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There are two political candidates, A and B. Each voter has a 51% chance of voting for A, and a 49% chance of voting for B (say they flip a slightly biased coin to choose). Out of a population of 100 voters, what's the chance that A wins if a majority of votes is necessary for a win?

My intuition immediately tells me that A has a 51% chance of winning because of the probabilities mentioned above.

But the true answer from the binomial distribution formula is (100 choose 51) * (0.51)^51 * (0.49)^49, which is around 8%.

What gives?

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    $\begingroup$ You need to compute the cumulative probability for $X \geq 51$ $\endgroup$ Commented Dec 31, 2022 at 7:22
  • $\begingroup$ "My intuition immediately tells me that A has a 51% chance of winning because of the probabilities mentioned above" This intuition is wrong! You'd expect that 51% of the population, approximately, will vote for candidate A. If the population is large, then the fraction that votes for candidate A is unlikely to deviate much from 51%. So, if the population is large, we're very confident that candidate A will win. $\endgroup$
    – littleO
    Commented Dec 31, 2022 at 7:28
  • $\begingroup$ @trueblueanil what would that number come out to? $\endgroup$
    – asdf1234
    Commented Dec 31, 2022 at 7:32
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    $\begingroup$ @littleO: Around 54% $\endgroup$ Commented Dec 31, 2022 at 7:37
  • $\begingroup$ @littleO Does me saying A has a 51% chance of winning mean the same thing as 51% of the population is expected to vote for A? Also, could you explain: "if the population is large, then the fraction that votes for candidate A is unlikely to deviate much from 51%" $\endgroup$
    – asdf1234
    Commented Dec 31, 2022 at 7:42

2 Answers 2

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As said by true blue anil you have to calculate $P(X\geq 51)=1-P(X\leq 50)$. Since the population is $n=100>30$ you can apply the central limit theorem. In this context $\Phi(z)$ denotes the cdf of the standard normal distribution. The mean and the standard deviation of the binomial distribution are $n\cdot p=100\cdot 0.51$ and $\sqrt{n\cdot p\cdot (1-p)}=\sqrt{100\cdot 0.51\cdot 0.49}$, respectively Then we have

$$P(X\geq 51)\approx 1-\Phi\left(\frac{50\color{blue}{+0.5}-100\cdot 0.51}{\sqrt{100\cdot 0.51\cdot 0.49}}\right)=1-\Phi(-0.1)$$

$\color{blue}{+0.5}$ is the continuity correction factor. The value of $\Phi(-0.1)$ can be looked up at wiki, for instance. Thus $P(X\geq 51)\approx 53.983\%\approx 54\%$

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Both intuitions are wrong.

My intuition immediately tells me that A has a 51% chance of winning because of the probabilities mentioned above.

No, that's the chance of winning if only one person votes. When many persons vote, the average tends to concentrate around the mean (law of large numbers) and hence we should expect that the chances of A increase.

But the true answer from the binomial distribution formula is (100 choose 51) * (0.51)^51 * (0.49)^49, which is around 8%.

No, that's the probability that A receives exactly 51 votes. But A wins not only when he receives 51 votes, also when he receives 52, or 53, or...

Hence, you need to sum over all those probabilities.

The result is $\approx 0.54$ (google sheet).

As mentioned by @callculus42, you can also resort to the CLT approximation.

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