How to find the probability that candidate $A$ wins

Question

enter image description here

In an election between two candidates, $A$ and $B$, one million individuals cast their votes. Among these, 2000 know candidate $A$ from her election campaign and vote unanimously for her. The remaining 998000 voters are undecided and make their decision independently of each other by flipping a fair coin. Approximate the probability $p_A$ that candidate $A$ wins up to 3 significant figures.

Answer 1

The answer is $0.977$!

The amount of votes candidate $A$ needs to win is $500001$. Since candidate $A$ already has $2000$ guaranteed votes, candidate $A$ needs $498001$ votes to win. To find this, we simply do $$1-\text{binomcdf}(998000,0.5,498001) \approx 0.977$$

Answer 2

Outline: A normal approximation would use $\mu = np$ and $\sigma = \ \sqrt{np(1-p)},$ where $n = 998000$ and $p = 1/2.$ If $X \sim \mathsf{NORM}(\mu, \sigma),$ then you seek $P(X \ge 49801),$ which you can evaluate by standardizing and using printed normal tables:

$$P(X > 498001) =P\left(\frac{X - \mu}{\sigma} > \frac{498001 - \mu}{\sigma}\right) = P(Z > z) \approx 0.977,$$ where the numerical value of $z$ is found by plugging in the values of $\mu$ and $\sigma$ from above. [I have omitted the continuity correction because of the size of the numbers.]

If you use R statistical software, you can get the answer $0.977$ (correct to three places) either using an exact binomial computation or an approximate normal one.

 1 - pbinom(498001, 998000, .5)
 ## 0.9771959
 1 - pnorm(498001, 499000, 499.4)
 ## 0.9772715

I'm not sure which statistical calculators ccould handle such large numbers without error messages or errors.

Note: Often normal approximations are only accurate to about two places--even when $n$ and $p$ obey various 'rules of thumb' for 'permitted' use of normal to approximate binomial probabilities.

This problem has two advantages toward a better approximation: (a) $n$ is very large and (b) $p = 1/2,$ making the binomial precisely symmetrical. Therefore, you get about four places of accuracy, where you are only asked for three.