Triangle Containing the Centre

Key idea. The centre lies inside the triangle exactly when the three points do not all lie in a common semicircle.

Why this criterion holds. Suppose all three points lie in a closed semicircle. The diameter bounding it has the whole triangle on one side, so the centre is on the boundary of the triangle or outside it. Conversely, suppose the centre is not in the closed triangle. Then some line separates the centre from the triangle. The parallel line through the centre is a diameter, and all three points lie in one of the semicircles it cuts off. Configurations where the centre lies on an edge need two antipodal points, so they have probability $0$.

The calculation. Fix the first point $A$; by rotational symmetry its position does not matter. Let $\theta\in[0,\pi]$ be the smaller central angle between $A$ and the second point $B$. Since $B$ is uniform on the circle, $\theta\sim\mathrm{Unif}(0,\pi)$.

Let $A'$ and $B'$ be the points antipodal to $A$ and $B$. The three points avoid every common semicircle exactly when the third point $C$ lands on the arc from $A'$ to $B'$ opposite the short arc $AB$. That arc has length $\theta$, the same angle as arc $AB$. If $C$ lands anywhere else, $A$, $B$ and $C$ fit in a semicircle bounded by the diameter through $A$ or through $B$.

Hence $\mathbb{P}(\text{centre inside}\mid\theta)=\frac{\theta}{2\pi}$, and averaging over $\theta$ gives

Alternative. Order the points around the circle. Let $E_i$ be the event that the other two points lie within the half-turn clockwise from point $i$. Then $\mathbb{P}(E_i)=\left(\tfrac12\right)^2$. The events $E_i$ are disjoint up to probability zero, and their union is the event “all three lie in a semicircle”. So that event has probability $3\cdot\frac14=\frac34$, and the answer is $1-\frac34=\frac14$.

Generalisation. The same argument shows that $n$ uniform points lie in a common semicircle with probability $n/2^{n-1}$.