Equal frequency, but A arrives first 80% of the time?

The trick is that equal frequency does not mean the trains are evenly spaced.

Example schedule
Both trains run once every 10 minutes.
B is next A is next Train A Train B Train A 0 min 2 min 10 min
Click the button to simulate arriving at a random time.
Arrive during minute 0–2 20% Train B arrives next.
Arrive during minute 2–10 80% Train A arrives next.

What’s going on?

Imagine the repeating schedule is: A at 0 minutes, B at 2 minutes, A again at 10 minutes.

Both trains still have the same frequency: each arrives once every 10 minutes. But they are clustered together.

After A arrives, there are only 2 minutes until B. After B arrives, there are 8 minutes until the next A. If you enter the station at a random time, you are four times more likely to land in that long 8-minute interval.

P(A next) = 8/10 = 80%
P(B next) = 2/10 = 20%

So the observation does not imply Train A runs four times as often. It only tells you something about how the two schedules are spaced relative to one another.

This is closely related to waiting-time bias / the inspection paradox: random observers are disproportionately likely to find themselves inside longer gaps.