Pick two points on the perimeter of a circle of radius 1. Find the expected value of the length of the shortest arc. Pick two points uniformly randomly on the perimeter of a circle of radius 1. This divides the circle into two pieces. Find the expected value of the length of the shortest piece.
I have no idea to this problem, does anyone could help me? Thank!
 A: Treat the circle as an interval with ends identified. You've then got a function $f$ on 
$$
X = [0, 2\pi] \times [0, 2\pi]
$$
which assigns to a pair $(s, t)$ the length of the smaller arc between $s$ and $t$. 
Note that $X$ is a probability space in a natural way. The interval $I = [0, 2\pi]$ is a probability space, with a uniform probability measure with density $\frac{1}{2\pi}$. The space $X$ is just the product $I \times I$ with the product density, $\frac{1}{4 \pi^2}$. To find the expected value of a random variable $f$ on $X$, we integrate $f$ times this density, i.e., we compute
$$
E[f] = \int_{0}^{2\pi}\int_{0}^{2\pi} f(x, y) \frac{1}{4 \pi^2} ~dx~dy.
$$
Returning to the function $(s, t) \mapsto f(s, t)$ itself: This function is piecewise linear in $s$ and $t$. You might want to make a graph of it. Once you do that, you can integrate it over the square, and divide by the area of the square ($4 \pi^2$) to get its average value as above.
I'll get you started. Let's look at $f(s, t)$ for the case $s = 0$. 
$$
f(0, t) = \begin{cases} t & 0 \le t \le \pi\\
2\pi - t & \pi \le t \le 2\pi
\end{cases}
$$
What about when $s = \pi/4$? 
$$
f(\pi/4, t) = \begin{cases} t-\pi/4 & \pi/4 \le t \le 3\pi/4\\
2\pi - (t-\pi/4) & \text{otherwise}
\end{cases}
$$
You might then ask "What happens when I integrate these with respect to $t$, and then integrate the result with respect to $s$?" (I'm using Fubini's theorem here, which says that to integrate over an area, it's OK to do the integral one variable at a time, provided that the function you're integrating is nice enough; $f$ is.)
Added after comments:
I pointed out that you want to compute
$$
\frac{1}{4\pi^2} \iint_0^{2\pi}  f(s, t) dt ds
$$
To do so, I propose that you compute
$$
\frac{1}{4\pi^2} \int_0^{2\pi} \left(\int_0^{2\pi} f(s, t) dt\right) ds
$$
The inner integral
$$
u(s) = \int_0^{2\pi} f(s, t) dt
$$
depends on $s$. But looking at the case where $s = 0$, it's clear that the value of $f$ increases from $0$ to $\pi$ on one half-circle, and decreases from $\pi$ to $0$ on the other, for an average value of $\pi/2$ over the whole circle (i.e., as $t$ goes from $0$ to $2 \pi$.). The value of $u(0)$ is therefore $(\pi/2) 2 \pi = \pi^2$. 
What about $u(\pi/4)$? On the interval from $0$ to $3\pi/4$, $f$ increases from $0$ to $\pi$; on the complementary portion of the circle, it decreases from $\pi$ to $0$. Again the average value is $\pi/2$, and the integral over the whole interval is $\pi^2$. 
The same argument holds for every $s$, so $u(s)$ is in fact the constant $\pi^2$. Thus our double-integral of $f$ reduces to 
\begin{align}
\frac{1}{4\pi^2} \iint_0^{2\pi}  f(s, t) dt ds &= 
\frac{1}{4\pi^2} \int_0^{2\pi}  u(s) ds \\
&= \frac{1}{4\pi^2} \int_0^{2\pi}  \pi^2 ds\\
&= \frac{1}{4\pi^2} \int_0^{2\pi}  \pi^2 ds\\
&= \frac{1}{4\pi^2} \pi^2 \int_0^{2\pi}   ds\\
&= \frac{1}{4} \int_0^{2\pi}   ds\\
&= \frac{1}{4} 2\pi\\
&= \frac{\pi}{2}.
\end{align}
A: Place your circle at the origin $(0,0)$ in $\mathbb{R}^2$.
After you've picked two points, you can always rotate the circle so that one of the points is located at $(-1,0)$, so suppose that we have done so.  Thus, one point is $(-1,0)$ and the other point is chosen randomly from the circle.
The randomly chosen point precisely corresponds to a choice of a number $\theta \in (-\pi /2,\pi /2)$, where $\theta$ is the angle between the $x$-axis and the chord from $(-1,0)$ to the other point.
Denote by $L(\theta )$ the length of the shortest piece for given theta.  Note that $L(\theta )=L(-\theta )$, and so
$$
E:=\frac{1}{\pi}\int _{-\pi /2}^{\pi /2}\mathrm{d}\theta \, L(\theta )=\frac{2}{\pi}\int _0^{\pi /2}\mathrm{d}\theta \, L(\theta ),
$$
where $E$ is of course the expected value, the number we would like to compute.  Thus, it suffices to calculate $L(\theta )$ for $\theta$ positive.
Draw two radii from the center, one to the $(-1,0)$ point and one to the other random point.  This should form an isosceles triangle with two angles $\theta$ and one angle $\pi -2\theta$.
As the radius of the circle is $1$ and the angle that cuts out the arc we are concerned about is $\pi -2\theta$, the length of the arc is $1\cdot (\pi -2\theta )=\pi -2\theta$
Hence,
$$
E=\frac{2}{\pi}\int _0^{\pi /2}\mathrm{d}\theta \, [\pi -2\theta ]=\frac{2}{\pi}\frac{\pi ^2}{4}=\frac{\pi}{2}.
$$
This, of course, makes perfect intuitive sense.
