Each day, I will catch either Bus A or Bus B on my journey home from work. Both take the same route, run every ten minutes, and take the same time to reach my destination. Neither is busier, cheaper or less comfortable, and I have no preference for one over the other.
I finish work at different times every day, somewhere between 4-6pm. I hustle to the bus stop and jump on whichever bus is there.
And yet 90% of the time, I take Bus B. Why?
It's probably due to scheduling.
If Bus A runs on the 0s, and bus B runs on the 9s, then, given a random arrival time at the stop, you have a 1 minute window to catch A, and a 9 minute window to catch B
0s and 9s mean? I don't understand this answer
$\endgroup$
How about
Bus B has its terminus at your stop so it tends to be sitting there waiting for its scheduled departure
Is it because
Bus B is a weekday bus and Bus A only runs starting on weekends. If Bus A starts running Friday afternoons, that would mean that Bus A would be caught 10% of the time (Friday evenings) while Bus B is caught the remaining 90% (M-F mornings, M-Th evenings).
The one thing that's not been mentioned so far:
Everything else in the question discounts there being a difference in the bus or the routes, therefore the difference must be in where you work. Should the two busses have the same route but be different then they are most likely running opposite directions to each other, and you tend to take the stop that is slightly closer to your work: on the same side of the street - but sometimes you'll cross the road and take the bus heading the other direction, either due to traffic (the crossing), or actually seeing it turn up just as you get to the empty stop on this side.
Maybe although you "finish work at different times every day, somewhere between 4-6pm",
90% of the time you finish at 4:05, when bus B arrives, and the other 10% of the time you finish at 4:11, when bus A arrives. You didn't preclude an extremely narrow distribution of departure times like this. (Of course, it works equally well for distribution between any two other times in that range.) Unless "different times every day" is meant to literally mean that no leaving time is ever repeated, in which case this solution could be modified to leaving at 4:05:00, 4:05:01, 4:05:02*, ... 90% of the time,
etc.
*Choose a smaller increment to fit, depending on the length of your career.
I think it's a matter of distance
My guess is that both busses arrive and leave at the same time. It just happens to be that Bus B is closer to your workplace so it's the first one you hop in. In case bus B is fully occupied, you walk over to the next bus.
Based on distribution:
Either they are more Bus B than Bus A (9 Bus B but only a single Bus A for example), like a Bus B every 10mn dispatched 1mn apart each, the 9th being a Bus A. The Bus B is likely the general route, and one out of ten bus make a different route after your home and, as such, are labelled A
Based on logic:
If there are 50% probability to get any bus, then you are the driver of Bus A. Thus to go home, you need to walk down your own bus, let the new driver jump in, and wait for the next bus (Bus B). The 10% of the time you take a Bus A is when you have to drive another bus (one day per your 5 days work week, thus 20% of your the times), and then you have 50% of chances to have a Bus A or Bus B.
My first guess would be
There's no reason, it's just a matter of coincidence. Though if the odds were perfectly distributed you'd ride 50% of the times on each, there's always a chance that you'll ride one more often than the other, whether it's 5% or 50% more times.
It must be
You have no preference on one buss over another but your annoying neighbor that also go to the same way prefers bus A and so you take bus B to avoid him.
There is another employee in another office who get Bus A 90% Of times. Tho keep balance. This is law of nature.
May be Bus A and B is there from home to work.
But from work to home only Bus B is there
