on how to continue (maybe using bayesian inference?) and need an help...

Just to clear this question from the unanswered queue:
This problem is known as the coupon collector's problem. Several question have already been asked about it on math.SE.
The coupon collectors problem
Coupon collectors problems with multiple coupons in a cereal box
https://math.stackexchange.com/search?q=coupon+collector+problem
Specifically, considering your problem is the basic coupon collector's problem, the expected number of trials is given by $n\cdot H_n$ in which $H_n$ is the n-th harmonic number and $n$ is the number of coupons to be collected.
