The parking problem riddle 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. 
 A: According to http://mathworld.wolfram.com/RenyisParkingConstants.html, if $M(n)$ is the expected number of cars of length 1 to fit onto a stripe of length $n$, you have $$
  m := \lim_{n\to\infty} \frac{M(n)}{n} \approx 0.7476 \text{.}
$$
In other words, the expected density (i.e., the total length of the parked cars divided by the length of the pavement) goes to $0.7476\ldots$ as the pavement length goes to infinity. In your case that leads to an estimate for the expected number $N$ of 3m long cars on a 300m long pavement of $$
  \frac{3N}{300} \approx 0.7476 \implies N \approx 74.76 \text{.}
$$
A more precise estimate from that page is $$
  M(n) \approx mn + m - 1 \text{,}
$$
again for the expected number of 1m long cars on a stripe of length $n$. To apply that to your question, one has to use that the expected number $N$ of  3m long cars on a 300m long stripe is the same as the expected number of 1m long cars on a 100m long stripe, which yields $$
  N \approx M(100) \approx 74.51 \text{.}
$$
A: 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.
