Asymptotics of $3n$ coin flips Suppose we throw a coin $2n$ times. The probability of $n$ times heads (and therefore $n$ times tails) is $$P(\text{"n times heads"}) = \frac{1}{4^n}\binom{2n}{n}.$$
We can use Stirling's formula to get the asymptotics $$\frac{1}{4^n}\binom{2n}{n} \sim \frac{1}{\sqrt{\pi n}} $$
as $n\to \infty$. Now I want to throw the coin $3n$ times and look at the probability of the event $$P(\text{"twice as often heads than tails"}) = \frac{1}{8^n}\binom{3n}{2n}.$$
Is there any "nice" asymptotics for this too? I tried using Stirling's formula again, but it does not seem to be as nice as it could get.

Using Stirling's approximation you get
$$\begin{align}
P(\text{twice as many heads})
&= {3n\choose 2n}(1/2)^{3n}\\
&= \frac{(3n)!}{(2n)! \ n!\ 2^{3n}}\\
&\approx {
(3n)^{3n}\ e^{-3n}\ \sqrt{6\pi n} \over 
(2n)^{2n}\ e^{-2n}\ \sqrt{4\pi n}\
(n)^{n}\ e^{-n}\ \sqrt{2\pi n}\
2^{3n}}\\
&= \sqrt{3\over 4\pi n}\ \left({27\over32}\right)^n.
\end{align}
$$
Numerical precision is quite good ; for $n=50$ for instance
$$
\begin{align}
P(\text{twice as many heads})&= {3n\choose 2n}(1/2)^{3n} =0.0000141031
\\
\quad\text{and}\quad \sqrt{3\over 4\pi n}\ \left({27\over32}\right)^n&=0.0000141306.
\end{align}
$$
