Randomly pick three points on a unit circle what is the probability that the resulting triangle contains the center of the circle? This is the same question as asked here. The intuition as explained is the following: fix a point $A$ on the circle, and another point $B$ on the circle, suppose that the arc length of $AB$ is $x$. if $x$ is strictly less than $\frac{1}{2}$, then the remaining point $C$ has to stay within an arc length of $x$. If $x > \frac{1}{2}$, then the remaining point $C$ has to stay within an arc length of $1 - x$. Thus the final probability can be calculated as $\int_{0}^{\frac{1}{2}} x + \int_{\frac{1}{2}}^{1} 1 - x$.
However I am a struggling to make this argument formal. Basically, I can start by setting up three independent uniformly distributed random variable $X_1, X_2$ where $X_1$ is the arc length of $AB$ and $X_2$ is the position of the point $C$. Then I think the above argument correspond to making a statement about the probability of the desired event conditioning on arc length of $AB = x$.
Mathematically $P(\text{the triangle contains the center of the circle} | AB = x) = x$ if $x < \frac{1}{2}$. However, what does it mean to condition on an event of probability $0$?
