# Simple really.

Here is how the problem from class works out.

We can approximate $k!$ using $k^{k+\frac12}e^{-k}\sqrt{2\pi}$.

The probability that one voter makes the decisive vote is

$\frac{(2k)!}{k!k!2^{2k}}$

Plugging rack factorial in to the above we get:

$~\frac{(2k)^{2k+\frac12}e^{-2k}\sqrt{2\pi}}{k^{2k+1}e^{-2k}(2\pi)2^{2k}}$
$=\frac{(2k)^{2k+\frac12}\sqrt{2\pi}}{k^{2k+1}(2\pi)2^{2k}}$
$=\frac{(2k)^{2k+\frac12}}{k^{2k+1}\sqrt{2\pi}2^{2k}}$
$=\frac{(2k)^{2k}(2k)^{\frac12}}{k^{2k+1}\sqrt{2\pi}2^{2k}}$
$=\frac{k^{2k}\sqrt{2k}}{k^{2k+1}\sqrt{2\pi}}$
$=\frac{k^{2k}\sqrt{2k}}{k^{2k}k\sqrt{2\pi}}$
$=\frac{\sqrt{2k}}{k\sqrt{2\pi}}$
$=\frac{\sqrt{k}}{k\sqrt{\pi}}$
$=\frac{1}{\sqrt{k\pi}}$

At last.