Assumptions necessary for the strong form of Olber's Paradox
For simplicity, let's pick a standard star, of radius $R_\odot$ (you can easily generalize this for stars of any radius by picking $R_\odot$ to be the lower bound of star radius). 

Let's split the universe into cubes. It would be much easier to use concentric shells, but later on it will be practical to have everything split into components of standard sizes. Each cube is of side $\ell$, and volume $\ell^3$. Within each cube, there is a probability distribution $P(N_\odot = n)$ to have $n$ stars within that cube. We're not requiring anything of that probability distribution, except

\begin{equation}
E[N_\odot] = \rho_\odot \ell^3
\end{equation}

There is an average stellar density $\rho_\odot &gt; 0$ that is the same everywhere. 

Now trace your line of sight from $0$ to infinity. This will go through an infinite number of such cubes. We can simplify the problem by having $\ell \approx R_\odot$, so that we don't have to worry about whether the line passes through a star. The fact that this line will encounter no star is equivalent to the experiment of each of those cubes to have zero stars. As each such probability is independent, this is

\begin{equation}
\prod_{n = 1}^\infty P(N_{n, \odot} = 0)
\end{equation}

We can set this as a Bernoulli trial. Zero stars is a failure with probability $p$, more than zero star is a success with probability $q$. The probability of zero star cannot be $1$, since we assumed a non-zero density. Therefore, the probability becomes

\begin{equation}
\lim_{n \to \infty} p^n = 0
\end{equation}

While it is not impossible, the probability of that line never hitting a star is of measure $0$, and the probability of hitting one is of measure $1$.