The gap you walk into
Buses arrive at a stop as a Poisson process, on average one bus every 10 minutes. The buses have been running for a very long time. You arrive at the stop at a time that has nothing to do with the buses.
Find the expected length, in minutes, of the time between the last bus before you arrive and the first bus after you arrive.
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
How long do you expect to wait for the next bus?
Run time backwards. How long ago, on average, did the last bus leave?
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s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. The gap you land in is the sum of two pieces: the time since the last bus and the time to the next one. Each has mean 10 minutes, so the gap has mean 20, twice the average gap between buses.
Step 1: forward. In a Poisson process with rate per minute, the time from any fixed moment to the next arrival is exponential with mean 10, whatever happened before (the process has no memory).
Step 2: backward. A Poisson process run backwards in time is again a Poisson process with the same rate. So the time since the last bus is also exponential with mean 10, and it is independent of the forward wait, because they depend on what happens before and after your arrival.
Step 3.
Why not 10? Gaps between buses do have mean 10, but you are more likely to arrive during a long gap than a short one: a gap of length is hit with probability proportional to . With exponential with mean 10, This is the inspection paradox. It is why a bus that comes every 10 minutes on average feels like it makes you wait longer than 5 minutes: here you wait 10 minutes on average.