The Inspector's Paradox

Why is the bus always late? Buses arrive at a stop with some average headway (gap between buses) of $\mu$ minutes. A passenger arrives at a uniformly random time and waits for the next bus. How long do they wait? The naive answer is $\mu / 2$: on average you land in the middle of a gap. The correct answer is almost always longer—sometimes much longer. The expected wait is not $\mu/2$ but: $$E[\text{wait}] = \frac{\mu}{2} + \frac{\sigma^2}{2\mu}$$ where $\sigma^2 = \text{Var}[\text{headway}]$. The second term is always non-negative, so higher variance always means longer expected waits, even when the mean headway is unchanged. Three Bus Schedules with Mean Headway $\mu = 10$ Schedule $\sigma^2$ Predicted wait Naive wait Regular 0 5.0 5.0 Exponential 100 10.0 5.0 Clustered 64 8.2 5.0 For exponentially distributed headways, $\sigma^2 = \mu^2$, so: $$E[\text{wait}] = \frac{\mu}{2} + \frac{\mu^2}{2\mu} = \mu$$ A passenger waits on average for an entire mean headway — twice the naive…