We look at three busses that are scheduled to arrive at 10min intervals, say, at the full hour, 10 past, and 20 past. We assume passengers arrive at a rate of 1 per minute, so if all three busses arrive on time, 10 will be sitting in the 00:10 bus and 10 in the 00:20 one.
Since 1 passenger arrives each minute, we have, in each group, 1 passenger that has just missed the earlier bus, so has to wait for 10 minutes; 1 passenger that has to wait for 9 minutes, and so on, meaning we have to sum up [ 1 .. n ] and divide by n to get the average waiting time,
avg = ( n ) -> ( [ 1 .. n ].reduce ( sum, a ) -> sum + a ) / n
This gives 5.5, not 5, minutes of average waiting time, so already a little over the naive expectation of dt/2 = 5. Since both sets of passengers are identical otherwise, for our 20 assumed riders, the average waiting time is still 5.5min.
But let's assume the 00:10 bus is late 5min. Late busses are often caused by busy streets, and busy streets mean a lot of people are moving. Regardless, now the busses arrive at 00:00, 00:15, 00:20, meaning there will be 15 people for the 00:10 tour and only 5 for the 00:20 tour, meaning that the first group has an average of avg(15) = 8 minutes to wait, and the second avg(5) = 3. The overall average waiting time is still 5.5min.
But will the 00:20 bus arrive in time? This is in reality a bit unlikely because it will probably experience very similar traffic conditions as the previous tour, meaning its arrival time will be delayed by a similar amount, so we should push its arrival time to 00:25, meaning there are now 10 people instead of 5 waiting for it, so the average waiting time is now closer to (avg(15)+avg(10))/2, which is 6.75, almost 7 minutes instead of the naive assumption of 5.
We look at three busses that are scheduled to arrive at 10min intervals, say, at the full hour, 10 past, and 20 past. We assume passengers arrive at a rate of 1 per minute, so if all three busses arrive on time, 10 will be sitting in the 00:10 bus and 10 in the 00:20 one.
Since 1 passenger arrives each minute, we have, in each group, 1 passenger that has just missed the earlier bus, so has to wait for 10 minutes; 1 passenger that has to wait for 9 minutes, and so on, meaning we have to sum up [ 1 .. n ] and divide by n to get the average waiting time,
This gives 5.5, not 5, minutes of average waiting time, so already a little over the naive expectation of dt/2 = 5. Since both sets of passengers are identical otherwise, for our 20 assumed riders, the average waiting time is still 5.5min.But let's assume the 00:10 bus is late 5min. Late busses are often caused by busy streets, and busy streets mean a lot of people are moving. Regardless, now the busses arrive at 00:00, 00:15, 00:20, meaning there will be 15 people for the 00:10 tour and only 5 for the 00:20 tour, meaning that the first group has an average of avg(15) = 8 minutes to wait, and the second avg(5) = 3. The overall average waiting time is still 5.5min.
But will the 00:20 bus arrive in time? This is in reality a bit unlikely because it will probably experience very similar traffic conditions as the previous tour, meaning its arrival time will be delayed by a similar amount, so we should push its arrival time to 00:25, meaning there are now 10 people instead of 5 waiting for it, so the average waiting time is now closer to (avg(15)+avg(10))/2, which is 6.75, almost 7 minutes instead of the naive assumption of 5.