Probability of selecting a red ball first An urn contains 3 red and 7 black balls. Players A
and B withdraw balls from the urn consecutively
until a red ball is selected. Find the probability that
A selects the red ball. (A draws the first ball, then
B, and so on. There is no replacement of the balls
drawn.)
How do I calculate this probability?
I tried using the total probability rule without success.
I used the $P(A)=\frac{3}{10}+P(A_2\mid B_1)$ and so on, where $B_i$=Player B doesn't get a red ball.
The answer should be $0.0888$
 A: Hint:
If $p(r,b)$ denotes the probability that the person drawing first selects a red ball, when there are $r$ red balls and $b$ black balls, then $p(r,0)=1$ and for $b>0$: $$p(r,b)=\frac{r}{r+b}+\frac{b}{r+b}(1-p(r,b-1))=1-\frac{b}{r+b}p(r,b-1)$$
Find $p(3,0),p(3,1),\dots,p(3,7)$ in this order.
A: In the following probability the marks $1$st, $2$nd etc. indicate the probability of Player $A$ selecting the red ball in the $1$st, $2$nd round etc.: $$\begin{align*}P(A)&=\underbrace{\frac{3}{10}}_{1st}+\frac{7}{10}\cdot\frac{6}{9}\cdot\left(\underbrace{\frac38}_{2nd}+\frac58\cdot\frac47\cdot\left(\underbrace{\frac36}_{3rd}+\frac36\cdot\frac25\cdot\left(\underbrace{\frac34}_{4th}+\frac14\cdot0\right)\right)\right)=\ldots=\\[0.3cm]&=0.583333\end{align*}$$

The answer cannot be as low as $0.0888$ because that would indicate a strong advantage of player B, whereas player A has an advantage by starting first. Thus a result with $P(A)>1/2$ should be expected.
A: Here's what I was thinking.
$$\color{RED}R+BB\color{RED}R+BBBB\color{RED}R+BBBBBB\color{RED}R$$
$$P(A)=\frac{3}{10}+\frac{7}{10}\frac{6}{9}\frac{3}{8}+\frac{7}{10}\frac{6}{9}\frac{5}{8}\frac{4}{7}\frac{3}{6}+\frac{7}{10}\frac{6}{9}\frac{5}{8}\frac{4}{7}\frac{3}{6}\frac{2}{5}\frac{3}{4}=\frac{7}{12}$$
