Independent of the lines dropped, there are eight (2n) ways to pick adjacent points and sixteen (n^2) different combinations of points.
There are four (n) adjacent points iif the points lie on the same half of the circle (proof by interactive visualization).
So the answer is eight sixteenths (2n/2^n).
This can be the case if the other point is exactly 180 degrees from the anchor point that works though? But I think this case occurs with probability 0 ("almost never") so can basically be ignored.
Also the "obvious (wrong) answer" that reasons "by symmetry" is interesting since the n=3 case _can_ actually be solved in this manner. It's well known that probability the center of the circle is contained in the triangle of 3 randomly chosen points can be computed is 1/4. (This can be found by placing 1 point arbitrarily then computing an integral that ranges over the arc length to the second point). For the n=4 case this can be done via a double integral considering two arcs but it's slightly more annoying.
I think there was a 3b1b problem about this that presents the more elegant approach the other commenter mentions.
I don't think the same argument works in higher dimensions. On a circle, we can canonically pick a semicircle corresponding to each point (we have two choices, let's say we pick the clockwise one).
In higher dimensions there's no canonical choice of half-sphere. In odd dimensions one could pick a canonical half-sphere per point but it might turn out that some other non-chosen half-sphere for that point contains all the other points. In even dimensions there isn't even a way to canonically pick a half-sphere for each point (this is a consequence of the Hairy Ball Theorem).
(For all I know the actual numbers might turn out to be the same, I don't know. I'm just saying that the argument doesn't work.)
At that point you've lost the ability to say that only one such half-sphere defined by a dropped point can be a valid solution, and you've also lost the ability to say that if a valid solution exists then there must be a valid solution defined by one of the points you want to include in the half-sphere, but you can define a canonical half-sphere for any point.
I was uncomfortable with the idea of picking "random points on a circle" to begin with, because of https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) , but the article doesn't even address whether the concept is well-defined. We can always choose a point on the perimeter deterministically from any chord (...that isn't a diameter), so the ill-definedness of the problem of choosing a random chord seems like it would infect the problem of choosing a random point on the perimeter.
Bertrand paradox just doesn't apply here, there's a natural measure on the circle and all higher dimensional spheres. I wouldn't expect an article on this subject to need to make that clarification unless it's dealing with chords or some other situation without a natural measure.
1. Your answer can be "no" when the true answer is "yes". Consider this process with a circle of perimeter "21":
--------------------- (unwrap the circle)
----------+---------- (bisect the line)
-**-------+--------** (drop four points)
2. Your answer of "1/2, because it's divided into two equal lengths" is completely wrong for the scenario that you specify.
Consider the case where we drop a single point. We can do the same procedure:
A. Unwrap the circle;
B. Bisect the line;
C. Drop one point.
But even though the line is still divided into two equal lengths, our one point has a 100% chance of falling either on one side of the bisection point, or on the other side.
For the case where we drop four points, the article already gives the correct answer for your method, which is 1/2^3 (because there are 3+1 points).
Woah no need to be mean