Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss
Re: the runner's problem
20 views
Skip to first unread message
RichD
Oct 9, 2012, 12:49:51 AM10/9/12
to
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
>> > 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
RichD
Oct 10, 2012, 12:26:36 AM10/10/12
to
On Oct 6, Willem <wil...@turtle.stack.nl> 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?
>
> Assuming that this is hell, where you have to run endlessly (or at least
> long enough to have, or be, passed by everyone a lot of times),
Right. That's why I stipulated "after while, you start
counting...".
Then we can assume the runners, and speeds, are uniformly
distributed around the circuit ('completely random'), a key
assumption. So you pass/are passed by a representative sample
of the field, no matter where or when.
> and also that everybody runs at a constant speed, then your speed
> should be the same as that of Average Al.
>
> Sketch of proof: The number of times that a given person passes (or is
> passed by) you in a given period is a linear function of their speed
> difference to you. (Negative numbers meaning that you pass them.)
> So if the average of those passes is 0, the average of speed differences
> must also be 0, and your speed must be the average speed.
>
> Suppose the track is 1km long, then someone running 1km/h faster will
> pass you 1 time per hour, and someone running 3km/h slower will be
> passed by you 3 times an hour.
Another way to see it, is to subtract the average speed
from everyone's speed. Which simplifies the problem,
but doesn't change the relative speeds or positions.
Now let's try another: You perform a random telephone
survey of local residents, to estimate population body
weights. You notice that the number who weigh more
than you, equals the number who weigh less, no matter
how many you call. What can you infer about your weight,
relative to the mean and median?
Why is this case different, given that in both problems,
you sample from a uniformly distributed population?
--
Rich
> ) 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?
>
> Assuming that this is hell, where you have to run endlessly (or at least
> long enough to have, or be, passed by everyone a lot of times),
Right. That's why I stipulated "after while, you start
counting...".
Then we can assume the runners, and speeds, are uniformly
distributed around the circuit ('completely random'), a key
assumption. So you pass/are passed by a representative sample
of the field, no matter where or when.
> and also that everybody runs at a constant speed, then your speed
> should be the same as that of Average Al.
>
> Sketch of proof: The number of times that a given person passes (or is
> passed by) you in a given period is a linear function of their speed
> difference to you. (Negative numbers meaning that you pass them.)
> So if the average of those passes is 0, the average of speed differences
> must also be 0, and your speed must be the average speed.
>
> Suppose the track is 1km long, then someone running 1km/h faster will
> pass you 1 time per hour, and someone running 3km/h slower will be
> passed by you 3 times an hour.
Another way to see it, is to subtract the average speed
from everyone's speed. Which simplifies the problem,
but doesn't change the relative speeds or positions.
Now let's try another: You perform a random telephone
survey of local residents, to estimate population body
weights. You notice that the number who weigh more
than you, equals the number who weigh less, no matter
how many you call. What can you infer about your weight,
relative to the mean and median?
Why is this case different, given that in both problems,
you sample from a uniformly distributed population?
--
Rich
Tim Little
Oct 10, 2012, 4:17:47 AM10/10/12
to
On 2012-10-10, RichD <r_dela...@yahoo.com> wrote:
> Why is this case different, given that in both problems, you sample
> from a uniformly distributed population?
The frequency of being passed by a given runner is proportional to how
> Why is this case different, given that in both problems, you sample
> from a uniformly distributed population?
much faster than you they are. Likewise for you passing other runners.
In the weight survey case, the number of times you call a person of
greater weight is not proportional to how much more mass they have.
--
Tim
0 new messages