Tricky Geometric Probability

First, note that the distance from any point to the center of the circle doesn't matter, but what does matter is the angle between any two adjacent points, as this determines whether or not the triangle will contain the center of the circle. We can, however, from the argument of the previous sentence, assume that all three points are picked from the circumference of $\omega$ . We can pick our first point with probability $1$ , due to the rotational symmetry of the circle. Call the first point $A$ . After fixing this point $A$ , we choose another point $B$ . We can draw our circle and the first point out, and then experiment repeatedly with picking a second point, and then given the constraints of the first and second points, determine the degrees of freedom provided to the third point, $C$ . After some experimentation, we find that the location of $C$ is always constrained between the minor arc of the diametric opposites of $A$ and $B$ , which we'll call $A'$ and $B'$ , respectively. Note that the only case where this argument doesn't work is when $A$ and $B$ are diametrically opposite points themselves, but this is degenerate, so we can ignore it. Now, with some constructive probability, we can compute our answer. First, we can pick a point $A$ with probability $1$ . Now, one thing to keep in mind is that the circle is symmetric with respect to $A$ , which means that it doesn't matter where we pick $B$ with respect to $A$ . Therefore, suppose we pick $B$ at random to the right of $A$ , and the shortest distance (minor arc) between $A$ and $B$ has length $x$ degrees (although radians work just as well here). However, due to the linearity of the probability, it is easy to see that the probability is symmetric around the average of the extreme cases, namely $x=0^\circ$ and $x=180^\circ$ , so the average probability occurs when $x=90^\circ$ . In this case, we notice that the angular measure of the minor arc formed by $A$ and $B$ is the same as that formed by $A'$ and $B'$ , due to the fact that vertical angles are equal. Therefore, we can choose $B$ from $90^\circ$ average degrees of freedom, meaning that our probability is $\frac{90}{360}=\frac{1}{4}$ . Keep in mind that the probability is same even if we extend our sample space from the circumference to the whole circle, as all that matters are the relative angles between the points, which will stay constant, so our desired answer is $\boxed{\frac{1}{4}}$ .