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 $\mu = 10$

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 expectation.

Why This Happens: Length-Biased Sampling

A passenger arriving at a random time is more likely to land inside a long gap than a short one, because long gaps occupy more time on the clock. This is called length-biased sampling. The interval containing your arrival is not a random headway: it is drawn from the length-biased distribution with density:

$$f^*(h) = \frac{h \cdot f(h)}{\mu}$$

The mean of this biased distribution is $\mu + \sigma^2/\mu$, and you arrive uniformly within it, giving expected wait $(\mu + \sigma^2/\mu)/2$.

The same phenomenon explains why the average class size experienced by a student exceeds the average class size reported by the university (large classes have more students to report them).

Why “Inspector’s Paradox”?

The name comes from quality control, where an inspector arrives at a random time to sample a production process and systematically encounters longer-than-average intervals. The paradox is that a random observer is more likely to land inside a long gap than a short one, so their experienced mean interval exceeds the true mean interval. It feels paradoxical because you’d expect a random arrival to see the average gap, but length-biased sampling guarantees they see worse-than-average gaps whenever there’s any variance at all.

Understanding the Math

Length-biased sampling

Suppose buses run on an irregular schedule where gaps between buses are either 2 minutes or 18 minutes, each with probability 1/2. The mean gap is $\mu = (2 + 18)/2 = 10$ minutes. Now ask: if you arrive at a completely random moment, which gap are you most likely to land inside?

A 2-minute gap occupies only 2 minutes on the clock, but an 18-minute gap occupies 18. Out of every 20 minutes of clock time on average, 2 minutes belong to a short gap and 18 to a long one. So a random arrival lands in a short gap with probability $2/(2+18) = 1/10$ and in a long gap with probability $18/20 = 9/10$. The expected gap length you experience is:

$$E[\text{gap experienced}] = \frac{1}{10} \cdot 2 + \frac{9}{10} \cdot 18 = 0.2 + 16.2 = 16.4 \text{ minutes}$$

That is far above the mean gap of 10 minutes. You are disproportionately likely to land inside a long gap simply because it takes up more time.

The wait formula

Once you are inside a gap, you arrive uniformly within it, so on average you land in the middle. Your expected wait is half the gap length you experience. The full formula is:

$$E[\text{wait}] = \frac{\mu}{2} + \frac{\sigma^2}{2\mu}$$

Here $\mu$ is the mean gap and $\sigma^2 = \text{Var}[\text{gap}]$ is the variance of gap lengths. The first term, $\mu/2$, is what you would get if every gap were exactly $\mu$ (deterministic buses — arrive in the middle every time). The second term, $\sigma^2/(2\mu)$, is the extra waiting from length-biased sampling. It is always non-negative, so irregular buses always make you wait longer than regular buses with the same mean headway.

Why variance matters

The variance $\sigma^2$ measures how spread out the gap sizes are. A perfectly regular bus schedule has $\sigma^2 = 0$ and gives the naive answer $\mu/2$. An exponentially distributed schedule has $\sigma^2 = \mu^2$, which doubles the expected wait to $\mu$. More irregular buses, higher penalty.

Connecting to expected values

The formula arises from a standard result: the expected length of the gap containing a random arrival is $\mu + \sigma^2/\mu$. You can think of this as the mean gap plus a correction term proportional to the variance divided by the mean. Dividing by 2 (uniform arrival within the gap) gives the wait formula above.

This article was originally written for marimo.io.