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?

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 $\frac 1{12.56}$