I think about this by unwrapping the circle to form a straight line. Then you draw an imaginary point in the middle of the line. Then what are the chances they will all fall on one side of the line or the other? 1/2 because it's divided into two equal lengths.

This approach implies the probability doesn’t depend on N. It only happens to be 1/2 for N=4 (the article goes into this). The trick is that you don’t know beforehand which semicircle all the points can land in, but your unwrapping step assumes you do.

There are two major problems with your idea:

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).