# Probability of candidate winning majority vote

I was trying to work out probability of a candidate winning the majority vote but I got a crazy z value, which i'm not sure is correct or not but it sure as hell seems crazy. Bear in the mind, that I've just started university and so my knowledge of statistics is limited. I will be working this out in a very simple way:

Lets go back a few months and assume that Donald Trump isn't elected yet. From the polls, Hillary has 51.5% chance of winning and Trump has 48.5%.

I'm going to assume that each person in America has a 48.5% chance of voting Donald Trump. There are 132,899,453 potential voters so:

$$n = 132,899,453\\ p = 0.485$$

Let X be the random variable modelling the number of people that will vote for Donald Trump:

$$X \sim B(132899453, 0.485)$$

Since n is extremely large, I will use a normal approximation which should fit well since p is close to 0.5 and n is very large.

$$E(X) = np = 132899453 * 0.485 = 64456235$$ $$Var(X) = np(1-p) = 132899453 * 0.485 * (1-0.485) = 331194961$$ $$Y \sim N(64456235, 331194961)$$

I want to work out the probability that Trump gets more than 50% of the votes: \begin{align*} &= P(Y \geq 132899453*0.5 = 66449727)\\ &= 1 - P(Y \leq 66449727)\\ &= 1 - P(Z \leq \frac{66449727-64456235}{\sqrt{331194961}}) \\ &= 1 - P(Z \leq 346) \end{align*}

Why is the Z value (346) I got so high? This would mean he has a near 0% chance of winning majority vote. This seems counter-intuitive, because each person in America is assumed to have a 48.5% chance of voting for Trump but somehow the z value suggests he has a near 0% chance of winning majority vote, I would've expected him to have around 40% chance of winning majority vote.

• This is a consequence of the Law of Large Numbers: since your sample size is so large, the distribution of possible outcomes is very tightly clustered around the most likely outcome (51.5% of people voting for Clinton). – liangjy Jan 19 '17 at 12:09
• As an addition to liangjy's comment: you assume a very high certainty by extrapolating to the entirety of the US voters. Basically you are saying that (whatever the source of this poll) it is not only representative of all voters, but that is somehow captures all uncertainty about people's voting behaviour. This would only be possible if the poll was the actual election and 100% of voters showed. You should therefore calculate your probability distribution based on the $n$ which compromised your source poll. On a sidenote, weren't there more than two parties you could vote for? ;) – IWS Jan 19 '17 at 12:34

Let $p(H)$ be the probability that Hillary wins, and $p(h)$ be the probability that any given person votes Hillary. The number of votes $k$ that Hillary receives then follows a binomial distribution: $$p(k) = \binom{n}{k}p(h)^k(1-p(h))^{n-k}$$ Where $n$ is the total number of votes cast. The probability that Hillary wins, i.e. $p(H)$, is equal to the probability that she gets at least $n/2+1$ votes: $$p(H) = \sum_{k=\frac{n}{2}+1}^{n}p(k)$$ What a poll tells you, or at least gives you an estimate of, is $p(H)$. As you can see from these equations, $p(H)\neq p(h)$, unless $n=1$. Working back from $p(H)$ to $p(h)$ when $n$ is large seems very hard, if not practically impossible.
Edit: I might add that the probability you were trying to work out is the one you actually started with. If the polls give Hillary a 51.5% chance of winning, that's your best estimate of her chances of getting the majority vote. In other words, you've already got (an estimate of) $p(H)$ and there's nothing more to be done with that information.