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.

A suitable discretization of the problem is the following: we count in how many ways we can choose $n-1$ integers $i_1,\ldots,i_{n-1}$ such that $0=i_0\leq i_1\leq i_2\leq\ldots\leq i_{n-1}\leq i_n=2mn$ and for every $k\in[1,n]$ we have $i_{k}-i_{k-1}< mn$. This number is equal to the coefficient of the monomial $x^{2mn}$ in the product 
$$(1+x+\ldots+x^{mn-1})^n = \left(\frac{1-x^{mn}}{1-x}\right)^n=$$
$$=\left(1-nx^{mn}+\binom{n}{2}x^{2mn}-\ldots\right)\cdot \sum_{j=0}^{+\infty}\binom{n+j-1}{j}x^j,$$
hence it is equal to:
$$\binom{n}{2}-n\binom{mn+n-1}{n-1}+\binom{2mn+n-1}{n-1},\tag{1}$$
while the number of choices that satisfy only $0=i_0\leq i_1\leq i_2\leq\ldots\leq i_{n-1}\leq i_n=2mn$ is equal to the coefficient of $x^{2mn}$ in the product $\frac{1}{(1-x)^n}$, i.e. the last term of $(1)$. So the probability that a sequence $0=i_0\leq i_1\leq i_2\leq\ldots\leq i_{n-1}\leq i_n=2mn$ has at least one index $k\in[1,n]$ such that $i_k-i_{k+1}\geq mn$ is equal to:
$$\frac{n\binom{mn+n-1}{n-1}-\binom{n}{2}}{\binom{2mn+n-1}{n-1}}.\tag{2}$$
Taking the limit of $(2)$ with $m$ that goes to infinity gives that the probability $H_n$ for $n$ randomly chosen points on a circle to lie in the same half-circle. We have:
$$ H_2 = 1, \quad H_3=\frac{3}{4},\quad H_4=\frac{1}{2},\quad H_5=\frac{5}{16},\quad H_n=\frac{2n}{2^n}.$$
