Puzzle : Birds on a circular wire The problem is taken from my course on randomized algorithms : 
There is a circle made of wire. n birds (assume n>2) occupy uniformly random position over it (visualize each bird occupying a point on the circumference of the circle). This will lead to partitioning of circle into n segments. We follow the following rule for painting these segments. A segment is painted if it is smaller than at least one of its neighboring segments. What is the expected fraction of the circle which gets painted? 
I am not able to frame it mathematically. Any hints?

Let an $(n+1)$-th bird perch randomly uniformly on the wire. Now the circle is divided into $n+1$ segments, and we're looking for the probability that the two segments next to the additional bird together are smaller than at least one of the segments adjacent to them. Let $a\le b\le c$ be an ordered triple uniformly randomly drawn from $[0,1]^3$; then with $x:=c-a$ the measure of the set in which $x\lt a$ or $x\lt1-c$ is
$$
\int_0^{1/3}x(1-x)\mathrm dx+2\int_{1/3}^{1/2}x(1-2x)\mathrm dx=\left[\frac12x^2-\frac13x^3\right]_0^{1/3}+2\left[\frac12x^2-\frac23x^3\right]_{1/3}^{1/2}=\frac7{108}\;,
$$
where the first factor $x$ measures the possibilities for $b$ and the second factor measures the possibilities for $a$ that satisfy the inequalities. The measure of all ordered triples is $1/3!$, so the desired fraction is $7/18$.
P.S.: I tested the result numerically; here's the code.
