A Probability problem on an Election result

A friend of mine recently posed the following problem to me: Given two contestants in an election $X$ and $Y$ (with equal popularity among the voters), what is the probability that the contestant leading the election after 80% of polling eventually loses the election?

I misunderstood the question in ways more than one and, not wanting to use paper-pencil, solved this problem with an answer of $(2/\pi)\tan^{-1}4$ which was wrong.

I enquired with him the source of the problem which also has a solution. But the solution there seems very convoluted and needs Wolfram Alpha to get a closed form. Seeing the solution, I realized that I have misunderstood the question and that my method is a lot easier.

Let $X_1,Y_1$ be the votes received by the contestants in after 80% of polling respectively and $X_2,Y_2$ be the votes respectively in the remaining 20% polling. For the sake of simplicity, let the total number of votes be $20n$ for a large $n$.

We know that, if $X\sim \text{Bin}(m,p)$, then for large $m$, $X$ is approximately distributed as $\mathcal{N}(mp,mpq)$. Therefore, for $p=q=1/2$,

$\displaystyle X_1,Y_1\sim \mathcal{N}\left(\frac{16n}{2}, \frac{16n}{4}\right)$ and $\displaystyle X_2,Y_2 \sim \mathcal{N}\left(\frac{4n}{2}, \frac{4n}{4}\right)$

Let $E$ be the event denoting the player trailing after 80% polling eventually wins the election. Then,

$\displaystyle \mathbb{P}(E)=2\cdot \frac{1}{2}\cdot \mathbb{P}(X_1+X_2 \leq Y_1+Y_2 \text{ and }X_1 \geq Y_1)$

We can rewrite the same to get

$\mathbb{P}(E)=\mathbb{P}(X_1-Y_1 \leq Y_2 - X_2 \text{ and }X_1-Y_1 \geq 0)$

We also know that if $U\sim \mathcal{N}(\mu_1,\sigma_1^2)$ and $V\sim \mathcal{N}(\mu_2,\sigma_2^2)$, then $aU+bV\sim \mathcal{N}(a\mu_1+b\mu_2,a^2\sigma_1^2+b^2\sigma_2^2)$

Therefore, $X_1-Y_1 \sim \mathcal{N}\left(0,8n\right)=\sqrt{8n}Z_1$ and $X_2-Y_2\sim \mathcal{N}\left(0,2n\right)=\sqrt{2n}Z_2$

where $Z_1$ and $Z_2$ are standard Normal variables.

Therefore,

$\begin{align}\displaystyle \mathbb{P}(E)&=\mathbb{P}(2\sqrt{2n}Z_1\leq \sqrt{2n}Z_2 \text{ and } 2\sqrt{2n}Z_1 \geq 0)\\ &=\mathbb{P}(2Z_1\leq Z_2 \text{ and } Z_1 \geq 0) \\ &= \mathbb{P}(Z_1 \leq Z_2/2 \text{ and }Z_1 \geq 0)\\ &=\mathbb{P}(0 \leq Z_1 \leq Z_2/2) \\ &= \mathbb{P}\left(0 \leq \frac{Z_1}{Z_2} \leq \frac{1}{2}\right)\\ &= \mathbb{P}\left(0 \leq W \leq \frac{1}{2}\right)\\ \end{align}$

where $W$ is the ratio of two standard Normal distributions and hence a Cauchy random variable. As the CDF of a Cauchy variable has a closed form in terms of arctangent function, we finally have

$\displaystyle \mathbb{P}(E)=\frac{1}{\pi}\tan^{-1}\left(\frac{1}{2}\right)\approx 0.1475$

If $E$ denotes the event that the contestant trailing when the fraction $a$ of votes remains to counted eventually wins the election. then

$\displaystyle \mathbb{P}(E)=\frac{1}{\pi}\tan^{-1}\left(\sqrt{\frac{a}{1-a}}\right)=\frac{1}{\pi}\sin^{-1}\sqrt{a}$

Hope you enjoyed the discussion. See ya in the next post.

Until Then

Yours Aye

Me