Georgia on my mind

The riddle of this week was inspired by the latest presidential elections when one State after another flipped the winner from Trump to Biden. Incl. Georgia.

On election night, the results of the 80 percent who voted on Election Day are reported out. Over the next several days, the remaining 20 percent of the votes are then tallied. What is the probability that the candidate who had fewer votes tallied on election night ultimately wins the race?

Assuming many votes, perfect balance between both candidates (p=½), and homogeneity between early and late ballots, the question boils down to the probability of a sum of two normals, X+Y, ending up being of the opposite sign from X, when the variances of X and Y are α and 1-α. Which writes as the expectation

2 \mathbb{E}_\alpha[\Phi(-X/\sqrt{1-\alpha})]

equal to

\frac{2}{2\pi}\left(\frac{\pi}{2} + \arctan\{\sqrt{\alpha/(1-\alpha)|}\}\right)

which returns a probability of about 0.14 when α=0.8. When looking at the actual data for Georgia, out of 5 million voters, at some point 235,000 ballots remained to be counted with Trump on the lead. This means an α about 0.05 and implies a probability of 7% (not accounting for the fact that the remaining mail-in-ballots were more favourable to Biden.)

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