Generalisation of an Interesting Problem - Probability This isn't a duplicate, I did find another link with the same question but wasn't clear about the solution there except for n = 3. 
This is an interesting problem, that I'd first solved with only 3 points - and I understood it then (My solution was different, a bit less rigorous). I'm not able to understand the generalisation for $n$ points. Please explain this solution, or any other possible - 
Problem: 
Given n points drawn randomly on the circumference of a circle, what is the probability they will all be within the same common semicircle?
Solution:
One might think that there are infinitely many semi-circle on the circumference of a circle. However, the beauty is that we need to consider only n semi-circle here.
If a semi-circle covering all n points, indeed exists, then, a semi-circle covering all n points and starting from one of the points in a clock-wise direction also exists.
So, given a semi-circle which starts at one of the point in clock-wise direction. The probability that the rest of the n-1 points will be in that semi-circle is $\dfrac{2}{2^n}$. So for n such semi-circle, the probability will be $\dfrac{2n}{2^n}$.
Please explain the above solution in detail.
Source: 
https://sbjoshi.wordpress.com/2009/10/04/puzzle-semi-circle-covering-n-points/
 A: When there is a new point in a circle, there is a 1/2 chance of that point to be in the semicircle relative to your first point, this applies for every point you add since there are infinitely many points in a circle. so if there are n points, there would be a $1/2^{n-1}$ chance, which is a $2/2^n$ chance. There are n starting points, so this function is repeated n times, which gives the total probability to be $2n/2^n$.
A: 1) There are $n$ clockwise semicircles that could be formed starting at each of the $n$ points.  For any other random point, the probability of it being in that particular semi-circle is $\frac{1}{2}$.  There are $n-1$ other points, so the probability of all of them being in that particular semicircle is $(\frac{1}{2})^{n-1}=\frac{2}{2^{n}}$.  Since there are $n$ such possible semicircles, the total probability is $n\cdot\frac{2}{2^n}=\frac{2n}{2^n}$.
2) If all the points are on a semicircle, then there is also an empty semicircle which contains no points.  The points can then be (well) ordered by starting at a position on this "empty" semicircle and then proceeding clockwise until you get to a first point, a second point, etc.  The first point you came to is therefore a point which, if you make a clockwise semicircle starting at that point will contain all the other points.
