Urn problem with balls Say you have $2n$ black balls and an additional blue and a red ball.
If there exist two urns and you randomly throw $n+1$ balls in each, then what is the probability that you do not have both the blue and the red balls in the same urn? 
 A: There are $\binom{2n+1}{n+1}$ ways to fill the first urn. But we only want those in which urn $1$ has both colored balls or none of them.
There are $\binom{2n-2}{n-1}$ which have both red and blue.
There are $\binom{2n-2}{n+1}$ which have neither.
Thus the answer is $$\frac{\binom{2n-2}{n-1}+\binom{2n-2}{n+1}}{\binom{2n+2}{n+1}}$$
A: Carry out the splitting and then color the balls.
Specifically, split $2n+2$ black balls between the urns and then randomly and independently color one ball blue and the other red.  The chance they are both in the same urn is the number of ways to choose two balls from one urn, plus the number of ways to choose them from the other urn, all divided by the number of ways in toto to choose two balls.  Subtracting this from $1$ gives the chance they are in different urns:
$$1 - \left(\binom{n+1}{2} + \binom{n+1}{2}\right) / \binom{2n+2}{ 2} = \frac{n+1}{2n+1} = \frac{1}{2} + \frac{1}{2(2n+1)}.$$
This formula, because it is simpler than others that might be derived from the original description, lends itself to easy simplification and clearly shows just how close to $1/2$ the chance is.
A: This looks like a hypergeometric distribution. We choose for Urn 1, which fixes Urn 2. There are $2n$ black balls, and we choose $n$. Then we choose $1$ of the $2$ remaining balls, and divide out by the number of ways to choose $n+1$ balls:
$$\frac{ \binom{2n}{n} \binom{2}{1} } { \binom{2n+2}{n+1} }$$
