Particles chasing one another around a circle Two particles start out at random positions on a unit-circumference circle.
Each has a random speed (distance per unit time) moving counterclockwise uniformly distributed
within $[0,1]$.  How long until they occupy the same position?  In the example below,
the red particle catches the green particle at $t=5.9$, i.e., nearly six times around the circle:

          


The distribution of overtake-times is quite skewed, indicating perhaps
the mean could be $\infty$.  For example, in one simulation run, it
took more than $3$ million times around the circle before one particle finally caught
the other. So I don't trust the means I am seeing (about $25$).

What is the distribution of overtake-times?

I was initially studying $n$ particles on a circle, but $n=2$ seems already somewhat interesting...
Update (2Dec12).
Alexandre Eremenko concisely established that the expected overtake-time (the mean)
is indeed $\infty$.
But I wonder what is the median, or the mode?  Simulations suggest the median is about $1.58$ and
the mode of rounded overtake-times is $1$, reflecting a distribution highly skewed toward rapid overtake. (The median is suspiciously close to $\pi/2$ ...)
Update (3Dec12). Fully answered now with Vaughn Climenhaga's derivation of the
distribution, which shows that the median is
$1 + \frac 1{\sqrt{3}} \approx 1.577$.
 A: Let the circle have length $1$ unit.
Let $\theta$ be the angle (anticlockwise) from the first particle to the second
at the initial position. Let $v_1,v_2$ be the speeds of the particles. I suppose they
move anti-clockwise, as in your movie.
If $v_1>v_2$, they collide in time $T(v_1,v_2,\theta)=\theta/(v_1-v_2).$
If $v_2>v_1,$ they collide in time $T(v_1,v_2,\theta)=(1-\theta)/(v_2-v_1)$.
The expectation of the time is
$$\int_Q T(v_1,v_2,\theta)dv_1dv_2d\theta,$$
where $Q=[0,1]^3$.
The integral is easy to evaluate by breaking $Q$ into two pieces.
But it is indeed $+\infty$, as you guessed:-)
A: To answer your questions about median and mode, one can take Alexandre's answer a little further and compute the exact distribution function for the overtake-times.
Note that the overtake-time doesn't depend on $v_1,v_2$ directly, but only on their difference.  Call the difference $v$.  Now $v$ is the difference of two uniformly distributed random variables on $[0,1]$, so it is supported on $[-1,1]$ with probability density function $1-|v|$.  Moreover, since $\theta$ is uniformly distributed we can without loss of generality identify the cases $(v,\theta)$ and $(-v,1-\theta)$ and reduce everything to the following set-up:


*

*$v$ is distributed on $[0,1]$ with density function $2(1-v)$.

*$\theta$ is uniformly distributed on $[0,1]$.

*The overtake-time is $t=\theta/v$.


Now we can compute the cumulative density function for the overtake-time.  Indeed, we have $P(t<T) = P(\theta/v<T) = P(\theta < Tv)$, which we can get by the following integral:
$$
P(t<T) = \int_0^1 2(1-v) P(\theta < Tv | v) \,dv.
$$
The probability $P(\theta < Tv | v)$ is given by the function $f(\theta,v) = \max(Tv,1)$.  Thus for $T\leq 1$, we have $f(\theta,v)=Tv$ for all $v\in[0,1]$, so integrating gives $P(t<T) = T/3$, while for $T\geq 1$, we integrate and find
$$
P(t<T) = \int_0^{1/T} 2(1-v)Tv\,dv + \int_{1/T}^1 2(1-v)\,dv = 1-\frac 1T + \frac 1{3T^2}.
$$
So in the end the cumulative density function for the overtake-time is
$$
P(t<T) = \begin{cases} 
\frac T3 & T\leq 1, \\
1 - \frac 1T + \frac 1{3T^2} & T \geq 1.
\end{cases}
$$
The term $1/T$ in the last expression will give you the infinite mean, since upon differentiating the CDF you'll get a term $1/T^2$, which upon multiplying by $T$ and integrating to get the mean you end up integrating $1/T$ from $1$ to $\infty$.
As for the median, it looks as though any proximity to $\pi/2$ is just a red herring, because solving for $P(t<T) = 1/2$ yields $T=1 + \frac 1{\sqrt{3}} \approx 1.57735\dots$.
