Solution:

Suppose $a$ is the number of black balls and $b$ is the number of white balls, and let $E_{a, b}$ denote the expected number of black balls Bob has once all the balls are removed with Alice going first. Then we want to find $E_{10,10}$. It is evident that if $E_{0, b}=0$. Also, since Bob chooses a black ball with probability $\frac{a-1}{a+b-1}$, if $a>0$ we have
$$\begin{aligned} E_{a, b} & =\frac{a-1}{a+b-1}\left(E_{a-2, b}+1\right)+\frac{b}{a+b-1} E_{a-1, b-1} \\ & =\frac{(a-1)\left(E_{a-2, b}+1\right)+b E_{a-1, b-1}}{a+b-1} \end{aligned}$$

We claim that $E_{a, b}=\frac{a(a-1)}{2(a+b-1)}$, which will yield an answer of $\frac{45}{19}$. To prove this, we use induction. In the base case of $a=0$ we find $\frac{a(a-1)}{2(a+b-1)}=0$, as desired. Also, for $a>0$ we have that by the inductive hypothesis
$$\begin{aligned} E_{a, b} & =\frac{(a-1)((a-2)(a-3)+2(a+b-3))+b(a-1)(a-2)}{2(a+b-1)(a+b-3)} \\ & =\frac{(a-1)(a-2)(a+b-3)+2(a-1)(a+b-3)}{2(a+b-1)(a+b-3)} \\ & =\frac{a(a-1)}{2(a+b-1)} \end{aligned}$$
as desired.