Sampling without Replacement and Non-uniform Distribution There are $N$ items, numbered $1 \ldots N$. The probability of selecting item $i$ in one draw is $p_i$. Items are drawn without replacement, so after each draw we need to do a re-normalization. Now, could you please tell me how to write down the probability of drawing $i$ in $k$ ($k \le N$) attempts? Even an approximate solution is good. Thanks!
 A: Sounds like you are looking for the multivariate Wallenius' noncentral hypergeometric distribution. ie weighted balls, multiple colors, multiple draws.wikipedia link
I did try to find something analytic, but even in simple cases this problem becomes increasingly complicated. I would advise looking at some of the packages in the link.
A: A long comment which I have have had to post as an answer. I  have
no objection if some kindly Moderator converts it into a series of
comments on the main question.
Your question is essentially unanswerable in one sense. 
You have told us that the probability of drawing the $i$-th object on the first draw is $p_i$.  In the general scheme of drawing with replacement, the probability of drawing the $i$-th object remains the same on all further draws. 
However, in the general scheme of drawing without replacement
(which you want to consider), there is no immediate answer to the question: what is the
probability of drawing the $i$-th object on the second draw? the third draw? etc. There is also the question whether we want the unconditional
probability of drawing the $i$-th object on the second draw, or the
conditional probability of drawing the $i$-th object on the second draw given that the first draw resulted in the $j$-th object being drawn?
One possibility to consider is that the (conditional) probabilities are obtained by re-normalizing the probabilities
of the remaining balls at each step.  That is, we write
\begin{align}
P(i_i) &= p_{i_1},\\
P(i_1, i_2) &= p_{i_1}\cdot \frac{p_{i_2}}{1-p_{i_1}}\\
P(i_1, i_2,i_3) &= p_{i_1}\cdot \frac{p_{i_2}}{1-p_{i_1}}
\cdot \frac{p_{i_3}}{1-p_{i_1}-p_{i_2}}\\
P(i_1, i_2,i_3,i_4) &= p_{i_1}\cdot \frac{p_{i_2}}{1-p_{i_1}}
\cdot \frac{p_{i_3}}{1-p_{i_1}-p_{i_2}}
\cdot \frac{p_{i_4}}{1-p_{i_1}-p_{i_2}-p_{i_3}}
\end{align}
and so on. But there could be other kinds of assumptions that might be
made about what happens on successive draws, and so unless you
specify what the probabilities are on successive draws, w=your 
question is unanswerable.
A: Since you asked for an approximation, I will propose one. It's not perfect: it requires additional reasonable assumptions; I won't prove it's accurate; I'm not sure it's a valid pmf. 
If $N$ is large, $K << N$, and the $p_j's$ are small, then drawing once gives a probability of $p_j$. The next draw will be only slightly better than $p_j$, etc. The draws are not independent, but are not too dependent. So I would approximate drawing the desired event in $K$ trials as drawing one sample in a poisson with $\lambda=K \times p_j$.
This gives you a probability of $K \times p_j \times  e^{-K p_j}$.
