91Ó°ÊÓ

A larger coin tossing experiment is underway. N=1000 coins are flipped. Stirling's approach may be used to calculate the likelihood of receiving exactly 500 head and 500 tails. The total number of possible outcomes is as follows:

$2^N=2^{1000}$

Where $n=500$heads as

$\begin{array}{l}\mathrm{Î\copyright }(N,n)=\left(\begin{array}{c}N\\ n\end{array}\right)=\frac{N!}{n!(N−n)!}\\ \mathrm{Î\copyright }(1000,500)=\left(\begin{array}{c}1000\\ 500\end{array}\right)\\ \mathrm{Î\copyright }(1000,500)=\frac{1000!}{500!(1000−500)!}\\ \mathrm{Î\copyright }(1000,500)=\frac{1000!}{500{!}^2}.\end{array}$

Using Stirling's approximation as

$\begin{array}{l}N!≈N^N{e}^{−N}\sqrt{2\mathrm{Ï€}N}\\ \mathrm{Î\copyright }(1000,500)=\frac{1000!}{500{!}^2}≈\frac{{1000}^{1000}{e}^{−1000}\sqrt{2\mathrm{Ï€}â‹\dots 1000}}{{\left({500}^{500}{e}^{−500}\sqrt{2\mathrm{Ï€}â‹\dots 500}\right)}^2}\\ \mathrm{Î\copyright }(1000,500)≈\frac{{1000}^{1000}×e^{−1000}×\sqrt{1000}}{{500}^{1000}×e^{−1000}×500×\sqrt{2\mathrm{Ï€}}}\end{array}$

From the above equation,

$\begin{array}{l}\mathrm{Î\copyright }(1000,500)≈\frac{(500×2{)}^{1000}×\sqrt{1000}}{{500}^{1000}×500×\sqrt{2\mathrm{Ï€}}}\\ \mathrm{Î\copyright }(1000,500)≈\frac{(2{)}^{1000}×\sqrt{1000}}{500×\sqrt{2\mathrm{Ï€}}}\end{array}$

Probability approximately getting exactly role="math" localid="1650333432268" $500$heads as

$\begin{array}{l}P\left(n\right)=\frac{\mathrm{Î\copyright }(N,n)}{2^N}\\ P\left(500\right)=\frac{\mathrm{Î\copyright }(1000,500)}{2^{1000}}\end{array}$

Substituting,we get

$\begin{array}{l}P\left(500\right)=\frac{\mathrm{Î\copyright }(1000,500)}{2^{1000}}≈\frac{1}{2^{1000}}\frac{(2{)}^{1000}×\sqrt{1000}}{500×\sqrt{2\mathrm{Ï€}}}\\ P\left(500\right)≈\frac{\sqrt{1000}}{500×\sqrt{2\mathrm{Ï€}}}\\ P\left(500\right)≈0.02523.\end{array}$

The number getting $n=500$heads as

$\begin{array}{l}\mathrm{Î\copyright }(N,n)=\left(\begin{array}{c}N\\ n\end{array}\right)\\ \mathrm{Î\copyright }(N,n)=\frac{N!}{n!(N−n)!}\\ \mathrm{Î\copyright }(1000,600)=\left(\begin{array}{c}1000\\ 600\end{array}\right)\\ \mathrm{Î\copyright }(1000,600)=\frac{1000!}{600!(1000−600)!}\\ \mathrm{Î\copyright }(1000,600)=\frac{1000!}{600!400!}.\end{array}$

Using Stirling's approximation as

role="math" localid="1650333713655" $\begin{array}{l}N!≈N^N{e}^{−N}\sqrt{2\mathrm{Ï€}N}\\ \mathrm{Î\copyright }(1000,600)=\frac{1000!}{600!400!}≈\frac{{1000}^{1000}{e}^{−1000}\sqrt{2\mathrm{Ï€}×1000}}{\left({600}^{600}{e}^{−600}\sqrt{2\mathrm{Ï€}×600}\right)\left({400}^{400}{e}^{−400}\sqrt{2\mathrm{Ï€}×â‹\dots 400}\right)}\\ \mathrm{Î\copyright }(1000,600)≈\frac{{1000}^{1000}×e^{−1000}×\sqrt{1000}}{{400}^{400}×{600}^{600}×e^{−1000}×\sqrt{600×400}×\sqrt{2\mathrm{Ï€}}}\end{array}$

From the above equation,

$\begin{array}{l}\mathrm{Î\copyright }(1000,600)≈\frac{2^{1000}×{500}^{1000}×\sqrt{1000}}{{400}^{400}×{600}^{600}×\sqrt{600×400}×\sqrt{2\mathrm{Ï€}}}\\ \mathrm{Î\copyright }(1000,600)≈\sqrt{\frac{1}{480\mathrm{Ï€}}}×\frac{2^{1000}×{500}^{1000}}{{400}^{400}×{600}^{600}}\end{array}$

Probability approximately getting exactly $500$heads as

$\begin{array}{l}P\left(n\right)=\frac{\mathrm{Î\copyright }(N,n)}{2^N}\\ P\left(600\right)=\frac{\mathrm{Î\copyright }(1000,600)}{2^{1000}}\end{array}$

Substituting,we get

$\begin{array}{l}P\left(600\right)=\frac{\mathrm{Î\copyright }(1000,600)}{2^{1000}≈\frac{1}{2^{1000}}}×\sqrt{\frac{1}{480\mathrm{Ï€}}}×\frac{2^{1000}×{500}^{1000}}{{400}^{400}×{600}^{600}}\\ P\left(600\right)≈\sqrt{\frac{1}{480\mathrm{Ï€}}}×\frac{{500}^{1000}}{{400}^{400}×{600}^{600}}\end{array}$

The ratio is not applicalable, so simplify as

$\begin{array}{l}\sqrt{\frac{1}{480\mathrm{π}}}×\frac{{500}^{1000}}{{400}^{400}×{600}^{600}}=\sqrt{\frac{1}{480\mathrm{π}}}×\frac{(1.25×400{)}^{1000}}{(1×400{)}^{400}×(1.5×400{)}^{600}}\\ →\sqrt{\frac{1}{480\mathrm{π}}}×\frac{(400{)}^{1000}×(1.25{)}^{1000}}{(400{)}^{400}×(1{)}^{400}×(400{)}^{600}×(1.5{)}^{600}}\end{array}$

But,

$\begin{array}{l}\frac{(400{)}^{1000}}{(400{)}^{600}×(400{)}^{400}}=\frac{(400{)}^{1000}}{(400{)}^{1000}}=1\\ →\sqrt{\frac{1}{480\mathrm{π}}}×\frac{(1.25{)}^{1000}}{(1{)}^{400}×(1.5{)}^{600}}\end{array}$

Where,

$\begin{array}{l}P\left(600\right)≈\sqrt{\frac{1}{480\mathrm{π}}}×\frac{(1.25{)}^{1000}}{(1{)}^{400}×(1.5{)}^{600}}\\ P\left(600\right)≈4.635×{10}^{−11}.\end{array}$

Where as Maple is working large exponents directly.