Consider a game with following rules : in initial state we have $m$ black and $n$ red balls in urn. We may randomly choose one ball from the urn and if it black, we return red ball, if it red ball - we take it back. So number of red balls non-decrease random value.
Let $\xi_k$ is a number of red balls after $k$ iterations. We want to determine $\mathbb{E}(\xi_k)$ and $\mathbb{Var}(\xi_k)$.
My attempts :
1) Let $$a(k,l) = \mathbb{P}(\xi_k = l) = \mathbb{P}(\xi_k=l, \xi_{k-1} = l-1) \mathbb{P}(\xi_{k-1} = l - 1) + \mathbb{P}(\xi_k=l, \xi_{k-1} = l) \mathbb{P}(\xi_{k-1} = l) = \frac{n+m-l + 1}{n+m}a(k-1,l-1) + \frac{l}{n+m} a(k-1,l)$$ So we have some reccurence relation. However I don't think it's possible to solve it (at least for me).
2) $\displaystyle \mathbb{P}(\xi_{k-1} = l) = \frac{m (m-1) \dots(m-l + n )}{(n+m)^k} \cdot \sum_{i_1 + \dots i_{l-n} = k+n-l} n^{i_1}\cdot(n+1)^{i_2} \dots l^{i_{l-n}}$ . The denominator always the same after $k$ repetitions, because total amounts of balls is the same, the part $m \dots(m-l + n)$ occurs because we need to pick $l - n$ times black ball. But how can we calculate the last sum? I thought it can be calculated with multinomial coefficients.
Any hints?
Equivalently, we can consider colouring those black balls when picked. Number those balls from $1-m$ (black balls) and from $m+1 - m+n$ (red balls)
Note that we're only interested in the number of black balls which were coloured red, because $\mathbb E[X(k)+n] = \mathbb E[X(k)] +n$, while $Var(X(k)+n) = Var(X(k))$, where $n-$number (before starting a game) of red balls and $X(k)$ number of red balls after $k'$th iteration which have replaced black balls.
At the beginning we have $m-$black balls, so let $X(k) = \sum_{j=1}^m Y_j(k)$, where $Y_j(k)$ is $\{0,1\}$ random variable, $1$ when $j'$th black ball is coloured red at $k'$th iteration (or before), and $0$ if it's still black after $k'$th iteration.
Then $\mathbb E[X(k)] = \sum_{j=1}^m \mathbb E[Y_j(k)] = m\mathbb E[Y_1(k)] = m \mathbb P(Y_1(k) = 1) = m - m\mathbb P(Y_1(k)=0) = m - m (\frac{m+n-1}{m+n})^k$
Second equality sign due to the same distribution, and the last one $\mathbb P(Y_1(k) = 0) = (\frac{m+n-1}{n+m})^k$, since $k-$ times we need to pick ball different from $1$.
So your $\mathbb E[\xi_k] = \mathbb E[X(k) + n] = n + m - m(1-\frac{1}{m+n})^k$
Now looking at variance, we have:
$Var(X(k)) = \mathbb E[X^2(k)] - (\mathbb E[X(k)])^2$
So all we are left with is to compute $\mathbb E[X^2(k)] = \sum_{j=1}^m \mathbb E[Y_j^2(k)] + 2\sum_{1 \le i < j \le m} \mathbb E[Y_i(k)Y_j(k)] = \mathbb E[X(k)] + 2{m \choose 2} \mathbb E[Y_1(k)Y_2(k)]$.
The last thing is $\mathbb E[Y_1(k)Y_2(k)] = 1 - \mathbb P(Y_1(k)Y_2(k) = 0) = 1 - (\frac{m+n-2}{m+n})^2$
I'll leave you with all those components to glue up together