How to calculate $\frac{100!}{(50!)^2 2^{100}}$ solely using mental arithmetic? Question: When $100$ coins are tossed, what is the probability that exactly $50$ are heads?
I mange to solve the question with answer 
$$\binom{100}{50} \left(\frac{1}{2}\right)^{100} = \frac{100!}{(50!)^2 2^{100}}.$$
But if I want to approximate the quantity above solely using mental arithmetic, how would one approach it? 
 A: Stirling is a reasonable approach here.  We have 
$$\frac{100!}{(50!)^2 2^{100}}\approx \frac {100^{100}e^{50}e^{50}\sqrt{2\pi 100}}{50^{50}50^{50}e^{100}2^{100}(2\pi 50)}=\frac 1{\sqrt{50\pi}}\approx \frac 1{7\cdot 1.8}=\frac 1{12.6}$$
Where I took $\sqrt{50} \approx 7$ and $\sqrt \pi \approx 1.8$ because $\sqrt 3 \approx 1.732$ and $\pi$ is a little greater than $3$ 
I did this without checking with Alpha, which shows it is about $\frac 1{12.56}$
A: Completely aside from the mental math estimation, you may see that the standard deviation of the number of heads is $\sqrt{100(0.5)(0.5)} = 5$, which means that the probability of being between $45$ and $55$ is approximately $68/10 = 6.8$ percent, which we can bump up based on the fact that the peak is higher than the average (duh).  This gives us a decent ballpark estimate.

This seems as good an opportunity as any to plug my version of the empirical rule:


*

*The probability of being outside one standard deviation is approximately one in $\pi$.

*The probability of being outside two standard deviations is approximately one in $7\pi$.

*The probability of being outside three standard deviations is approximately one in $16e^\pi$.

A: Hint :
For large factorials you can use stirling's formula
$$n! \approx (\sqrt{2\pi}) n^{n+0.5}e^{-n}$$
But if you want to be more accurate you can use Ramanujan's factorial formula
$$n! \approx \sqrt{\pi}\left(\frac{n}{e}\right)^n \left(8n^3+4n^2+n+\frac{1}{30}\right)^{\frac{1}{6}}$$
