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the runner's problem
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RichD
Oct 5, 2012, 8:56:29 PM10/5/12
to
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? Do the clues
given provide sufficient information to solve the case,
Watson?
--
Rich
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? Do the clues
given provide sufficient information to solve the case,
Watson?
--
Rich
Eric Sosman
Oct 5, 2012, 9:07:41 PM10/5/12
to
On 10/5/2012 8:56 PM, RichD 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.
ITYM "who pass youm."
> 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? Do the clues
> given provide sufficient information to solve the case,
> Watson?
P
O
I
L
A
G
E
You're running faster than those you pass and slower than
those who pass you, and since these numbers are equal you're
running at about the same speed as Median Mel.
Nothing can be said about your speed vis-a-vis Average Al.
The leader in the race might be Lightspeed Lenny, who got round
the track all umpty-mumble times before the rest of you clods
got your first blister; in this case Al is second and beats all
but Lenny by a wide margin. Then again, the race's trailer might
be Myocardial Mort, who keeled over ten paces into the first lap
and hasn't moved since; Al would then be second-to-last. Or, of
course, Al could be anywhere in between.
--
Eric Sosman
eso...@comcast-dot-net.invalid
Ted Schuerzinger
Oct 5, 2012, 9:38:03 PM10/5/12
to
On Fri, 5 Oct 2012 17:56:29 -0700 (PDT), RichD wrote:
> How fast are you, relative to Al or Mel? Do the clues
> given provide sufficient information to solve the case,
> Watson?
Insufficient information. Everybody runs at a uniform (for each
> How fast are you, relative to Al or Mel? Do the clues
> given provide sufficient information to solve the case,
> Watson?
individual) speed, with the fastest runner running at 4.00m/s, and the
slowest runner running at 3.90m/s, which means the fastest runner gains
360m on the slowest runner per hour, and nobody passes anybody on the
much longer (~5km) loop until after about 13 hours.
--
Ted S.
fedya at hughes dot net
Now blogging at http://justacineast.blogspot.com
RichD
Oct 5, 2012, 9:43:02 PM10/5/12
to
On Oct 5, Eric Sosman <esos...@comcast-dot-net.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.
>
> ITYM "who pass youm."
I was trying to do my Harvard grad imitation.
> > 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.
>
> ITYM "who pass youm."
Did I counter-exemplify?
> > 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?
> You're running faster than those you pass and slower than
> those who pass you, and since these numbers are equal you're
> running at about the same speed as Median Mel.
>
> Nothing can be said about your speed vis-a-vis Average Al.
> The leader in the race might be Lightspeed Lenny, who got round
> the track all umpty-mumble times before the rest of you clods
> got your first blister; in this case Al is second and beats all
> but Lenny by a wide margin. .
> those who pass you, and since these numbers are equal you're
> running at about the same speed as Median Mel.
>
> Nothing can be said about your speed vis-a-vis Average Al.
> The leader in the race might be Lightspeed Lenny, who got round
> the track all umpty-mumble times before the rest of you clods
> got your first blister; in this case Al is second and beats all
If someone passes you twice, you count him twice.
That's why I stipulated, it's around a track, many laps,
with a large field.
--
Rich
Mark Brader
Oct 5, 2012, 11:33:30 PM10/5/12
to
Rich Delaney:
Schuerzinger's response. Marathons aren't customarily run in
laps around a closed track.
> 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? Do the clues
> given provide sufficient information to solve the case,
> Watson?
Nowhere near sufficient. For one thing, we don't know anything
about how anybody's speed varies during the race. And even if
everybody's speed was constant, implying that multiple laps are
being run, it could be that most of the field is moving at almost
the same field -- either just ahead of me or just behind me --
and none of us have lapped each other yet.
--
Mark Brader | "...he entertained the notion that I was cribbing from
Toronto | other [students' exams] until it was pointed out that
m...@vex.net | I often had the only correct answer..." --Lars Eighner
My text in this article is in the public domain.
> You've entered a road race. It's around a loop, a lakeside
> race, but very long, like a marathon.
I assumed this meant *once* around the loop until I read Ted
> race, but very long, like a marathon.
Schuerzinger's response. Marathons aren't customarily run in
laps around a closed track.
> 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? Do the clues
> given provide sufficient information to solve the case,
> Watson?
about how anybody's speed varies during the race. And even if
everybody's speed was constant, implying that multiple laps are
being run, it could be that most of the field is moving at almost
the same field -- either just ahead of me or just behind me --
and none of us have lapped each other yet.
--
Mark Brader | "...he entertained the notion that I was cribbing from
Toronto | other [students' exams] until it was pointed out that
m...@vex.net | I often had the only correct answer..." --Lars Eighner
My text in this article is in the public domain.
Tad Perry
Oct 6, 2012, 3:47:22 AM10/6/12
to
Can we restate it that you've been consigned to a sort of hell and that you
have to run around the track endlessly and that this is what you discover?
And also, can we assume a constant speed for everyone, although some people
might be running at the same speed?
tvp
Clave
Oct 6, 2012, 4:28:21 AM10/6/12
to
"Tad Perry" <tadp...@comcast.net> wrote in message
news:k4onng$kdv$1...@dont-email.me...
my head.
Jim
Willem
Oct 6, 2012, 5:49:30 AM10/6/12
to
RichD 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? Do the clues
) given provide sufficient information to solve the case,
) Watson?
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), 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.
Perhaps a better way to place the puzzle would be on a racetrack, like
the 24h of Le Mans. After the race, everybody has their average speed
posted (including pit stops), and they get points for passing and minus
points for being passed. Of course, you're the one with 0 points.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
) 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? Do the clues
) given provide sufficient information to solve the case,
) Watson?
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), 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.
Perhaps a better way to place the puzzle would be on a racetrack, like
the 24h of Le Mans. After the race, everybody has their average speed
posted (including pit stops), and they get points for passing and minus
points for being passed. Of course, you're the one with 0 points.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
Dave Baker
Oct 6, 2012, 5:54:01 AM10/6/12
to
average can be far apart. The median speed runner will " at some point" pass
every slower runner and be passed by every faster one but how often this
happens will not be the same hourly total for each.
Only the runner running at exactly the average speed will, on average, pass
and be passed by the same number of people per hour regardless of the
distribution of speeds.
--
Dave Baker
Dave Baker
Oct 6, 2012, 6:05:09 AM10/6/12
to
speed 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. In column A list out a variety of
runner's speeds varying from say 1 mph to anything you like, in ascending
order. Use an odd number of runners so there's a clear median value. In
column B the equation is (X - value in adjacent cell in column A) where X is
your speed. Assuming a 1 mile track this value is the number of times per
hour you will pass a slower runner (positive number) or the number of times
you are passed by a faster runner (negative number). The total of column B
is zero only when X is the average speed in column A.
--
Dave Baker
Dave Baker
Oct 6, 2012, 6:06:25 AM10/6/12
to
Dave Baker wrote:
> RichD 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? Do the clues
>> given provide sufficient information to solve the case,
>> Watson?
Edit
> RichD 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? Do the clues
>> given provide sufficient information to solve the case,
>> Watson?
>
> Yes. Regardless of the distribution of runner's speeds only the
> average speed runner will, on average, be passed by the same number
> of runners (PER HOUR) as he passes himself.
> Yes. Regardless of the distribution of runner's speeds only the
> average speed runner will, on average, be passed by the same number
Clave
Oct 6, 2012, 6:10:26 AM10/6/12
to
"Dave Baker" <Nu...@null.com> wrote in message
news:k4ov3q$rmv$1...@news.datemas.de...
Jim
Jason Pawloski
Oct 6, 2012, 10:51:51 AM10/6/12
to
constant?
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.
--
WILLIAM COLEMAN PHOTO GALLERY! See here:
http://www.youtube.com/watch?v=fGg-mbTtlJ4&feature=youtu.be
William Coleman of Las Vegas Nevada is a dangerous, violent convicted
criminal. He has recently committed felonies in his harassment of an RGP
user.
"As a result of my encounter with the Henderson Police SWAT team, I
eventually pled guilty to Conspiracy to Commit a Crime, a gross
misdemeanor." - William Coleman admits that he is a convicted criminal
(cite: http://recgroups.com/a/1/127894/)
"I have killed many dogs in my life, and I will kill many more." William
P. Coleman boasting about his sociopathic tendencies (cite:
http://recgroups.com/a/1/128390/)
"For years, when I see one or more teenaged girls walking down the
street, I pull up beside them and ask them if they want to go smoke
dope and fuck. Amazingly, this technique gets good results fairly
often. Otherwise, I wouldn't continue to use it." Convicted criminal
William Coleman (cite: http://recgroups.com/a/1/123930/)
WARNING! 67-year old convicted criminal William Coleman describes how he
uses Facebook to meet minor girls. See:
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Tad Perry
Oct 6, 2012, 8:35:59 PM10/6/12
to
Willem wrote:
> RichD 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? Do the clues
> ) given provide sufficient information to solve the case,
> ) Watson?
>
> 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),
> and also that everybody runs at a constant speed, then your speed
> should be the same as that of Average Al.
Even if my assumptions are used, you're not necessarily traveling the same
> RichD 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? Do the clues
> ) given provide sufficient information to solve the case,
> ) Watson?
>
> 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),
> and also that everybody runs at a constant speed, then your speed
> should be the same as that of Average Al.
speed as Average Al. There could be someone running at near light speed,
everyone else running at human speeds, and "Average Al" could be in second
place and one of the runners that's constantly passing you. Or there's an
extremely slow runner that brings the average down so low that Average Al is
someone you keep passing.
tvp
Beldin the Sorcerer
Oct 6, 2012, 10:45:14 PM10/6/12
to
there must be the same number running faster as there is running slower....
this makes you median mel, or running at the same speed
Eric Sosman
Oct 6, 2012, 11:40:35 PM10/6/12
to
On 10/6/2012 10:45 PM, Beldin the Sorcerer wrote:
> [...]
out my mistake -- and it *was* a mistake; they're right. For a
concrete example, consider a race with five runners:
- Sluggish Sam, 1 cm/s
- Torpid Tommy, 1 cm/s
- Median Mel, 1 cm/s
- You, 3 m/s
- Lightspeed Lenny, 9 m/s
Each time you circle the track you'll pass the three slowpokes,
but Lightspeed Lenny will pass you three times[*]. Does this mean
you're running at Median Mel's speed? No: You're running a whole
lot faster, and he's one of the slowpokes you keep overtaking.
[*] Approximately. If we knew the length of the track, we
could concoct speeds that would work out perfectly.
For an example "from the other direction," consider the hour,
minute, and second hands of an analog clock. In half a day the
hour hand makes one circuit while the minute hand makes twelve,
so the minute hand passes the hour hand eleven times. Meanwhile,
the second hand makes 720 complete trips, passing the minute hand
708 times. Quite a discrepancy between eleven and 708; would you
say the minute hand's speed is not the median?
--
Eric Sosman
eso...@comcast-dot-net.invalid
> [...]
> In order to pass, and be passed by, the same number of people ON AVERAGE,
> there must be the same number running faster as there is running slower....
> this makes you median mel, or running at the same speed
That's what I supposed at first, but others have pointed
> there must be the same number running faster as there is running slower....
> this makes you median mel, or running at the same speed
out my mistake -- and it *was* a mistake; they're right. For a
concrete example, consider a race with five runners:
- Sluggish Sam, 1 cm/s
- Torpid Tommy, 1 cm/s
- Median Mel, 1 cm/s
- You, 3 m/s
- Lightspeed Lenny, 9 m/s
Each time you circle the track you'll pass the three slowpokes,
but Lightspeed Lenny will pass you three times[*]. Does this mean
you're running at Median Mel's speed? No: You're running a whole
lot faster, and he's one of the slowpokes you keep overtaking.
[*] Approximately. If we knew the length of the track, we
could concoct speeds that would work out perfectly.
For an example "from the other direction," consider the hour,
minute, and second hands of an analog clock. In half a day the
hour hand makes one circuit while the minute hand makes twelve,
so the minute hand passes the hour hand eleven times. Meanwhile,
the second hand makes 720 complete trips, passing the minute hand
708 times. Quite a discrepancy between eleven and 708; would you
say the minute hand's speed is not the median?
--
Eric Sosman
eso...@comcast-dot-net.invalid
Randy Hudson
Oct 6, 2012, 11:51:40 PM10/6/12
to
In article <k4qqc0$5pi$1...@dont-email.me>,
of unique individuals, then Median Mel has equal numbers. But if you take
it to count each occasion of passing or being passed, regardless of whom by,
then Average Al will have equa numbers.
Beldin the Sorcerer <Beld...@verizon.net> wrote:
> In order to pass, and be passed by, the same number of people ON AVERAGE,
> there must be the same number running faster as there is running slower....
> this makes you median mel, or running at the same speed
If you take "number of runners [passed / passed by]" as meaning the number
> In order to pass, and be passed by, the same number of people ON AVERAGE,
> there must be the same number running faster as there is running slower....
> this makes you median mel, or running at the same speed
of unique individuals, then Median Mel has equal numbers. But if you take
it to count each occasion of passing or being passed, regardless of whom by,
then Average Al will have equa numbers.
Beldin the Sorcerer
Oct 6, 2012, 11:55:04 PM10/6/12
to
Eric Sosman wrote:
> On 10/6/2012 10:45 PM, Beldin the Sorcerer wrote:
>> [...]
>> In order to pass, and be passed by, the same number of people ON
>> AVERAGE, there must be the same number running faster as there is
>> running slower.... this makes you median mel, or running at the same
>> speed
>
> That's what I supposed at first, but others have pointed
> out my mistake -- and it *was* a mistake; they're right. For a
> concrete example, consider a race with five runners:
>
> - Sluggish Sam, 1 cm/s
> - Torpid Tommy, 1 cm/s
> - Median Mel, 1 cm/s
> - You, 3 m/s
> - Lightspeed Lenny, 9 m/s
>
> Each time you circle the track you'll pass the three slowpokes,
> but Lightspeed Lenny will pass you three times[*]. Does this mean
> you're running at Median Mel's speed? No: You're running a whole
> lot faster, and he's one of the slowpokes you keep overtaking.
Ok, you're an idiot
> On 10/6/2012 10:45 PM, Beldin the Sorcerer wrote:
>> [...]
>> In order to pass, and be passed by, the same number of people ON
>> AVERAGE, there must be the same number running faster as there is
>> running slower.... this makes you median mel, or running at the same
>> speed
>
> That's what I supposed at first, but others have pointed
> out my mistake -- and it *was* a mistake; they're right. For a
> concrete example, consider a race with five runners:
>
> - Sluggish Sam, 1 cm/s
> - Torpid Tommy, 1 cm/s
> - Median Mel, 1 cm/s
> - You, 3 m/s
> - Lightspeed Lenny, 9 m/s
>
> Each time you circle the track you'll pass the three slowpokes,
> but Lightspeed Lenny will pass you three times[*]. Does this mean
> you're running at Median Mel's speed? No: You're running a whole
> lot faster, and he's one of the slowpokes you keep overtaking.
IN ORDER FOR THE STATEMENT TO BE TRUE, that has to be the case
IT DOES NOT MATTER if there are cases where that is NOT the case and you are
the median speed runner.
IF you are passing and being passed in equal numbers THEN you must be
traveling at the median speed
That is NOT the same thing as saying IF you are traveling the median speed,
you must be passing and being passed by the same number of runners.
It isn't possible UNLESS you are travelling the median speed. That doesn't
mean it always happens.
Go reread logic, so you can tell statements, converse, inverse, and
contrapositives apart, and understand which pairings mean the same thing and
which don't.
<Snip converse argument, which only shows poster failed logic>
Eric Sosman
Oct 7, 2012, 8:47:15 AM10/7/12
to
On 10/6/2012 11:55 PM, Beldin the Sorcerer wrote:
> Eric Sosman wrote:
>> On 10/6/2012 10:45 PM, Beldin the Sorcerer wrote:
>>> [...]
>>> In order to pass, and be passed by, the same number of people ON
>>> AVERAGE, there must be the same number running faster as there is
>>> running slower.... this makes you median mel, or running at the same
>>> speed
>>
>> That's what I supposed at first, but others have pointed
>> out my mistake -- and it *was* a mistake; they're right. For a
>> concrete example, consider a race with five runners:
>>
>> - Sluggish Sam, 1 cm/s
>> - Torpid Tommy, 1 cm/s
>> - Median Mel, 1 cm/s
>> - You, 3 m/s
>> - Lightspeed Lenny, 9 m/s
>>
>> Each time you circle the track you'll pass the three slowpokes,
>> but Lightspeed Lenny will pass you three times[*]. Does this mean
>> you're running at Median Mel's speed? No: You're running a whole
>> lot faster, and he's one of the slowpokes you keep overtaking.
>
> Ok, you're an idiot
"I love you just the way you are."
> Eric Sosman wrote:
>> On 10/6/2012 10:45 PM, Beldin the Sorcerer wrote:
>>> [...]
>>> In order to pass, and be passed by, the same number of people ON
>>> AVERAGE, there must be the same number running faster as there is
>>> running slower.... this makes you median mel, or running at the same
>>> speed
>>
>> That's what I supposed at first, but others have pointed
>> out my mistake -- and it *was* a mistake; they're right. For a
>> concrete example, consider a race with five runners:
>>
>> - Sluggish Sam, 1 cm/s
>> - Torpid Tommy, 1 cm/s
>> - Median Mel, 1 cm/s
>> - You, 3 m/s
>> - Lightspeed Lenny, 9 m/s
>>
>> Each time you circle the track you'll pass the three slowpokes,
>> but Lightspeed Lenny will pass you three times[*]. Does this mean
>> you're running at Median Mel's speed? No: You're running a whole
>> lot faster, and he's one of the slowpokes you keep overtaking.
>
> Ok, you're an idiot
> IN ORDER FOR THE STATEMENT TO BE TRUE, that has to be the case
>
> IT DOES NOT MATTER if there are cases where that is NOT the case and you are
> the median speed runner.
>
> IF you are passing and being passed in equal numbers THEN you must be
> traveling at the median speed
circle the track, yes? How many times are you yourself passed in
that same interval? Three times. You are "passing and being
passed in equal numbers," but not running at the median speed.
--
Eric Sosman
eso...@comcast-dot-net.invalid
Willem
Oct 7, 2012, 2:18:13 PM10/7/12
to
Eric Sosman wrote:
) On 10/6/2012 11:55 PM, Beldin the Sorcerer wrote:
)> Eric Sosman wrote:
)>> On 10/6/2012 10:45 PM, Beldin the Sorcerer wrote:
)>>> [...]
)>>> In order to pass, and be passed by, the same number of people ON
)>>> AVERAGE, there must be the same number running faster as there is
)>>> running slower.... this makes you median mel, or running at the same
)>>> speed
)>>
)>> That's what I supposed at first, but others have pointed
)>> out my mistake -- and it *was* a mistake; they're right. For a
)>> concrete example, consider a race with five runners:
)>>
)>> - Sluggish Sam, 1 cm/s
)>> - Torpid Tommy, 1 cm/s
)>> - Median Mel, 1 cm/s
)>> - You, 3 m/s
)>> - Lightspeed Lenny, 9 m/s
)>>
)>> Each time you circle the track you'll pass the three slowpokes,
)>> but Lightspeed Lenny will pass you three times[*]. Does this mean
)>> you're running at Median Mel's speed? No: You're running a whole
)>> lot faster, and he's one of the slowpokes you keep overtaking.
)>
)> Ok, you're an idiot
)
) "I love you just the way you are."
)
)> IN ORDER FOR THE STATEMENT TO BE TRUE, that has to be the case
)>
)> IT DOES NOT MATTER if there are cases where that is NOT the case and you are
)> the median speed runner.
)>
)> IF you are passing and being passed in equal numbers THEN you must be
)> traveling at the median speed
)
) In the example I gave, you pass three runners each time you
) circle the track, yes? How many times are you yourself passed in
) that same interval? Three times. You are "passing and being
) passed in equal numbers," but not running at the median speed.
Yes, but that is the same runner that passed you three times,
whereas there are three different runners that you pass.
That's a different number of people.
See where the interpretation can differ?
) On 10/6/2012 11:55 PM, Beldin the Sorcerer wrote:
)> Eric Sosman wrote:
)>> On 10/6/2012 10:45 PM, Beldin the Sorcerer wrote:
)>>> [...]
)>>> In order to pass, and be passed by, the same number of people ON
)>>> AVERAGE, there must be the same number running faster as there is
)>>> running slower.... this makes you median mel, or running at the same
)>>> speed
)>>
)>> That's what I supposed at first, but others have pointed
)>> out my mistake -- and it *was* a mistake; they're right. For a
)>> concrete example, consider a race with five runners:
)>>
)>> - Sluggish Sam, 1 cm/s
)>> - Torpid Tommy, 1 cm/s
)>> - Median Mel, 1 cm/s
)>> - You, 3 m/s
)>> - Lightspeed Lenny, 9 m/s
)>>
)>> Each time you circle the track you'll pass the three slowpokes,
)>> but Lightspeed Lenny will pass you three times[*]. Does this mean
)>> you're running at Median Mel's speed? No: You're running a whole
)>> lot faster, and he's one of the slowpokes you keep overtaking.
)>
)> Ok, you're an idiot
)
) "I love you just the way you are."
)
)> IN ORDER FOR THE STATEMENT TO BE TRUE, that has to be the case
)>
)> IT DOES NOT MATTER if there are cases where that is NOT the case and you are
)> the median speed runner.
)>
)> IF you are passing and being passed in equal numbers THEN you must be
)> traveling at the median speed
)
) In the example I gave, you pass three runners each time you
) circle the track, yes? How many times are you yourself passed in
) that same interval? Three times. You are "passing and being
) passed in equal numbers," but not running at the median speed.
Yes, but that is the same runner that passed you three times,
whereas there are three different runners that you pass.
That's a different number of people.
See where the interpretation can differ?
Eric Sosman
Oct 7, 2012, 2:56:28 PM10/7/12
to
On 10/7/2012 2:18 PM, Willem wrote:
> Eric Sosman wrote:
>[...]
posting the problem itself, writing
"If someone passes you twice, you count him twice."
As I write, the problem is forty-two hours old, and the ambiguity
you mention was resolved forty-one hours ago. How many more times
will the runners circle the track before they get the news?
--
Eric Sosman
eso...@comcast-dot-net.invalid
> Eric Sosman wrote:
>[...]
> ) In the example I gave, you pass three runners each time you
> ) circle the track, yes? How many times are you yourself passed in
> ) that same interval? Three times. You are "passing and being
> ) passed in equal numbers," but not running at the median speed.
>
> Yes, but that is the same runner that passed you three times,
> whereas there are three different runners that you pass.
>
> That's a different number of people.
>
> See where the interpretation can differ?
The O.P. posted a clarification less than one hour after
> ) circle the track, yes? How many times are you yourself passed in
> ) that same interval? Three times. You are "passing and being
> ) passed in equal numbers," but not running at the median speed.
>
> Yes, but that is the same runner that passed you three times,
> whereas there are three different runners that you pass.
>
> That's a different number of people.
>
> See where the interpretation can differ?
posting the problem itself, writing
"If someone passes you twice, you count him twice."
you mention was resolved forty-one hours ago. How many more times
will the runners circle the track before they get the news?
--
Eric Sosman
eso...@comcast-dot-net.invalid
Tad Perry
Oct 7, 2012, 2:59:36 PM10/7/12
to
you more than once in a given hour? Wouldn't you, yourself, be traveling
faster than average in a case like that?
tvp
BillB
Oct 7, 2012, 3:07:30 PM10/7/12
to
Tad Perry
Oct 7, 2012, 3:07:58 PM10/7/12
to
generally clear up the ambiguity if you meant "unique individual runners"
and leave the ambiguity alone if you meant "any given runner." I'd have to
say that "any given runner" is the only reasonable interpretation, because
no one can reasonably be expected to know what the questioner meant and the
more open interpretation then comes into play.
tvp
Ted Schuerzinger
Oct 7, 2012, 7:06:04 PM10/7/12
to
On Sun, 7 Oct 2012 11:59:36 -0700, Tad Perry wrote:
> What about the fact it's a circular course and the very same person
> can pass you more than once in a given hour? Wouldn't you, yourself,
> be traveling faster than average in a case like that?
In theory, the loop could be small enough (or everybody runs fast
> What about the fact it's a circular course and the very same person
> can pass you more than once in a given hour? Wouldn't you, yourself,
> be traveling faster than average in a case like that?
enough) such that everybody can complete two loops in one hour.
Willem
Oct 8, 2012, 2:29:45 AM10/8/12
to
Eric Sosman wrote:
)> ) In the example I gave, you pass three runners each time you
)> ) circle the track, yes? How many times are you yourself passed in
)> ) that same interval? Three times. You are "passing and being
)> ) passed in equal numbers," but not running at the median speed.
)>
)> Yes, but that is the same runner that passed you three times,
)> whereas there are three different runners that you pass.
)>
)> That's a different number of people.
)>
)> See where the interpretation can differ?
)
) The O.P. posted a clarification less than one hour after
) posting the problem itself, writing
)
) "If someone passes you twice, you count him twice."
)
) As I write, the problem is forty-two hours old, and the ambiguity
) you mention was resolved forty-one hours ago. How many more times
) will the runners circle the track before they get the news?
I've no idea, I was just pointing out the most likely reason that the
person you were discussing with was coming up with a different answer.
) On 10/7/2012 2:18 PM, Willem wrote:
)> Eric Sosman wrote:
)>[...]
)> Eric Sosman wrote:
)> ) In the example I gave, you pass three runners each time you
)> ) circle the track, yes? How many times are you yourself passed in
)> ) that same interval? Three times. You are "passing and being
)> ) passed in equal numbers," but not running at the median speed.
)>
)> Yes, but that is the same runner that passed you three times,
)> whereas there are three different runners that you pass.
)>
)> That's a different number of people.
)>
)> See where the interpretation can differ?
)
) The O.P. posted a clarification less than one hour after
) posting the problem itself, writing
)
) "If someone passes you twice, you count him twice."
)
) As I write, the problem is forty-two hours old, and the ambiguity
) you mention was resolved forty-one hours ago. How many more times
) will the runners circle the track before they get the news?
I've no idea, I was just pointing out the most likely reason that the
person you were discussing with was coming up with a different answer.
Beldin the Sorcerer
Oct 8, 2012, 9:03:33 AM10/8/12
to
Tad Perry wrote:
>
> I agree they can be interpreted differently, but in an ambiguous case
> you'd generally clear up the ambiguity if you meant "unique
> individual runners" and leave the ambiguity alone if you meant "any
> given runner." I'd have to say that "any given runner" is the only
> reasonable interpretation, because no one can reasonably be expected
> to know what the questioner meant and the more open interpretation
> then comes into play.
Assuming someone posting to usenet will follow the style sheet's
>
> I agree they can be interpreted differently, but in an ambiguous case
> you'd generally clear up the ambiguity if you meant "unique
> individual runners" and leave the ambiguity alone if you meant "any
> given runner." I'd have to say that "any given runner" is the only
> reasonable interpretation, because no one can reasonably be expected
> to know what the questioner meant and the more open interpretation
> then comes into play.
recommendation is NOT a good bet.
Beldin the Sorcerer
Oct 8, 2012, 9:05:47 AM10/8/12
to
Multiple times or not, its one runner passing you in that given hour
Beldin the Sorcerer
Oct 8, 2012, 9:07:36 AM10/8/12
to
in my computer, I must state that it should have been cited.
Tad Perry
Oct 8, 2012, 8:10:53 PM10/8/12
to
definition of what was said (even though there are two possibilities) and
then use the definition that proves a point that depends on that definition.
Then there's your "commonly accepted definition" thing. Well, there's no
commonly accepted definition here. It's not even a common statement or
situation. There were avenues of stating this that could make it clearly one
or the other, and those avenues were not taken. There's no way to say that
one of the possible meanings is actually impossible.
Meanwhile, RichD hasn't bothered to answer my questions, so I'm not going to
bother trying to answer.
tvp
Beldin the Sorcerer
Oct 8, 2012, 11:57:49 PM10/8/12
to
All things come down to definitions. If people differ on the definition,
they aren't discussing the same problem
You only accept
> one definition of what was said (even though there are two
> possibilities) and then use the definition that proves a point that
> depends on that definition. Then there's your "commonly accepted
> definition" thing. Well, there's no commonly accepted definition
> here. It's not even a common statement or situation. There were
> avenues of stating this that could make it clearly one or the other,
> and those avenues were not taken. There's no way to say that one of
> the possible meanings is actually impossible.
I am happy to accept a clarification as to what the guy actually meant
>
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?
>
>> > 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.
>
> > 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.
> runner is constant?
> 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.
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
Tad Perry
Oct 9, 2012, 3:16:43 AM10/9/12
to
tvp
Patrick Powers
Oct 9, 2012, 3:35:13 AM10/9/12
to
On Oct 6, 8:56 am, RichD <r_delaney2...@yahoo.com> wrote:
> You've entered a road race. It's around a loop, a lakeside
> 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? Do the clues
> given provide sufficient information to solve the case,
> Watson?
>
> --
> Rich
Insufficient. I could be in first place, last place, or any other
place.
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?
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
> 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),
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.
>
> 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
mo_ntresor
Oct 10, 2012, 11:34:11 AM10/10/12
to
On Oct 7 2012 1:07 PM, BillB 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? Do the clues
> > given provide sufficient information to solve the case,
> > Watson?
>
> > 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? Do the clues
> > given provide sufficient information to solve the case,
> > Watson?
>
> I think this question was on my LSAT. The answer is (B).
you can't attend mcgill, but southwestern saskatchewan province has a
slot. i predict a life of alimony shakedowns and leftwing whining.
mo_ntresor
Phil Carmody
Oct 10, 2012, 4:12:26 PM10/10/12
to
m...@vex.net (Mark Brader) writes:
> Rich Delaney:
> Schuerzinger's response. Marathons aren't customarily run in
> laps around a closed track.
Marathons being two circuits of a half-marathon route is
certainly quite common. I believe that every town where
I've lived, and there's been a marathon, has used that
configuration. (Though not every metropolis where I've
lived.) People certainly do get lapped in those races.
Phil
--
Regarding TSA regulations:
How are four small bottles of liquid different from one large bottle?
Because four bottles can hold the components of a binary liquid explosive,
whereas one big bottle can't. -- camperdave responding to MacAndrew on /.
> Rich Delaney:
> > You've entered a road race. It's around a loop, a lakeside
> > race, but very long, like a marathon.
>
> I assumed this meant *once* around the loop until I read Ted
> > race, but very long, like a marathon.
>
> Schuerzinger's response. Marathons aren't customarily run in
> laps around a closed track.
Marathons being two circuits of a half-marathon route is
certainly quite common. I believe that every town where
I've lived, and there's been a marathon, has used that
configuration. (Though not every metropolis where I've
lived.) People certainly do get lapped in those races.
Phil
--
Regarding TSA regulations:
How are four small bottles of liquid different from one large bottle?
Because four bottles can hold the components of a binary liquid explosive,
whereas one big bottle can't. -- camperdave responding to MacAndrew on /.
Phil Carmody
Oct 10, 2012, 4:20:18 PM10/10/12
to
Willem <wil...@turtle.stack.nl> writes:
> RichD 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? Do the clues
> ) given provide sufficient information to solve the case,
> ) Watson?
number you pass". This can only sensibly be interpreted without
changing its literal meaning as "the number of runners who pass you"
and "the number of runners you pass". The same person overtaking you
twice, which you would tally as 2, is only one person passing you, and
so I think by the literal wording of the question should tally as just
1. And therefore I think you're going to be Median Mel, as the number
you are faster than and lap must equal the number you are slower than
and are lapped by.
Yes, I'm guess this wasn't what he intended to ask, but this is
rec.puzzles, dammit!
> RichD 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? Do the clues
> ) given provide sufficient information to solve the case,
> ) Watson?
>
> 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), and also
> Assuming that this is hell, where you have to run endlessly (or at least
> 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.
I disagree. We were asked for "the number who pass you" and "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.
number you pass". This can only sensibly be interpreted without
changing its literal meaning as "the number of runners who pass you"
and "the number of runners you pass". The same person overtaking you
twice, which you would tally as 2, is only one person passing you, and
so I think by the literal wording of the question should tally as just
1. And therefore I think you're going to be Median Mel, as the number
you are faster than and lap must equal the number you are slower than
and are lapped by.
Yes, I'm guess this wasn't what he intended to ask, but this is
rec.puzzles, dammit!
BillB
Oct 11, 2012, 3:14:49 PM10/11/12
to
On 10/10/2012 8:34 AM, mo_ntresor wrote:
>> I think this question was on my LSAT. The answer is (B).
>
> you can't attend mcgill, but southwestern saskatchewan province has a
> slot. i predict a life of alimony shakedowns and leftwing whining.
My combined GPA and LSAT score would have put me very near the top of
>> I think this question was on my LSAT. The answer is (B).
>
> you can't attend mcgill, but southwestern saskatchewan province has a
> slot. i predict a life of alimony shakedowns and leftwing whining.
the application pile at any law school in North America, bar none. The
only constraining factor for me was cost.
According to the Princeton Review (whose books I used to "crack the
LSAT"), students admitted to study law at McGill have an average LSAT of
162 and an average GPA of 3.7. Where I went (U of M), the average LSAT
is 160 and the average GPA is 3.9.
There is no law school in "southwestern saskatchewan." There isn't even
a major city in southwestern Saskatchewan. The only law school in
Saskatchewan is in northern Saskatchewan at the University of
Saskatchewan, which has similarly onerous entrance requirements (albeit
slightly relaxed) to the two I previously mentioned.
As usual, you simply have no idea what you are talking about, and are
not the least bit ashamed to broadcast that fact.
mo_ntresor
Oct 11, 2012, 3:28:19 PM10/11/12
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want to whine about the way your life turned out, blame yourself then him.
mo_ntresor
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