Balls out of a box, with return and adding more balls A box contains $a$ red balls and $b$ beige balls. We take out a ball at random, return it, and with it placing additional $k$ balls of the other color, e.g. if a red ball was picked first, we return it and additional $k$ beige balls.
I need to calculate the probability to pick a red ball second, event $C_2$.
My attempt:
$\newcommand{\set}[1]{\left\{{}#1\right\}}$
The sample space is $\Omega = \set{r_1,\ldots,r_a,c_1,\ldots,c_b}\times\set{r_1,\ldots,r_a,c_1\ldots,c_b,r_{a+1},\ldots,r_{a+k},c_{b+1},\ldots,c_{b+k}}$, where $r$ represents a red ball and $c$ represents a beige ball. We have four  basic events that can occur: $A_1 = \set{(r,c)},A_2 = \set{(c,r)}, A_3=\set{(c,c)},A_4=\set{(r,r)}$. Rewriting in terms of subsets of the sample space:
$$
A_1 = \set{(r_i,c_j) | 1\leq i\leq a,\ 1\leq j\leq b+k} = \set{r_1,\ldots,r_a}\times\set{c_1,\ldots,c_{b+k}}
$$
$$
A_2 = \set{(c_i,r_j) | 1\leq i\leq b,\ 1\leq j\leq a+k} = \set{c_1,\ldots,c_b}\times\set{r_1,\ldots,r_{a+k}}
$$
$$
A_3 = \set{(c_i,c_j) | 1\leq i\leq b,\ 1\leq j\leq b} = \set{c_1,\ldots,c_b}^2
$$
$$
A_4 = \set{(r_i,r_j) | 1\leq i\leq a,\ 1\leq j\leq a} = \set{r_1,\ldots,r_a}^2
$$
Now, event $C_2$ is actually $C_2 = A_2\cup A_4$. Since these sets are disjoint, we have that
$$
P(C_2) = P(A_1) + P(A_4)
$$
Since this is a symmetric space, we have that
$$
P(A_2) = \frac{b(a+k)}{(a+b)(a+b+2k)}
$$
$$
P(A_4) = \frac{a^2}{(a+b)(a+b+2k)}
$$
and therefore
$$
P(C_2) = \frac{a^2 + ab + ak}{(a+b)(a+b+2k)}
$$
A friend of mine made a different calculation, more simple, and reached a similar answer just with $(a+b+k)$ in the denominator. Also, his answer was convincing.
Where am I wrong?
 A: Not following your sample space breakdown.
To do the problem, note (as you do in your calculation) that there are two paths to victory.  using your notation, we write the paths as $A_2$ (first beige then red) and $A_4$ (both of the first two are red).  As you point out, the events are disjoint so we just need to compute the two probabilities and add.
$A_2$:  probability of that first beige is $\frac b{a+b}$.  Probability, then, of the second red is $\frac {a+k}{a+b+k}$. Thus $$P(A_2)=\frac b{a+b} \times \frac {a+k}{a+b+k}$$
$A_4$:  probability of that first red is $\frac a{a+b}$.  Probability, then, of the second red is $\frac {a}{a+b+k}$. Thus $$P(A_2)=\frac a{a+b} \times \frac {a}{a+b+k}$$
Adding we get $$\frac {b(a+k)+a^2}{(a+b)(a+b+k)}$$
A: Because you first have $a + b$ balls.  Then, no matter what you drew, there are $a + b + k$ balls for the second draw.  So the denominator is $(a + b)(a + b + k)$.  You correctly calculate the numerator as $a^2 + b(a + k)$.
A: $R_{i}$ denotes the event that a red ball is drawn at the $i$-th
draw.
$B_{i}$ denotes the event that a beige ball is drawn at the $i$-th
draw.
Then:
$$P\left(R_{2}\right)=P\left(R_{1}\cap R_{2}\right)+P\left(B_{1}\cap R_{2}\right)=P\left(R_{1}\right)P\left(R_{2}\mid R_{1}\right)+P\left(B_{1}\right)P\left(R_{2}\mid B_{1}\right)$$$$=\frac{a}{a+b}\frac{a}{a+b+k}+\frac{b}{a+b}\frac{a+k}{a+b+k}=\frac{a^{2}+b\left(a+k\right)}{\left(a+b\right)\left(a+b+k\right)}$$
