TITLE: What is the probability that the center of the circle is contained within a triangle formed by choosing three random points on the circumference?
QUESTION [31 upvotes]: Consider the triangle formed by randomly distributing three points on a circle. What is the probability of the center of the circle be contained within the triangle?

REPLY [2 votes]: Here's a rigorous solution that doesn't require calculus or reasoning about distributions.
Let's pick the first point $A_0$ at random, then pick two diameters $B$ and $C$ (instead of two points) at random, and then pick which endpoint of each diameter at random by flipping a coin for each.  That gives the same distribution as picking three points at random.
Going clockwise from $A_0$, label the endpoints of the two diameters by $B_0$, $C_0$, $B_1$, and $C_1$ in order.  Now, there are four ways the coin flips could land to pick the endpoints of B and C, yielding the four possible triangles $(A_0, B_0, C_0), (A_0, B_0, C_1), (A_0, B_1, C_0)$, or $(A_0, B_1, C_1)$. Each of these outcomes is equally likely (probability $1/4$) after picking $A_0$.  
Of these, the only one whose triangle contains the center is $(A_0, B_1, C_0)$ — and I found this interesting — regardless of the relative angles of the diameters.  Triangle $(A_0, B_0, C_0)$ cannot contain the center, because all three points lie on the same semicircle starting at $A_0$ and going clockwise, which means that $B_0$ is an obtuse angle.  Likewise, $(A_0, B_1, C_1)$ is out, because all three points lie on the same semicircle ending at $A_0$, which means that $B_1$ is an obtuse angle.   Finally, $(A_0, B_0, C_1)$ all land in the semicircle centered at $A_0$, meaning that $A_0$ is an obtuse angle.  But $(A_0, B_1, C_0)$ contains the center, because the minor arc from $C_0$ to $B_1$ is less than $180^{\circ}$, making the angle at $A_0$ less than $90^{\circ}$, i.e., acute.  
Thus, the answer is $1/4$.