Asymptotics of $3n$ coin flips

Question

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.

Answer 1

Simply use that $$ \log {3n \choose 2n} = 3n\log(3n)-3n -2n\log(2n)+2n-n\log(n)+n +O(\log(n)) = n\log\left(\frac{27}{4}\right) $$ implies that

$$ P("\text{twice as often heads than tails}") \sim \frac{1}{8^n} \frac{27^n}{4^n} = \frac{27^n}{32^n} \ll 1 $$

Answer 2

Suppose that we toss a fair coin $N$ times. What is the probability that we get $p N$ heads and $(1-p)N$ tails? It is $2^{-N}{N \choose pN}$. Now, $${N \choose pN} \approx 2^{-H(p) N},$$ where $H(p)$ is the binary entropy function, defined as $H(p) = p \log_2\frac1p + (1-p) \log_2 \frac{1}{1-p}$ (see also https://en.wikipedia.org/wiki/Binary_entropy_function). Thus, the probability of the desired event is approximately $2^{(-1 + H(p)) N}$.

Plugging in $N = 3n$ and $p = 1/3$, we get that the probability is approximately $(27/32)^n$.

Answer 3

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 $$ \begin{align} P(\text{twice as many heads})&= {3n\choose 2n}(1/2)^{3n} =0.0000141031 \\ \quad\text{and}\quad \sqrt{3\over 4\pi n}\ \left({27\over32}\right)^n&=0.0000141306. \end{align} $$