A parking lot consists of an infinite row of bays. Cars arrive at random intervals (mean interval $T_a$) and stay for a random time (mean stay $T_s$). The time intervals are memoryless (negative exponential distribution). An arriving car parks in bay $n$, which is the first available bay. What is the distribution of $n$?
The question is motivated by the following annoyance: why is it that, when I meet someone at the airport, I always end up driving to the far end of the lot?
Here is a bit more information, based on simulation. The ratio of the times, rather than their actual values, determines the behaviour. Thus we can write $k=T_s/T_a$ and use $k$ as the single parameter. If $k=1$, the parking lot can be small (two bays are usually enough), and the distribution is interesting only for large values of $k$. E.g., cars arrive every 2 minutes and stay for 2 hours.
The first bay is the one most likely to be free (!). However, for large $k$, the distribution is almost flat, falling off rapidly after $k$ bays and therefore having a mean of $k/2/$. In fact, I conjecture that $\mu \rightarrow k/2$ as , as $k \rightarrow \infty$.