Expected length of arc in a randomly divided circle Choose, at random, three points on the unit circle. Interpret them as cuts that
divide the circle into three arcs. Compute the expected length of the arc that contains the point (1, 0).
Generalise for N points.
 A: The length of the arc from $(1,0)$ to the next point is the minimum $M$ of three (or $n$) independent uniform random numbers $X_1,\ldots, X_n$ from $[0,2\pi)$.
Since $P(M>x)=\prod P(X_i>x)=\left(1-\frac x{2\pi}\right)^n$, we find $$E(M) = \int_0^{2\pi}P(M>x)dx =\left. -\frac{2\pi}{n+1}\left(1-\frac x{2\pi}\right)^{n+1}\right|_0^{2\pi}=\frac{2\pi}{n+1}.$$
The same holds for the arc length to the the other arc end, hence the expected length of the full arc containing $(0,1)$ is 
$$\frac{4\pi}{n+1}.$$
A: You can cut the circle at (1,0) point and make it a line. The both side of the line corresponds to point (1,0) in the unit circle. 
We also know that the answer for the circle is $2\pi$ times the answer of the uniform[0,1] case. 
For uniform[0,1] case with the cutting method mentioned above, The expectation is
$E(min(X_1,...,X_n)+1-max(X_1,...,X_n))$
We know that in this case $E(min(X_1,...,X_n))=\dfrac{1}{n+1}$ and $E(max(X_1,...,X_n)) = \dfrac{n}{n+1}$, so we have the final answer as
$2\pi (1+\dfrac{1}{n+1}-\dfrac{n}{n+1})=\dfrac{4\pi}{n+1}$
