Bus wait when some gaps include an hour-long refuel

Buses are scheduled $10$ minutes apart. Before each gap, independently, the driver stops to refuel with probability $10\%$, and a refuel adds $60$ minutes to that gap (making it $70$ minutes). You arrive at the stop at a uniformly random time of day. What is your expected wait, in minutes, for the next bus?

$\tfrac{145}{8}$ minutes (18.125). You are more likely to land in a long gap (inspection paradox). Your expected wait is $\tfrac{\mathbb{E}[L^2]}{2\,\mathbb{E}[L]}$ for gap length $L$: $\mathbb{E}[L]=16$ and $\mathbb{E}[L^2]=0.9\cdot100+0.1\cdot4900=580$, giving $\tfrac{580}{32}=\tfrac{145}{8}$.