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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 $E(L)$.

Kouloubou Dimitra
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  • I guess the intention was that the variables be iid and not just independent. – David K Jul 04 '19 at 10:34
  • Yes of course!!! Thank you very much!!! – Kouloubou Dimitra Jul 04 '19 at 11:00
  • other questions with a similar set-up: https://math.stackexchange.com/questions/1033468/expected-number-of-cluster-of-cars https://math.stackexchange.com/questions/201807/probability-problem-cars-on-the-road https://math.stackexchange.com/questions/2151206/combinatoric-task-about-flock-of-sheep – Henry Jul 04 '19 at 12:48

2 Answers2

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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 $1$
Henry
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  • Nope. For the first convoy to be of length L=k, you need car 2 is slower than car 1, car 3 be slower than car 2,... , car k-1 be slower than car k-2 and car k be faster at least one of cars 1,2,...,k-1. Convoy can't be of length L=1. Convoy to be of length L=2 you need car 2 be faster than car 1 – Kouloubou Dimitra Jul 04 '19 at 12:03
  • @KouloubouDimitra If car 2 is slower than car 1 then the first convoy will definitely be of length 1, since car 2 will never reach car 1 and no other car will pass car 2 – Henry Jul 04 '19 at 12:43
  • Thank you Very much!! Now I understand the problem... – Kouloubou Dimitra Jul 04 '19 at 21:37
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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 $$ P(L=n)={(n-1)!\over n!}={1\over n}. $$

A.G.
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  • This calculates $P(L \geq n)$, not $P(L = n)$. So $P(L = n) = P(L \geq n) - P(L \geq n+1) = 1/n - 1/(n+1)$. It was obvious your answer could not be correct because $\sum_n P(L = n)$ must be $1$! – Anonymous Jun 23 '20 at 10:05