Probability that $n$ points lie within a semicircle - different approach The main question is that what is the probability that $n$ points on a circle lie on the same semi circle. I was trying to solve the question by thinking about the following equivalent question:
What is the probability of selecting $n$ $\textit{Uniform(0,1)}$ numbers such that their range is less than $0.5$.
I have attached an image with the derivation of the range of $N$ uniform random variables. 
If you plugin $d=0.5$ in the cdf for the range of uniform distribution, we should get the answer but the answers don't match. What is wrong here? The answer should be $\frac{n}{2^{n-1}}$ but using the cdf the answer is $\frac{n+1}{2^n}$.
Image showing the derivation of range of n uniform random variables
 A: The main point is that a circle is round and 'connected' at it's 'ends', while an interval is not. Your reformulation does not catch the cases where some of the random points are say between $0$ and $0.25$ while the remaining points are between $0.75$ and $1$. 
If you imagine that you take the interval and make a circle by connecting the ends at $0$ and $1$, then the random points will now all be in a semi-circle centered around the connection point. But your range calculation gives a values larger than $0.5$ in this case (neglecting the case where all points are either $0.25$ or $0.75$).
By 'cutting' the circle open at a point and transforming it into an interval, you've determined the probability that the random points lie in a semi-circle that does not pass throuh that cutting point.
As should be clear now, your interval calculation gives an answer that is too small because by using the transformation it misses some point configurations that would count as "all points in one semi-circle".
