Consider an urn with $r$ red and $b$ black balls in it. We start drawing balls (without replacement) from it with following rules:
If the drawn ball is red, we stop the draw with probability $p$. Conversely, we continue the draw with probability $1-p$.
If the drawn ball is black, we stop the draw.
What is the expected number of drawn balls?
For $p=0$ the question was already solved here. It doesn't seem clear to me how to incorporate the random stop into the solution of $p=0$ case.
You will draw the first ball with probability $1.$ The probability that you stop after drawing the first ball is $$ \frac{b + pr}{b + r}. $$ The probability that you draw the second ball is $$ 1 - \frac{b + pr}{b + r} = \frac{qr}{b + r}.$$ The probability that you draw the $n$th ball, given that you drew the $(n-1)$st ball, is $$ 1 - \frac{b + p(r + 2 - n)}{b + r + 2 - n} = \frac{q(r + 2 - n)}{b + r + 2 - n}. $$
So if $N$ is the total number of balls drawn, \begin{align} \mathbb P(N\geq1) &= 1, \\ \mathbb P(N\geq2) &= \frac{qr}{b + r}, \\ \mathbb P(N\geq3) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1}, \\ \mathbb P(N\geq4) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1} \cdot \frac{q(r - 2)}{b + r - 2}, \\ \mathbb P(N\geq k) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1} \cdots \frac{q(r + 2 - k)}{b + r + 2 - k}, \\ \mathbb P(N\geq k + 1) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1} \cdots \frac{q(r + 1 - k)}{b + r + 1 - k}, \\ \mathbb P(N\geq r + 1) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1} \cdots \frac{q}{b + 1}, \\ \mathbb P(N\geq m) &= 0 \quad\text{if}\quad m \geq r + 2. \end{align}
The expectation of $N$ is $$ \mathbb E(N) = \sum_{k=1}^{r+1} \mathbb P(N \geq k). $$
We can write the probabilities a little more compactly in one of the following forms:
$$ P(N \geq k + 1) = \frac{r!\, (b+r-k)!\, q^k}{(r-k)!\,(b + r)!} = \frac{\displaystyle \binom rk q^k}{\displaystyle \binom{b + r}k} = \frac{r^{(k)} q^k}{(b + r)^{(k)}} $$
where $a^{(k)} = a(a-1)(a-2)\cdot(a-k+1)$ is the falling factorial. So for example, substituting $k = j + 1$ in the formula for $\mathbb E(N)$ above, we can write
$$ \mathbb E(N) = \sum_{j=0}^r \mathbb P(N \geq j + 1) = \sum_{j=0}^r \frac{r!\, (b+r-j)!\, q^j}{(r-j)!\,(b + r)!}, $$
but I have not been able to simplify this or any of the other forms further.
We can imagine that before any draws happen, all of the $r+b$ balls are randomly lined up, and then drawn in that order. Let $X$ be the number of red balls that occur before the earliest black ball in random that order. Then for each $x\in \{0,1,\dots,r\}$, $$ P(X = x)=\frac{r}{r+b}\cdot \frac{r-1}{r+b-1}\cdots \frac{r-x+1}{r+b-x+1}\cdot \frac{b}{r+b-x}=\frac{r!/(r-x)!\cdot b}{(r+b)!/(r+b-x-1)!} $$ Now, let $N$ be the total number of draws. First, we find the conditional expectation of $N$ given each possible value of $X$. For this part, I will use the conditional version of $E[X]=\sum_{k=0}^\infty P(X>k)$. $$ P[N>n\mid X=x]= \begin{cases}(1-p)^n & n\le x \\ 0 & n> x \end{cases} $$ Letting $q:=1-p$ to make things look neater, $$ E[N\mid X=x] =\sum_{n=0}^{x} P[N>n\mid X=x] =\sum_{n=0}^{x} q^n=\frac{1-q^{x+1}}{1-q} =\frac{1-q^{x+1}}{p} $$ Finally, we conclude $$ \begin{align} E[N] &=\sum_{x=0}^{r}E[N\mid X=x]P(X=x) \\&=\sum_{x=0}^r\frac{1-q^{x+1}}{p}\cdot \frac{r!/(r-x)!\cdot b}{(r+b)!/(r+b-x-1)!} \\&=\boxed{\frac1p-\frac {1-p}p\sum_{x=0}^r(1-p)^x\cdot \frac{r!/(r-x)!\cdot b}{(r+b)!/(r+b-x-1)!}} \end{align} $$ I do not know if this can be simplified any further.
