A colored ball problem Say you have $2n+2b$ balls where $2n$ balls are colored white, $b$ balls are colored blue and $b$ balls are colored red.
You have two urns. You randomly choose $n+b$ balls and throw in urn $1$ while you place the remaining $n+b$ balls in urn $2$. 
What is the probability that the blue balls and red balls are in separate urns?
I am most interested in case $\frac{n}b\rightarrow\infty$ such as $b=n^{\frac1c}$ with $c>1$ being fixed and in case $\frac{n}b\rightarrow c$ such as $b={\frac nc}$ with $c>1$.
 A: There are $\binom{2n}n$ ways of selecting $n$ of the $2n$ white balls to go with the red balls. Thus, for $b\gt0$ there are $2\binom{2n}n$ ways of separating the red and blue balls, where the factor $2$ occurs because the red balls can be in either of the two urns. There are $\binom{2n+2b}{n+b}$ ways to select balls for one of the urns, so this is the total number of outcomes. Since the selections are equiprobable, the probability for a separating selection is
$$
\frac{2\binom{2n}n}{\binom{2n+2b}{n+b}}\;.
$$
For $b=0$, the factor of $2$ should be omitted, and the probability of (trivial) separation is $1$.
A: Let's pretend for a moment that the balls are numbered (therefore distinct), the total number of possible choices is ${2n+2b}\choose{n+b}$. In how many of these do you get all of the blue balls and none of the red balls in the first urn? You have to pick $n$ of the $2n$ white balls to place in the first urn and then add all pf the blue ones (no choice), that is ${2n}\choose n$.
Now you only have to multiply this by two (for symmetry between the urns or the colors, depending on your point of view) and end up with $2\frac{{2n}\choose n}{{2n+2b}\choose{n+b}}$.
