# Expected number of draws to get at least N red balls among a box of red and black balls

I have found this topic (Average number of trials until drawing $k$ red balls out of a box with $m$ blue and $n$ red balls) which gives the answer but I'd like to know what is wrong with my reasoning (see below).

Here is the problem: let's have $R$ red balls and $B$ black balls in a box, and let $N=R+B$ be the total number of balls.

The balls are drawn one by one without replacement. What is the expected number of draws needed to have $M \leq R$ red balls?

I have had the following reasoning, but it is obviously false since the sum of probabilities is not 1.

Let $L=R-M$ be the number of red balls left over in the box when the success condition is reached.

Let $n$ be the number of draws and $p(n)$ be the probability of reaching the success condition after these draws.

If $n<M$ (not enough draws) or $n>N-L$ (even in the worst case where only red balls would be left in the box, we would still have drawn enough of them), $p(n)=0$

Otherwise, we succeed after $n$ draws if there is left in the box exactly $L$ red balls and $N-n-L$ black balls, which I think gives:

$$p(n) = pr^L * pb^{N-L-n} * \binom {N-n} {L}$$

with $pr = R/N$ (respectively $pb = B/N$) the probability of drawing a red (respectively black) ball.

So the question is: what is wrong with my reasoning? And what would be the correct answer to get the expected number of draws needed?

For $n\in\{M,\dots,M+B=N-L\}$ we succeed if the $n$-th ball drawn is red and exactly $M-1$ of the preceding draws result in a red ball.
$$p(n)=\Pr(n\text{-th draw red}\mid M-1\text{ of preceding draws red})\Pr(M-1\text{ of preceding draws red})$$
$$\Pr(M-1\text{ of preceding draws red})=\frac{\binom{R}{M-1}\binom{B}{n-M}}{\binom{N}{n-1}}$$ and:$$\Pr(n\text{-th draw red}\mid M-1\text{ of preceding draws red})=\frac{R-M+1}{N-n+1}=\frac{L+1}{N-n+1}$$
The expectation can be found now as $\sum np(n)$ but there is a much nicer way. For that see this answer on the linked question.