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More options Oct 9 2012, 12:49 am
Newsgroups: rec.gambling.poker, rec.puzzles, alt.math.recreational
From: RichD <r_delaney2...@yahoo.com>
Date: Mon, 8 Oct 2012 21:49:51 -0700 (PDT)
Local: Tues, Oct 9 2012 12:49 am
Subject: Re: the runner's problem
On Oct 6,  "Jason Pawloski" <a679...@webnntp.invalid> wrote:

>> > You've entered a road race.  It's around a loop, a lakeside
> > > race, but very long, like a marathon.

> > > It contains a large field of competitors, including Median
> > > Mel, who's the median speed runner.  And Al Average,
> > > who runs at the average speed of the entire field.

> > > After a while, you get bored, so you start to count the
> > > runners whom you pass, and the runners whom pass
> > > you.  You notice something funny:  the number who pass
> > > you equals the number you pass, on average, per hour.
> > > How fast are you, relative to Al or Mel?

> > Regardless of the distribution of runner's speeds only the
> > averagespeed runner will, on average, be passed by the same
> > number of runners as he passes himself.
> > It's easy to prove this in a spreadsheet.

> Is there a hidden assumption somewhere that the speed of each
> runner is constant?

In math puzzles like this, one makes the simplest assumptions.

> You can get in trouble pretty quickly if half of the field is running at a
> constant speed and the other half is oscillating at a fixed period so that
> they are running faster and slower than you.

That's an interesting question.

If a runner's speed oscillates, both faster and slower than
you, he'll overtake you when he accelerates, and verse
visa when he slows.  These cancel, thus the solution is
unaffected.

--
Rich