How many times do I have to draw until I get $s$ red objects and $t$ white objects? We have red objects and white objects. The probability of drawing a red object or a white object is uniform, ie $0.5$. If when we draw an object we return it, then what is the average number of times we need to draw objects until we have $s$ red objects and $t$ white objects?
I understand that this is similar to the coupon collector problem. If $s,t$ are both $1$ then it is obvious that the average is $3$ since it is the random variable consisting of a sum of $G(1)$ and $G(\frac{1}{2})$. For $s=t$ in general (and $s\neq t$) I am having a bit of trouble. It doesn't matter what we get in the first $min(s,t)$ tries, we will always draw something. But then, the other tries depend on the results of the first ones, and proceeding to calculate it this way is in-feasible.
Will be glad for a hint on how to go about this. This is a homework question so please do not supply a full solution.
 A: At some point you will end up in one of the following situations:


*

*The last ball drawn was red and exactly $s$ red balls are drawn in total. Next to that $q\in\{0,\dots, r-1\}$ white balls have been drawn. Let's call this event $R_q$.

*The last ball drawn was white and exactly $r$ white balls are drawn in total. Next to that $p\in\{0,\dots, s-1\}$ white balls have been drawn. Let's call this event $W_p$.
If you are in the situation first sketched then $s+q$ balls are drawn and a geometric process now starts. You can find the corresponding mean.
If you are in the situation secondly sketched then $r+p$ balls are drawn and a geometric process now starts. You can find the corresponding mean.
It comes to finding the probabilities $P(R_q)$ and $P(W_p)$ to come to a full answer.
Can you take it from here?

edit: 
As an example to make things more clear let it be that $s=3,r=4,q=2$. 
Then $R_q=R_2$ is the event: $$RRWWR\cup RWRWR\cup RWWRR\cup WRRWR\cup WRWRR\cup WWRRR$$
and it has probability $\binom422^{-5}=6\cdot2^{-5}$ to occur. Here e.g. $RRWWR$ stands for the event that first two red balls are drawn, then two white balls and then a red ball. In this situation we are ready with the red balls ($s=3$ are drawn) but yet $2$ white balls must be drawn. The average waiting time for that to happen is $2\cdot2=4$.
If $X$ denotes the waiting time in this example then  $\mathbb E(X\mid R_2)=5+2\cdot2=9$
What we are really after is finding: $$\mathbb EX=\sum_{i=0}^3\mathbb E(X\mid R_i)P(R_i)+\sum_{i=0}^2\mathbb E(X\mid W_i)P(W_i)$$
A: Given $s$ and $t$ the expected number $E(s,t)$ of draws is given by
$$E(s,t)=s+t+E_{\rm futile}(s,t)\ ,$$ whereby $E_{\rm futile}(s,t)$ denotes the expected number of futile draws. The full history can be viewed as a random walk in the first quadrant of the lattice. The walk starts at $(s,t)$ and for each useful draw moves one step west or one step south until it reaches the origin. As soon as the walk hits the boundary $t=0$ or $s=0$ we observe the onset of futile draws: We have to expect one futile draw per step $(k,0)\to(k-1,0)$, resp., $(0,k)\to(0,k-1)$. If the walk hits the boundary for the first time at $(r,0)$ the expected number of futile draws therefore is $r$.
We therefore have to determine the probability that the walk hits the boundary $t=0$ for the first time at the point $(r,0)$,  $\>1\leq r\leq s$. This event happens iff the walk reaches the point $(r,1)$ and then makes a step south. There are $(s+t)-(r+1)$ steps from $(s,t)$ to $(r,1)$. The probability in question therefore is
$${1\over 2^{(s+t)-(r+1)}}{(s+t)-(r+1)\choose t-1}\>\cdot{1\over2}\ .$$
Since the walk may as well hit the boundary $s=0$ we therefore obtain
$$E_{\rm futile}(s,t)={1\over 2^{s+t}}\left(\sum_{r=1}^s {(s+t)-(r+1)\choose t-1}r\>2^r+\sum_{r=1}^t {(s+t)-(r+1)\choose s-1}r\>2^r\right)\ .$$
