This is a comment not an answer, although I hope this can be used to create an answer.

There is a natural martingale you can associate with this process. When there are $r$ red and $b$ blue balls, bet $r$ against $b$ that you will draw a red ball (and eliminate one of the blue balls). The nice thing about this procedure is that the value doesn't depend on the path. To go from $(n,n)$ to $(r,b)$, you must win $n-b$ times, gaining $n, n-1, n-2, ..., n-b+1$. You must lose $n-r$ times, losing $n, n-1, ..., n-r+1$, so that you net $\frac{r(r+1)}{2}-\frac{b(b+1)}{2}$. If you end up with $r$ red balls left, the value of this martingale is $\frac{r(r+1)}{2}$.

This gives a quick way to see that if there is a scaling law, it must be $n^{3/4}$. When you make a fair bet of $r$ to win $b$, the variance is $rb$. The total variance along any path from $(n,n)$ to $(r,0)$ or $(0,b)$ is $\Theta(n^3)$. For this to be $\Theta(n^c(n^c+1)/2)^2$, $4c = 3$ so $c=3/4$.

left in the bagare the same color. – Joel Reyes Noche May 17 '13 at 6:28