TITLE: The parking problem riddle
QUESTION [7 upvotes]: Assume a street of 300 meters, that you can park your car alongside the pavement. Assume that there is a big parking problem in the area. Assume that the pavement is continuous, without interruptions, and that you can park alongside all of it. Assume that the length of a car is 3 meters long. Assume, for simplicity, that cars can park without space between them (bumper to bumper). Assume, that when a car comes to the street if chooses an equally random parking space (please try to express this randomness) from the free spaces left. Therefore, it may "ruin" parking places for other cars. Please try to determine what is the expectancy of cars the can park alongside the street.

REPLY [2 votes]: Let $f(r)$ be the expected number of cars that will park in an interval of length $r$.
By considering what happens when the first car comes along, we should have
$$ f(r) = \cases{ 0 & if $r < 1$\cr
                 1 + \frac{1}{r-1} \int_0^{r-1} ds\; (f(s) + f(r-1-s))\cr
                 \ = 1 + \frac{2}{r-1} \int_0^{r-1} ds \; f(s) & otherwise}$$
So:
If $1 \le r < 2$, $f(r) = 1$.
If $2 \le r < 3$, $f(r) = 1 + \frac{2 (r-2)}{r-1} = 3 - \frac{2}{r-1} $.
etc.
Unfortunately, by the time we get to $6 < r < 7$, the integrals can't be done in closed form.