Use Stirling's approximation to evaluate the probability $\lim_{n\to \infty}\binom{2n}{n}\left(\frac{1}{4}\right)^n$ As part of a some probability problem (probability of gettin $n$ heads and $n$ tails in $2n$ trials with a fair coin) I have derived the formula: $$p_n=\binom{2n}{n}\left(\frac{1}{4}\right)^n$$
I want to evaluate:
$$\lim_{n\to \infty} p_n$$
for $n\in \mathbb N$, using the following bound:
$$e^{\frac{1}{12n+1}}<\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}<e^{\frac{1}{12n}}$$
This is a "special" case of Stirling's approximation from Herbert Robbins that can be found here. I am stuck because I am not sure if my calculation is correct and the result makes no sense to me. When I calculate the limit I get zero. Does this not contradict the law of large numbers? Wouldn't I expect to get a distribution of $50 \%$ heads and $50\%$ tails if I toss a fair coin an infinite amount of times which then should imply $\lim_{n \to \infty} p_n=1$?
What I have tried so far:
$$\begin{equation*}\begin{split}\lim_{n\to \infty} p_n&=\lim_{n\to \infty} \binom{2n}{n} \left(\frac{1}{4}\right)^n \\[10pt] &=\lim_{n \to \infty} \frac{(2n)!}{(n!)^24^n} \end{split}\end{equation*}$$
Rewriting the approximation:
$$ \begin{equation*}\begin{split} &\phantom{\iff} \, \, \, \,e^{\frac{1}{12n+1}}<\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}<e^{\frac{1}{12n}} \\[10pt] & \iff e^{\frac{1}{12n+1}}  \sqrt{2\pi n}\left(\frac{n}{e}\right)^n < n!<e^{\frac{1}{12n}} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\end{split}\end{equation*}$$
Replacing $n!$ in $p_n$
$$\begin{equation*}\begin{split} &\phantom{iff}\frac{(2n)!}{\left(e^{\frac{1}{12n+1}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n}<\frac{(2n)!}{(n!)^2 4^n}<\frac{(2n)!}{\left(e^{\frac{1}{12n}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n} \\[10pt] &\iff\underbrace{\frac{1}{\left(e^{\frac{1}{12n+1}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n}}_{:=g(n)}<\underbrace{\frac{1}{(n!)^2 4^n}}_{:=f(n)}<\underbrace{\frac{1}{\left(e^{\frac{1}{12n}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n}}_{:=h(n)}\end{split}\end{equation*}$$
From the squeeze theorem: If $\lim_{n\to \infty} g(n)=\lim_{n \to \infty} h(n)=L$ and $g(n) < f(n) < h(n)$ then $\lim_{n \to \infty} f(n)=L$
$$\begin{equation*}\begin{split}\lim_{n \to \infty} g(n) &= \lim_{n \to \infty} \frac{1}{\left(e^{\frac{1}{12n+1}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n} \\[10pt] &=\frac{1}{2 \pi}\lim_{n \to \infty} \frac{1}{e^{\frac{2}{12n+1}} e^{(-2n)} n^{(2n+1)}4^n} \\[10pt] &= \frac{1}{2\pi} \lim_{n \to \infty }\frac{e^{-\frac{24n}{12n+1}}}{n^{2n+1}4^n}=0\end{split}\end{equation*}$$
Analogous $\lim_{n \to \infty} h(n)=0 \implies \lim_{n \to \infty } f(n)=0$. Therefore,
$$\lim_{n \to \infty}p_n=0$$
 A: The correct estimations are as follows:
$$
\frac{{(2n)!}}{{n!^2 }} \le \frac{{(2n)^{2n} e^{ - 2n} \sqrt {2\pi 2n} e^{\frac{1}{{24n}}} }}{{\left( {n^n e^{ - n} \sqrt {2\pi n} e^{\frac{1}{{12n + 1}}} } \right)^2 }} = \frac{{4^n }}{{\sqrt {\pi n} }}e^{\frac{1}{{24n}} - \frac{2}{{12n + 1}}} 
$$
and
$$
\frac{{(2n)!}}{{n!^2 }} \ge \frac{{(2n)^{2n} e^{ - 2n} \sqrt {2\pi 2n} e^{\frac{1}{{24n + 1}}} }}{{\left( {n^n e^{ - n} \sqrt {2\pi n} e^{\frac{1}{{12n}}} } \right)^2 }} = \frac{{4^n }}{{\sqrt {\pi n} }}e^{\frac{1}{{24n + 1}} - \frac{1}{{6n}}} .
$$
Thus,
$$
\frac{1}{{\sqrt {\pi n} }}e^{\frac{1}{{24n + 1}} - \frac{1}{{6n}}}  \le p_n  \le \frac{1}{{\sqrt {\pi n} }}e^{\frac{1}{{24n}} - \frac{2}{{12n + 1}}} .
$$
In particular,
$$
p_n  \sim \frac{1}{{\sqrt {\pi n} }}
$$
as $n\to +\infty$.
A: $0$ is correct for the limit.
Suppose you flipped a coin $2n$ times.  You are asking for the probability exactly $n$ are heads.
For this binomial distribution, the mean is $n$ and variance is $\frac n2$.  A normal distribution with variance $\sigma^2$ would have a peak density of $\frac{1}{\sqrt{2\pi\sigma^2}} = \frac{1}{\sqrt{\pi n}}$ here so we can use a Central Limit Theorem argument to suggest $\sqrt{\pi n}\, p_n \to 1$
Here are some values:
        n      p_n     1/sqrt(pi n)                 
        1 0.5000000000 0.5641895835
       10 0.1761970520 0.1784124116
      100 0.0563484790 0.0564189584
     1000 0.0178390111 0.0178412412
    10000 0.0056418253 0.0056418958
   100000 0.0017841219 0.0017841241
  1000000 0.0005641895 0.0005641896

A: To see why the limiting probability is zero, consider that as $n$ becomes extremely large, the random variable $$Z = \frac{X - 2np}{\sqrt{2np(1-p)}} \sim \operatorname{Normal}(0,1)$$ where $$X \sim \operatorname{Binomial}(2n, p = 0.5).$$  That is to say, $Z$ is approximately standard normal, with the approximation improving as $n$ increases.  Then with continuity correction, $$\Pr[X = n] = \Pr[n - 1/2 \le X \le n + 1/2] = \Pr\left[-\sqrt{2/n} \le \frac{X - n}{\sqrt{n/2}} \le \sqrt{2/n} \right] \approx \Pr\left[ |Z| \le \sqrt{2/n}\right].$$  As $n \to \infty$, it follows that $\Pr[X = n] \to 0$.
This should also illustrate on an intuitive level why the LLN is not violated:  the specific event $X = n$ becomes less likely simply because there are many more possible outcomes clustered around it.  In other words, if I toss a coin $2$ times, the probability of getting exactly $1$ head is actually quite large, but as I toss the coin $200$ times, the chance of getting exactly $100$ heads is naturally much smaller because you have many other possible outcomes in which a fair coin can give "close to" but not exactly $100$ heads.
