I’ve run across an interesting problem. On my trip home from my internship, I take a train followed by a bus. However, more than one bus will get me close enough to home so that walking the rest of the trip is reasonable.

Buses A, B, C: These three buses use the same stop that I get off at; in fact, the stop serves as a transfer point to switch from one of these buses to another. The obvious pro is that there are three such buses.

Bus D: This is only one bus, but this stop is closest to my house.

I have to cross a somewhat busy intersection in both instances; the only difference between ABC and D is where.

The train station I get off at serves as an end point for all four buses. As a result, the buses will idle for a bit and let people getting off the train get on the bus. Excluding today, for the past two days two buses–D and one of A, B, and C–were available. By some coincidence I chose the bus that left later both times, resulting in a hungry Sushi by the time I arrived home. Unfortunately it rained one of those days, making me grumpy as well as hungry.

So here’s the problem. Given no knowledge of a bus schedule, and given a choice between bus D and one of buses A, B, and C, what is the probability that the bus you choose will leave first? For my situation I could probably solve it by using common sense and looking at a bus schedule, but what about an ordinary day? What about multiple buses without three buses using the same stop? What if only two buses are available at your stop?

This bears pondering.

It’s more of a Murphy’s law type of thing than a true probability problem. No matter which bus you choose, the one you choose will either leave last or break down on the way home. Best to keep food with you at all times. ðŸ™‚