While following a recent election, I wanted to estimate population proportion of people who voted for a certain candidate knowing the sample proportion, sample size (and population size).
I first thought of using a simple confidence interval using the standard error of sample proportion given by:
$$ \text{SE}_p = \sqrt{\dfrac{p(1-p)}{n}} $$
The assumption behind this estimate is that the samples are randomly drawn from the population, which is clearly violated in an election. The consequence of this assumption is that the standard error gets small pretty quickly, even when say only $1000$ out of $10^6$ votes have been counted. On the other hand, the result should be quite uncertain at that point, so it's easy to understand intuitively that the assumption does not hold indeed.
My next attempt was Bayesian estimation of the resulting Bernoulli distribution by using the Beta distribution as a prior (knowing it is Bernoulli's conjugate prior). I stumbled at setting the parameters $\alpha$ and $\beta$ of this $\text{Beta}(\alpha, \beta)$ distribution, however.
An "uninformed" Beta distribution of $\text{Beta}(1, 1)$ suffers from a similar problem as my first approach, in that the data overwhelm this prior knowledge very quickly. I considered using a higher number for $\alpha$ and $\beta$, but couldn't figure out what would be an appropriate value. For example, setting $\alpha = \beta = N$ to the population size seemed to be too large. I thought that perhaps there ought to be some relationship between, $\alpha$, $\beta$, $n$, and $N$.
Question: how can I estimate the sample proportion uncertainty, either by a confidence interval or a highest density interval, when the sample proportion ($p$), sample size ($n$), and population size ($N$) are given, but random sampling cannot be assumed?
In other words: how confident can I be that a certain candidate will win the election if I know a count of a minority of the votes?
(For simplicity, you can assume only two candidates in the election and that without any count we can guess they have a 50-50 chance of winning.)
