What you have is reasonable. The order of the points does not matter. So, what you are saying is: pick A, B and C in that order.

The probability that A and B, do not share a semicircle is zero (obviously).

So, we have to wait for C. For A, B and C to not share a semicircle, C has to be such that neither B nor C lie on the same side of the diameter AA' and neither A nor C lie on the same side of the diameter BB'.

This makes sense, however, we can't go anywhere from here.

I would do the following:

Let x be the smaller of the distance between A and B. Then 0 \leq x \leq \pi r .

Then the probability density function of x , f(x) = \frac{1}{\pi r} .

Then, the probability that C will not be in the same semicircle as A and B, given the distance between A and B is x , g(\text{not semi-C}|x)=\frac{x}{2\pi r}.

Therefore, the total probability density function that A, B and C will not share a semicircle is given by \frac{x}{2 \pi^2 r^2} . I.e. C has to fall in the unbolded portion of the circle whose length is also x .

Therefore, total probability that A, B and C do not lie on the same semicircle is given by \int_0^{\pi r} \frac{x}{2 \pi^2 r^2} dx = \frac{1}{4} .

I am sure there is a much more elegant argument given that the answer has turned out to be this simple.