Overtaking Probability On a uni-directional circular road where all vehicles travel in the same direction with varying speeds and overtaking is allowed on the entire road, what is the expected number of times 
(a) a particular vehicle will overtake others?
(b) a particular vehicle will be overtaken by others?
In particular, what I want to find out: Is (a) = (b)?
Thanks,
 A: Certainly not unless the speed of the vehicle is the mean speed of the traffic.  I think in that case you can make a symmetry argument-go to a coordinate system that rotates at the mean speed.
A: Let the speeds be $s_1 \le s_2 \le \cdots \le s_n$ (in laps per unit time).  Relative to vehicle $k$, vehicle $j$ moves at a rate of $s_j - s_k$.  Assuming the vehicles are presently uniformly and independently distributed around the road, the expected number of overtakings in a period of time $t$ therefore is $t|s_j - s_k|$.  Vehicles $1, 2, \ldots, k-1$ will be overtaken and vehicles $k+1, k+2, \ldots, n$ will overtake vehicle $k$.  Therefore:
(a) expected number of times vehicle $k$ overtakes others = $t\sum_{j=1}^{k-1}(s_k-s_j)$;
(b) expected number of times vehicle $k$ is overtaken = $t\sum_{j=k+1}^{n}(s_j-s_k)$.
(Empty sums equal zero, of course.)
When, as Ross Millikan suggests, vehicle $k$ moves at the mean speed of traffic, the sums (a) and (b) must be equal, because $s_k = \frac{1}{n}\sum_{j=1}^{n}s_j$ is algebraically equivalent to (b) - (a) equalling zero.
Clearly the implication works in the other direction, too.
