# Ask a binomial distribution problem in the voting setup

Two candidates were running for a post. The voting machines recorded 520,000 votes for the first candidate and 480,000 votes for the second one. Afterwards it became apparent that the voting machines were defective— they randomly and independently switched each vote for the opposite one with probability of 45%. The losing candidate asked for a re-vote. Is there a basis for a re-vote?

Source: Theory of probability and random processes / leonid b. koralov, yakov g. sinai. Section 2.4 Question 7

My attempt: I know p=0.45 is the parameter in the binomial distribution. But I don't know how to build such a binomial distribution. I am thinking a basis for a re-vote can be there are 2000 votes from the first candidate should belong to the second candidate. Then I am studying $$P(X \geq 2000)$$ where X is Binom(520000, 0.45). Is this thinking correct?

• Unclear if I am interpreting the problem correctly. $$[520 \times 0.55] + [480 \times 0.45] = 502 > 500.$$ The presumption is that with such large numbers involved, the number of votes that need to be switched is very close to $(0.45)$ of the $(520000)$ votes in favor of the winner and $(0.45)$ of the $(480000)$ votes in favor of the loser. Assuming that these estimates are close enough, the original winner still wins, because $(502) > (500).$ Commented Sep 18, 2022 at 4:22
• @user2661923 Yes, this also makes sense to me. But I guess the point is I need to use normal approximation. Commented Sep 18, 2022 at 14:41
• I suspect that if you told a court that the machine result was a majority of $40000$ but should have been a majority about $90\%$ smaller, the judges might not be too impressed with being told that was still a big enough majority to avoid the need for at least a hand recount - and then they may not trust the $45\%$ and independent claim Commented Sep 19, 2022 at 1:38

The defective voting machines can switch votes both ways -- a vote for candidate 1 can be switched to a vote for candidate 2, and vice versa. So the number of intended votes for candidate 2 equals the random variable $$X+Y$$, where $$X$$ is the number of recorded votes for candidate 1 that in fact were switched from candidate 2, and $$Y$$ is the number of recorded votes for candidate 2 that were not switched to candidate 1. Then $$X$$ and $$Y$$ are independent binomial variables; note they have different success probabilities.
You want the probability $$P(X+Y>500,000)$$ since $$\{X+Y>500,000\}$$ is the event that candidate 2 received more intended votes than candidate 1. You should use the normal approximation to compute this probability.
• Right. To do a normal approximation for $X+Y$, note that $X$ is approximately normal, and $Y$ is approximately normal. Then use the fact that the sum of independent normal variables is also normally distributed. In your situation, what would be the mean and variance of the sum $X+Y$? Commented Sep 18, 2022 at 19:41