TITLE: Probability problem: cars on the road
QUESTION [20 upvotes]: I heard this problem, so I might be missing pieces. Imagine there are two cities separated by a very long road. The road has only one lane, so cars cannot overtake each other. $N$ cars are released from one of the cities, the cars travel at constant speeds $V$ chosen at random and independently from a probability distribution $P(V)$. What is the expected number of groups of cars arriving simultaneously at the other city? 
P.S.: Supposedly, this was a Princeton physics qualifier problem, if that makes a difference.

REPLY [2 votes]: This is not drastically different from the answers above but a slightly different way to arrive at the recurrence relation.
Let $E(n)$ be the expected number of clusters in the case where we begin with $n$ cars in the beginning. Now, if we add one more car from the behind (i.e behind the last car of the slowest cluster), two cases arise :
1) It catches up with the slowest cluster and joins it; or
2) It is too slow to catch up and forms a singleton cluster of its own
The probability of event 2 is $\dfrac 1 {n+1}$. We have $n+1$ in the denominator because it could lie in any interval between and outside the n initial speeds (imagine them on a number line and you'll see $n+1$ possible interval) of the $n$ cars we had in the start.
Therefore, the probability of event 1 is $\dfrac n {n+1}$.
Thus $E(n+1) = \big( E(n)+1 \big) \dfrac 1 {n+1} + E(n) \dfrac n {n+1}$, so $E(n+1) = E(n) + \dfrac 1 {n+1}$ i.e. $E(n)$ is the $n$th harmonic number.