Discrete approximation to continuous normal Im reading Gelman's "Regression and other stories" book and I'm finding it difficult to wrap my head around one idea: in chapter 3 (p. 46) he approximates a discrete density using a normal density function. The question he's asking is what is the probability of tied election if we assume that the election outcome is forested to be N(0.49, 0.04). He than goes on and, assuming a population of voters of n=200000, calculated that P(tied election) = 1/n * f, where f is the PDF of this forecast. Where does the 1/n comes from in this formula? He uses the same rationale in his " Estimating the Probability of Events That Have Never Occurred: When Is Your Vote Decisive?" Gelamn et al. (1998) paper (J of the American Statistical Association):
Pr(your vote is decisive in your state)
= Pr(np = .5n) = Pr(p = .5), which is approx.  f(.5)/n
 A: Conclusion
The division by n takes into account the fact that there are other voters in the state. If you were the only voter in your state, the probability of your vote being decisive would be $f_v(.5)$, the density/probability of the state being at .5 democratic share. If there are 100 voters, certainly the probability of your vote being decisive is not the same as in a state with just one person, because any voter can be the “decisive vote” hence the division by n. The authors distract the reader by including misleading details like “n is the number of voters excluding yourself”.
tl;dr
It’s talking about the probability that a randomly selected persons vote will be decisive in a state.
Shoddy derivation of the formula (page 3)
From the paper there are 3 places where formula is discussed. They depend on $n_i=\text{number of voters in your state excluding yourself}$, and $v_i=\text{democratic share of the two party vote in the state excluding you}$. The first is $P(\text{your vote is decisive in your state})=P(nv =.5n)=P(v=.5)\approx f(.5)/n$. A more correct formula would be $P(\text{your vote is decisive in your state})=P(v=.5)/n=f_v(.5)/n$.
Using model parameters (page 5)
The second hint that a Normal approximation is used for binomial vote counts is that for each simulated parameter vector, using data from previous elections, this expression is modeled to be $\frac 1 n  N(.5, X\beta+\gamma+\delta, n^{-\theta}\sigma^2)$
Using some made-up values (page 6)
The third is a discussion of the actual values. Suppose n people in your state vote and the forecast has mean $\mu$ and std $\tau$. Then the probability a single vote will be decisive is $(\sqrt{2\pi}\tau n)^{-1}exp (-(\mu-.5)^2/2\tau^2)$. The probability of a tie is maximized at $\mu=.5$ (i.e. the model predicts a tie). A minimum bound on $\tau$ is about 2%, since it’s hard to forecast an election to more accuracy than that. This gives 20/n as an upper bound that a randomly selected persons vote will be decisive in a relatively close election. A typical value of n is 200,000 according to Congress, giving an upper bound of 1 in 10,000 that a randomly selected vote will be decisive. The authors also says another way to look at it is it’s not possible to forecast an election to closer than 10,000 votes.
etc.
Your vote has much higher chance of being decisive if you come from a small state, say n=2, than if n=10,000. The f is the probability that the state will be decisive, namely there are an equal number of democratic and republican ballots cast. And the probability that any particular vote is the decisive one is that divided by the number of votes.
A: This is from the approximating Binomial distribution to a normal when $n$ is large enough, as the skew of the distribution is not too great. In this case a reasonable approximation to $B(n, p)$ is given by the
$ \mathcal{N}(np,\,np(1-p))$
