Suppose we live in a country with an interesting electoral system: each electorate has exactly 3 voters. 2 parties run for office, and each voter has a 50/50 chance of voting for each party. Whoever wins the majority of electorates wins the election overall. Given an arbitrarily large number of electorates, what is the probability that the party that won the election lost the popular vote? (This is a more specific version of my earlier question Probability of winning an election while losing the popular vote)

My 'brute force' computer model yields an answer of very close to 1/6th. Does anyone have ideas for how to solve this problem analytically?

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    $\begingroup$ Let see if I understand your question correctly. Let $(X_0, X_1, X_2, X_3)$ be the number of electorate in which party $A$ receive $0, 1, 2, 3$ votes respectively. Then they jointly follows $\displaystyle \text{Multinomial}\left(n; \frac {1} {8}, \frac {3} {8}, \frac {3} {8}, \frac {1} {8}\right)$, assuming there are $n$ electorates and all voters are mutually independent. Then the probability you asked should be $\Pr\{X_1 + 2X_2 + 3X_3 < 3X_0 + 2X_1 + X_2|X_2 + X_3 > X_0 + X_1\}$ $\endgroup$
    – BGM
    Jan 12, 2018 at 9:41
  • $\begingroup$ Thanks BGM, that's exactly right. $\endgroup$ Jan 13, 2018 at 4:47

1 Answer 1


The limiting probability is indeed $\frac{1}{6}$.

Call the two candidates $A$ and $B$. For $0\leq i\leq 3$, define random variables $X_i$ depending on a random district $$ X_i=\begin{cases}1&:\text{exactly $i$ votes for $A$ in that district},\\ 0&:\text{otherwise}. \end{cases} $$ The multivariate central limit theorem implies that as the number of districts $n$ goes to infinity, the numbers $N_i$ of districts for which $X_i=1$ are distributed like a multivariate Gaussian distribution. We are interested in the probability that $N_0+N_1-N_2-N_3$ and $3N_0+N_1-N_2-3N_3$ have the opposite sign. The probability is equal to $1/\pi$ times the arccosine of the correlation between $X_0+X_1-X_2-X_3$ and $3X_0+X_1-X_2-3X_3$. The correlation can be computed using $P(X_i=1)={3\choose i}/8$, and the correlation is $\sqrt{3}/2$. So the desired probability is $$ \frac{\cos^{-1}(\sqrt{3}/2)}{\pi}=\frac{1}{6}. $$

  • $\begingroup$ Nicely done! Btw, my calculations indicate that it's also a 1/6 probability in the case of 3 electorates, each with n voters in them, as n approaches infinity. Can you also prove that? $\endgroup$ Apr 7, 2021 at 14:50

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