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By central limit theorem, \[ \dbinom nk p^k q^{n-k} = \dfrac1{ \sqrt{2\pi n pq}} \cdot \exp \left [- \dfrac{(k-np)^2}{2npq} \right ] \]

Then, \[ \dfrac1{2^{n_1}} \dbinom {n_1}{A_1} = \dfrac1{\sqrt{\pi n_1 /2}} \cdot \exp \left [ - \dfrac{2(A_1 - \tfrac{n_1 }2)^2 }{n_1}\right ] \]

And \[ \dfrac1{2^{n_1}} \dbinom {n_1}{A_1} \cdot \dfrac1{2^{n_2}} \dbinom {n_2}{A_2} 
= \sqrt{\dfrac4{\pi^2 n_1 n_2}} \cdot \exp \left [ - \dfrac{2(A_1 - \tfrac{n_1 }2)^2 }{n_1}\right ] 
\exp \left [ - \dfrac{2(A_2 - \tfrac {n_2 }2)^2 }{n_2}\right ]  \]

We have \[ \large \begin{array} { r l } 
& \dfrac2\pi \cdot \dfrac1{\sqrt{n_1 n_2} } 
\displaystyle \int \limits_{1 + \frac N2 - n_2}^{\frac {n_1}2 - 1} \, d A_1
\int \limits_{1 + \frac N2 - A_1}^{n_2} \, d_2
\cdot 
\exp \left [ - \dfrac{2(A_1 - \tfrac{n_1 }2)^2 }{n_1}\right ] 
\exp \left [ - \dfrac{2(A_2 - \tfrac {n_2 }2)^2 }{n_2}\right ] \\ \phantom0 \\
& \approx 
\dfrac2\pi \cdot \dfrac1{\sqrt{n_1 n_2} } 
\displaystyle \int \limits_{\frac{n_1 - n_2 }{2}}^{\frac {n_1}2 } \, d A_1
\int \limits_{\frac{n_1 + n_2 }{2}  - A_1}^{n_2} \, d_2
\cdot 
\exp \left [ - \dfrac{2(A_1 - \tfrac{n_1 }2)^2 }{n_1}\right ] 
\exp \left [ - \dfrac{2(A_2 - \tfrac {n_2 }2)^2 }{n_2}\right ]
\end{array}
\]

Apply the substitutions \(A_2 ' = A_2 - \dfrac12 n_2\) and \( A_1 ' = A_1 - \dfrac12 n_1, \)

The expression above simplifies to \[ \dfrac2\pi \cdot \dfrac1{\sqrt{n_1 n_2}} \int_{-\frac {n_2} 2}^0 
 \, dA_1 ' \int_{-A_1 '}^{\frac12 n_2} \, d A_2 ' \cdot e^{-2 (A_2 ') ^2 / n_2 }  \cdot e^{-2(A_1 ') ^2 /n_1 } \]

Evaluating this double integral and doubling this value (because \(n_1 \) and \(n_2 \) are interexchangeable), we have \[ \dfrac{\pi - 2 \tan^{-1} \sqrt{ \dfrac{n_1}{n_2}} }{2\pi } \]

For the original question of \((n_1, n_2) = (80\%, 20\% ) \), this simplifies to \[ \dfrac{\pi - 2 \tan^{-1}(2) }{2\pi} = \dfrac1{2\pi \cdot \tan^{-1}(2)} \approx 14.75836\% \]

We can simplify the general expression by denoting \(f\) as the the fraction of the voters on election night, that is, \(f = n_1 , \), this further simplifies the formula to \[ \boxed{\dfrac{ \cos^{-1} (\sqrt f)}{\pi}} \]

Sat, 24 Apr 2021 15:09 GMT