Probability to draw $r$ Red balls before drawing $g$ Green balls I have a problem that I have reduced to the following but have not found a question on here with quite the same nuance: 
Given a bag of $T$ marbles, $R$ red ones and $G$ green ones, find the probability that I will draw exactly  $r$ Red marbles before drawing $g$ Green marbles.  I only stop drawing once I have pulled out $g$ green marbles and I draw without replacing any marbles.
For $r=0$,  I think I would have a probability of $(\dfrac{G}{T}*\dfrac{G-1}{T-1}*\dfrac{G-2}{T-2}*...*\dfrac{G-g+1}{T-g+1})$ or $\dfrac{G!}{T!}*\dfrac{(T-g)!}{(G-g)!}$ 
For greater $r$ I have tried a few things but they are certainly wrong (Sum over $r$ from $0$ to $R$ does not add to 1).  Any ideas on how to calculate these probabilities?

It helps to imagine that the marbles are numbered, $R_1,R_2, \cdots, R_R$ and $G_1,G_2,\cdots, G_G$.
We know the $(r+g)^{th}$ draw must be green.  Prior to that one, we must draw exactly $r$ reds and exactly $g-1$ greens (in any order).  There are $\binom Rr\times \binom G{g-1}$ ways to do that.  There are $\binom {R+G}{r+g-1}$ unrestricted ways to choose $r+g-1$ marbles.  Having chosen a good collection for the first $r+g-1$ we then need to choose a green, the probability of that is $\frac {G-(g-1)}{R+G-r-(g-1)}$.  Thus the answer is $$\boxed {\frac {\binom Rr\times \binom G{g-1}}{\binom {R+G}{r+g-1}} \times \frac {G-(g-1)}{R+G-r-(g-1)}}$$
Sanity checks:
I. If $R=r,G=g$ then of course the answer is just the probability that the last marble drawn is green, which is clearly $\frac G{R+G}$.  The formula gives $$\frac {\binom RR\times \binom G{G-1}}{\binom {R+G}{R+G-1}} \times \frac {G-(G-1)}{R+G-R-(G-1)}=\frac {G}{R+G}\times 1$$ as desired.
II.  If $r=0$ the formula becomes $$\frac {\binom {G}{g-1}}{\binom {R+G}{g-1}}\times \frac {G-(g-1)}{T-(g-1)}=\frac {G!}{T!}\times \frac {(T-g)!}{(G-g)!} $$ as you already demonstrated.
