find a chance that all N points lie on the half circle. We are given a circle with N randomly allocated points on it. Task is to find a chance that all N points lie on the one half of circle.
I have drafted some solution:
1. Since there are no way to put two points on circle, so that they were not on the same half-circle, $P_1$ and $P_2$ picked randomly and didn't affect the chance. So, required probability is:
$P(P_3) \cdot P(P_4)\: \cdot ... \cdot P(P_n)$, where $P(P_i)$ is the chance that i-th point lays on the proper half of circle.
2. Let's visualize what's $P(P_3)$, $P(P_4)$ look like:
Grey sector highlights forbidden part of circle.
It's obvious from pictures, that $P(P_i)$ approaching 0.5 as point amount increases


For this specific example, we could write:
$P(P_3) = 1 - \frac {\Delta(\theta_2,\theta_1)} {2\pi}$ $P(P_4) = 1 - \frac {\Delta(\theta_3,\theta_1)} {2\pi}$.
3. Then if we generalize, $P(P_i) = 1 - \frac {\Delta_i} {2\pi}$, where $\Delta_i$ is a difference of angles of the most distant points.
And I'm understand that here I should introduce some generalized formula but I don't see it and don't want to make guesses. So I would appreciate any help.

This is not correct if which half circle is not specified at the start.  Say the next point is at $\frac {2\pi}3$ radians.  Your calculation says it has failed, but there is a half circle that contains both points-the one from $0$ to $\pi$, for example.  The second point will always be within some half circle of the first.  The hard part comes with the third point.  Continuing my example, it could be anywhere from $-\frac \pi 3$ to $\pi$ radians, giving a chance of $\frac 23$ that the three points are in a half circle.  Leaving the example, the third point is disallowed a range that matches the distance between the first two points.  That is general: each point is disallowed an amount of the circle that is the range of the existing batch of points.  This doesn't solve the problem.
