Three points, one half

Idea. Split the event according to which point starts the semicircle, going clockwise. These cases are disjoint and each has an easy probability.

Step 1: the events. For $i=1,2,3$, let $A_i$ be the event that the other two points lie in the half-circle that starts at $P_i$ and runs clockwise for half a turn. The three points lie in some semicircle if and only if at least one $A_i$ holds: if the points fit in some semicircle, turn that semicircle clockwise until its starting end reaches one of the points; it still contains all three.

Step 2: one event. Given $P_i$, each other point falls in its clockwise half-circle with probability $\frac{1}{2}$, independently. So $\mathbb{P}(A_i)=\frac{1}{4}$.

Step 3: disjointness. If $A_i$ and $A_j$ both held, $P_j$ would be within half a turn clockwise of $P_i$ and $P_i$ within half a turn clockwise of $P_j$. That forces them to be exactly opposite, an event of probability $0$. So the events are disjoint (up to probability zero) and $$\mathbb{P}(\text{some semicircle})=3\cdot \frac{1}{4}=\frac{3}{4}.$$

In general. With $n$ points the same argument gives $\frac{n}{2^{n-1}}$. For $n=4$ that is $\frac{1}{2}$.