Mean of the Multivariate Wallenius Non-Central Hypergeometric Distribution An urn contains $N$ balls where ball $i$ is of size $w_i$. We draw $n$ times without replacement. Let $x_i$ be the random variable indicating whether the ball $i$ has been drawn ($x_i=1$) or not ($x_i=0$) after $n$ iterations. Note that $\sum_{i=1}^Nx_i=n$. For every ball $j$ that is still in the urn, the probability to draw $j$ in the next iteration is $w_j/\sum_{i=1}^N(1-x_i)w_i$. 
For example, we have balls of sizes 1, 2, 3, 4, and 5 in the urn. The probability to draw the ball of size $5$ is $\frac{5}{1+2+3+4+5}=\frac{1}{3}$ while drawing the ball of size $1$ is only $\frac{1}{15}$. However, after the ball of size $5$ has been drawn, the probability to draw the ball of size $1$ in the next iteration increases to $\frac{1}{10}$. The probability to draw balls of sizes 1 and 5 in $n=2$ draws is $\frac{1}{3}\frac{1}{10}+\frac{1}{15}\frac{5}{14}=\frac{2}{35}$.
The resulting distribution is best described by a multivariate Wallenius' hypergeometric distribution. However, this is a special case where we have only one ball of each type. So, the following should be the formula for the underlying distribution:
$$
P(X=\{x_i\}_{i=1}^N) = s*\int_0^1u^{s-1} \cdot \prod_{i|x_i=1}(1-u^{w_i})\mathrm{d}u
$$
where
\begin{align}
s&=\sum_{i|x_i=0}w_i
\end{align}
For our example, the probability that balls of size $1$ and $5$ have been drawn is $
P(X=\{1,0,0,0,1\})=9*\int_0^1u^{8} (1-u^1)(1-u^5)\mathrm{d}u =\frac{2}{35}$.
What is the expected combined size of the balls drawn after $n$ iterations?
 A: I know I'm two years late here, but I just stumbled on this problem in my research. The trick is to rewrite the product in the integral as a sum, from which it's easy to compute the antiderivative and simplify.
Here is a cleaner simplification of the Wallenius distribution for our special case of 1 ball of each type:
$$Pr(\mathbf{X}=X)=\int_{t=0}^1\prod_{b\in{B_X}}(1-t^\frac{w_b}{s_X})dt$$
where $X=\{x_i\}_{i=1}^N$ and $B_X=\{b\in[1,N]:x_b=1\}$ is the set of balls in selection $X$. (I'm changing your notation a bit; here, $\mathbf{X}$ is the random variable and $X$ [composed of the $x_i$'s] is the value it takes on. Also, since $s$ depends on $X$, I added $X$ as a subscript.)
Let $\mathbf{P}(S)$ denote the power set of S, and let $W=\{{w_1,\dots,w_{N}\}}$ be the weight vector. The product can be rewritten as:
$$\prod_{b\in{B_X}}(1-t^\frac{w_b}{s_X})=\sum_{A\in\mathbf{P}(W_{B_X})}(-1)^{|A|}*t^{\frac{\underset{a\in A}{\sum}a}{s_X}}$$
(To see why this works, just expand the product out for a few terms.) Taking the antiderivative and evaluating it at $t=1$ minus $t=0$, we get:
$$Pr(\mathbf{X}=X)=\sum_{A\in\mathbf{P}(W_{B_X})}(-1)^{|A|}\frac{s_X}{s_X+\underset{a\in A}{\sum}a}$$
Now, let $Weight(X)=\sum_{b\in{B_X}}w_b$. The expected weight is:
$$E[Weight(\mathbf{X})]=\sum_{X'\in\{0,1\}^N:\sum_{j=1}^NX_j=n}Weight(X)*Pr(X)\\
=\sum_{X\in\{0,1\}^N:\sum_{j=1}^NX_j=n}\sum_{A\in\mathbf{P}(W_{B_X})}(-1)^{|A|}\frac{s_X*\sum_{b\in{B_X}}w_b}{s_X+\underset{a\in A}{\sum}a}$$
This closed-form answer was good enough for me. Since $s_X=1-Weight(X)$, you might be able to simplify further, but I didn't go down that road.
