Probability the three points on a circle will be on the same semi-circle

The first two points must be on the same semicircle. They are separated by an angle between $0^{\circ}$ and $180^{\circ}$ with uniform probability.

If the first two points are the same, then the third point must lie on the same semicircle as the first two (probability $1$). If the first two points approach defining a diameter, then the probability that the third lies on the same semicircle approaches $1/2$. The probability decreases linearly from $1$ to $1/2$ as the separation of the points goes from $0^{\circ}$ to $180^{\circ}.$

So the probability averaged over all angles is $3/4$.

Given a semi circle, the probability that a uniform random point will lie on that semi circle is $\frac{1}{2}$. If three points lie on a semi circle then they definitely lie on a semicircle starting at exactly one of the points. There are 3 points, thus three semi circles and the events that the all the points lie on a semi circle starting at each of the points are mutually exclusive. Therefore the probability is $3\times\frac{1}{2^2}$. This is easily extended to $n$ points as $\frac{n}{2^{n-1}}$.

P.S: I read the solution on this blog by Saurabh Joshi

Without loss of generality, we can assume the circumference of the circle to be equal to $1$. Cut the circle at the first point $A$ and spread it out as a line. Let the other two points $B$ and $C$ be located at distances of $x$ and $y$ from $A$. The midpoint of this line is $M$ and $0 \leq x,y \leq 1$.

Points $B$ and $C$ can be both located on same side of $M$ or on either side of $M$. If both points lie on the same side of $M$, then all $3$ points lie on the same semi-circle.
$$ P(\text{same side}) = 2 * \frac{1}{4} = \frac{1}{2} $$ We are multiplying by 2 as the points can lie on both sides of $M$.

They can also be located on the either side of $M$ and in this case, the conditions for all 3 points to lie on the same semi-circle are $$ x + 1 - y < 0.5 \Rightarrow y > x + 0.5 \text{ if } y > x\\ y + 1 - x < 0.5 \Rightarrow x > y + 0.5 \text{ if } x > y $$ Representing these two regions graphically, we get the following area:

The area of the shaded region is $\frac{1}{4}$. Combining all results, we get the probability that $A$, $B$ and $C$ all lie on the same semi-circle is $$\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$$.