I often shop downtown in the afternoon, leave at a random time, and take the bus home.

probability/bus.packs New

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Problem

I often shop downtown in the afternoon, leave at a random time, and take
the bus home.  The average time I need to wait for a bus is 9 minutes.
However, if I see a bus leaving just before I arrive at the bus stop, my
average waiting time for the next bus is only 6 minutes.  How can this
happen?

-- David Joseph Grabiner <grabiner@sunset.ma.huji.ac.il>

Show solution

Because, as is well known to all commuters, buses travel in packs. Thus
if you see a bus just leaving, there is a good chance there is another
one right behind it.

For example, if buses travel in packs of three every 18 minutes, then if
you see one leave there is a 2/3 chance there is another right behind
(0 wait) and a 1/3 chance of a 18 minute wait, giving your 6 minute
average wait.  If you arrive at another time, your expected wait is 1/2
of 18 minutes or 9 minutes.  (Packs of two don't seem to produce this
effect, however, as it's 50-50 whether the bus you see leaving is the
first or second of the pack).

This reminds me of the old problem of the man with two girlfriends in
different towns served by trains from the same station.  Both trains are
scheduled every 10 minutes, but because one is scheduled right after the
other, he finds himself visiting the one girlfriend much more often than
the other, even though he arrives at the station at a random time.

-- Peter Suzman <suzman@netcom.com>