Probability for getting $n$ black balls 
In a jar there are $ 5 $ white and $ 7 $ black balls. each time we choose a ball, it is returned with addition of two balls in the same color.Find the probability that the n first chosen balls are black.

For 2 balls the probability is $\dfrac{7\cdot 9}{12 \cdot 14}=\dfrac 3 8$, 
For 3 balls the probability is $\dfrac{7 \cdot 9 \cdot 11}{12 \cdot 14 \cdot 16}=\dfrac {33} {128}$
$\vdots$
For $n$ balls the probability is defined to be $\frac{\displaystyle\prod_{k=1}^{n}(5+2k)}{\displaystyle\prod_{j=1}^{n}(10+2j)}$.
Even though on first look it seems a final solution, when is n is divided by 4 some elements from the counter and the denominator are reducible. My question is if we can either determine which elements are reducible or simplify the result to a more "friendly" one. 
 A: If $P_n$ is the desired probability of choosing $n$ black balls in the first $n$ draws then $$P_n=\frac{\displaystyle \prod_{k=1}^n (5+2k)}{\displaystyle \prod_{j=1}^n (10+2j)}= \frac{\displaystyle \prod_{k=1}^n (5+2k)}{\displaystyle 2^n \prod_{j=1}^n (5+j)}=\frac{\displaystyle \prod_{k=\lceil\frac{n}{2}\rceil}^n (5+2k)}{\displaystyle 2^n \prod_{j=1}^{\lceil \frac{n}{2}\rceil} (5+2j-1)}$$
The result is not much simpler, but of course a little friendlier.
Another simple observation:
$$\frac{\ln (2^nP_n)}{n}=\frac{\displaystyle \sum_{k=1}^n \ln \left(1+\frac{k}{5+k}\right)}{n}\xrightarrow{n\rightarrow \infty}\ln 2\quad(\mbox{By Cesaro's theorem})$$
A: Using the double factorial and the fact that $\prod_{k=1}^n2k=2^nn!$ we can rewrite the fraction as
$$\frac{(2n+5)!!}{5\cdot3}\cdot\frac{2^5\cdot5!}{2^{n+5}(n+5)!}=\frac{(2n+5)!!}{2^{n-3}(n+5)!}\;,$$
where all of the odd factors in the denominator will cancel with factors in the numerator. It’s possible to do this cancellation symbolically, but the resulting expression will be somewhat uglier.
