# Statistical methods for determining probability of a 1 from a data set of 0's

I have been thinking recently if we assume that there is a finite population and one was taking some sort of survey. Something where it was a yes or no question which for these purposes, I am going to assign a 0 and 1. So, in this case, we have a binary system of possible responses. There are a large number of other cases where you could have this where a fractional average might mean more than what I've mentioned, but let's say we have two options: a response of 0 and a response of 1 and the population has to fall into either of these categories.

What I am wondering is: In this scenario, what can you get out of a sufficiently large data set that comes back as either all 0's or all 1's? Obviously, any sample that is not the entire population, if we assume it is 100% truthful responses, you don't know for certain the true value of probability of getting a 1 in the case that you have all 0's, but it also seems wrong to conclude that this value is 0 with an infinitely large confidence interval of 0.

So with that how does one determine what the probability that someone would respond with a 1 from a non-comprehensive set of data comprised of entirely 0's?

This is a famous statistical problem going back to Laplace, called the sunrise problem: Given that it has risen every day in the past, what is the probability that the sun will rise tomorrow?

Laplace modelled this problem using a Bayesian analysis, where he assumed that the observable values have distribution $X_1, X_2, X_3, ... \sim \text{IID Bern}(p)$, and with a uniform prior for the unknown probability $p$. With this approach, if $\sum_{i=1}^n x_i = s$ then the posterior density is:

$$\pi(p | x_1, ..., x_n) = \text{Beta} (p | s+1, n-s+1 ),$$

and the posterior mean is:

$$\mathbb{E}(p | x_1, ..., x_n) = \frac{s+1}{n+2}.$$

Hence, given $n$ observed times that the sun has risen (and none that it hasn't) the probability that the sun rises tomorrow is:

$$\mathbb{P}(X_{n+1} = 1 | x_1 = ... = x_n = 1) = \mathbb{E}(p | s = n) = \frac{n+1}{n+2}.$$

When $n$ is large, this probability is close to unity, as we would expect. It is worth noting that this method assumes that there is no other prior information about the sun, other than the fact it has been observed to rise $n$ times out of $n$ observations. Laplace correctly noted that in reality one has other prior information that bears on this question, so that his model is not necessarily a good representation of optimal probabilistic reasoning.

Application to your question: Using this same model, if you observed $x_1 = ... = x_n = 0$ then the probability that the next value is a one is $\mathbb{P}(X_{n+1} = 1 | s = 0) = 1 /(n+2)$.