Car Convoy Problem

Question

Cars travel down a one-way single-track road. The nth driver would like to drive at speed $V_n$ , where $ V_1,V_2, ..., V_n$ are iid random variables. Cars will get bunched into convoys. If $V_2>V_1>V_3$ , then the first convoy will consists of cars 1 and 2 , and will be of length 2. Let L be the length of the first convoy. Find the probability $P(L=n)$ and the expectation $E(L)$.

Answer 1

Hint:

• For the first convoy to be of length $L=k$, you need cars $2,3,\ldots,k$ to all be faster than car $1$, and car $k+1$ to be slower than car $1$
• As special cases, for the first convoy to be of length $L=1$ you need car $2$ to be slower than car $1$, and for the first convoy to be of length $L=n$ you need all the other cars $2,3,\ldots,n$ to be faster than car $1$

Answer 2

I will assume that the distribution of speeds is continuous, so that the probability of two identical speeds is zero.

What matters is that the distribution of speeds is iid, therefore if you assign a speed rank to each driver (1 being the fastest, $n$ the slowest) all permutations of ranks have the same probability: $1/n!$.

(For example if $n=3$, the permutation [1,3,2] means that $V_1>V_3>V_2$).

Now, $L=n$ iff no car is slower than car #1, i.e. car #1 is the slowest. There are $(n-1)!$ permutations where this happens, therefore $$ P(L=n)={(n-1)!\over n!}={1\over n}. $$