I'll continue on the answer by Ross Millikan, and give an exact solution (I'll just replace $R$ and $W$ by lower case $r$ and $w$, which intimidates me less). So for $r,w\in\mathbb N$, not both zero, let $P(r,w)$ denote the probability of leaving a white ball as last one when starting with $r$ red and $w$ white balls. One obviously has $P(r,0)=0$ for any $r>0$ and $P(0,w)=1$ for any $w>0$; moreover by the argument Ross gives one has the recurrence relation
$$
P(r,w) = \frac{r^2}{(r+w)^2}P(r-1,w) + \frac{w^2+2rw}{(r+w)^2}P(r,w-1) \quad\text{for all } r,w>0,
$$
since the first step reduces the problem defining $P(r,w)$ either to the one defining $P(r-1,w)$ or to the one defining $P(r,w-1)$, with the indicated factors as probablilities for the first step.
Now this recurrence has the (surprisingly simple) solution
$$
P(r,w)=\frac{w}{(r+1)(r+w)}\quad\text{for }(r,w)\in\mathbb N^2\setminus\{(0,0)\}.
$$
The proof is by a simple induction on $r+w$, expanding the recurrence relation, factoring out $\frac w{(r+w)^2(r+w-1)(r+1)}$ which leaves $r(r+1)+(w+2r)(w-1)$ as other factor, and rewriting that factor as $(r+w)(r+w-1)$ permits some cancelling and arriving at the desired result.
I did not guess the formula just like that of course. Rather I calculated a number of values $P(1,w)$ with exact rational arithmetic, noticing substantial simplifications and easily guessing $P(1,w)=\frac{w}{2(w+1)}$. I proceeded similarly for $P(2,w)$, again with significant simplfications, after which I guessed the general form.
For the concrete problem one gets $P(10,90)=\frac9{110}\approx0.08181818$, a chance of a bit less than $1$ in $12$ to be left with a white ball. In fact one sees that throwing in any number of white balls (with $10$ red balls) initially never even raises the chance to $1$ in $11$. And if there are any red balls at all initially, it is always more likely to be left with one of them than with a white ball!
Let $b_1$ be the event drawing a black ball the first draw; and $b_2$ the event of drawing a black ball the second draw.
We want $P(b_2|b_1)$. That is, we want $\frac{P(b_2\cap b_1)}{P(b_1)}$.
We have $P(b_1)=\frac12\cdot\frac12+\frac12\cdot\frac34=\frac58$.
To figure out $P(b_1\cap b_2)$ we note that there are four ways this can happen:
(i) Bin A, Black, Bin A, Black, with probability $\frac12\cdot\frac12\cdot\frac12\cdot\frac13=\frac1{24}$
(ii) Bin A, Black, Bin B, Black, with probability $\frac12\cdot\frac12\cdot\frac12\cdot\frac34=\frac3{32}$
(iii) Bin B, Black, Bin A, Black, with probability $\frac12\cdot\frac34\cdot\frac12\cdot\frac12=\frac3{32}$
(iv) Bin B, Black, Bin B, Black, with probability $\frac12\cdot\frac34\cdot\frac12\cdot\frac23=\frac18$
So $P(b_1\cap b_2)=\frac{1}{24}+\frac3{32}+\frac3{32}+\frac18=\frac{34}{96}=\frac{17}{48}$
Then the overall conditional probability $P(b_2|b_1)=\frac{\frac{17}{48}}{\frac58}=\frac{17}{30}$
Best Answer
Think of the order you draw the balls out as a sequence, and consider the very last ball in that sequence - whichever colour it is will be the colour left in the bag when you reach the given stopping condition (i.e. if the last ball to be drawn would be white, then you must have drawn all the red balls already, and vice versa). Since we only care about what that last ball is, we can assign its colour first and then ignore the rest of the sequence, and that will be red with probability equal to $\frac{\# \mbox{ of red balls}}{\# \mbox{ of total balls}} = \frac{r}{w+r}$.