What's a boy to do?

On Thu, 2 Nov 2006 10:38:41 -0600, Kevin Vang wrote:

In article ,
says...
According to your theory, that dart can easily
and readily strike any point on the disk or any point outside of the
circumference created by the selection of the first and second points,
up to and including "un-measurably" close to the inside or the outside
of the circumference, but can never actually strike a point on the
circumference. IOW, the third point (Dart C) can only create a second
radius that must be less than or greater than the first radius. With
not being able to select a second point on the circumference, arcs, in
such a world, don't exist. If arcs don't exist, geometry, trig, etc.
begins to break down. In the failure cascade of interrelated bits , it
takes all math down with it.



It's not that the arc doesn't exist, and we cannot choose points on
that arc. The point is that the probability of hitting that arc with
a dart is 0.

Intuitive explanation: Suppose your dartboard has radius 1. Throw a
dart at the dartboard, and let r1 = radius from the center of the
dartboard to the dart. Now throw a second dart, and let r be the
radius. Then the probability that r = r1 is

number of values of r for which r = r1 1
------------------------------------------- = ------------ = 0.
number of possible values for r infinity


More technical (and more correct) explanation: If we assume that every
point on the dartboard is equally likely to be hit, then the probability
that r = r1 is:

measure of the set for which r = r1 0
-------------------------------------- = ------------ = 0
measure of the dartboard pi * 1^2

because the dartboard is a 2-dimensional surface, the appropriate
measure is area. The measure of the entire dartboard is the area of
a circle with radius 1, so the area is pi*1^2 = 1. The set of points
for which r = r1 is the circle with radius r1. Since the circle is
just a curve with width 0 on the plane, it has area 0.

Slightly more technical (and more correct): Not every point on the
dartboard is equally likely to be hit.

Apparently. The word on the street is that at least some are completely
unhittable, what with the probability of doing so being zero or
infinitely small or pi-r-square or, well, something all dangerously full
of symbols and greek letters and ****...

If p(r,theta) is the
probablility density function giving the probability that the dart hits
point (r,theta) in polar coordinates, then the probability that r = r1
is:

/ r1
| p(r,theta) dA
/ r1 0
------------------------- = --- = 0
/ 1 1
| p(r,theta) dA
/ 0

because we are integrating with respect to area, and the top integral
is done over a region with area 0, so the value of the integral is 0.

SEE! SEE! I WARNED YA, BUT NOO-O-O-O-O...

IAC, three answers, each "and more correct" than the previous one.
Interesting. Is this progression going to lead to something infinitely
correct (or something to at least stick a fork in and call "done"), or
is the probability of hitting that target zero, too?


HTH,
Kevin
And I'm pretty certain that mathematics doesn't all disappear if
somebody doesn't understand one bit of it.

Hey, go easy on me, I'm learning...for example, I've already learned
that when 2 math whiz-types and a rat-gutter answer a question, the odds
that they will come up with the correct answer is like one in a
gazillion or bazillion or some other REALLY big ol' number...

And right back at ya, Pythagoras
R

wrote in message
...
On Thu, 2 Nov 2006 10:38:41 -0600, Kevin Vang wrote:

In article ,
says...
According to your theory, that dart can easily
and readily strike any point on the disk or any point outside of the
circumference created by the selection of the first and second points,
up to and including "un-measurably" close to the inside or the outside
of the circumference, but can never actually strike a point on the
circumference. IOW, the third point (Dart C) can only create a second
radius that must be less than or greater than the first radius. With
not being able to select a second point on the circumference, arcs, in
such a world, don't exist. If arcs don't exist, geometry, trig, etc.
begins to break down. In the failure cascade of interrelated bits , it
takes all math down with it.



It's not that the arc doesn't exist, and we cannot choose points on
that arc. The point is that the probability of hitting that arc with
a dart is 0.

Intuitive explanation: Suppose your dartboard has radius 1. Throw a
dart at the dartboard, and let r1 = radius from the center of the
dartboard to the dart. Now throw a second dart, and let r be the
radius. Then the probability that r = r1 is

number of values of r for which r = r1 1
------------------------------------------- = ------------ = 0.
number of possible values for r infinity


More technical (and more correct) explanation: If we assume that every
point on the dartboard is equally likely to be hit, then the probability
that r = r1 is:

measure of the set for which r = r1 0
-------------------------------------- = ------------ = 0
measure of the dartboard pi * 1^2

because the dartboard is a 2-dimensional surface, the appropriate
measure is area. The measure of the entire dartboard is the area of
a circle with radius 1, so the area is pi*1^2 = 1. The set of points
for which r = r1 is the circle with radius r1. Since the circle is
just a curve with width 0 on the plane, it has area 0.

Slightly more technical (and more correct): Not every point on the
dartboard is equally likely to be hit.

Apparently. The word on the street is that at least some are completely
unhittable, what with the probability of doing so being zero or
infinitely small or pi-r-square or, well, something all dangerously full
of symbols and greek letters and ****...

If p(r,theta) is the
probablility density function giving the probability that the dart hits
point (r,theta) in polar coordinates, then the probability that r = r1
is:

/ r1
| p(r,theta) dA
/ r1 0
------------------------- = --- = 0
/ 1 1
| p(r,theta) dA
/ 0

because we are integrating with respect to area, and the top integral
is done over a region with area 0, so the value of the integral is 0.

SEE! SEE! I WARNED YA, BUT NOO-O-O-O-O...

IAC, three answers, each "and more correct" than the previous one.
Interesting. Is this progression going to lead to something infinitely
correct (or something to at least stick a fork in and call "done"), or
is the probability of hitting that target zero, too?


HTH,
Kevin
And I'm pretty certain that mathematics doesn't all disappear if
somebody doesn't understand one bit of it.

Hey, go easy on me, I'm learning...for example, I've already learned
that when 2 math whiz-types and a rat-gutter answer a question, the odds
that they will come up with the correct answer is like one in a
gazillion or bazillion or some other REALLY big ol' number...

And right back at ya, Pythagoras

Hee, hee, hee.

Wolfgang
is it just me or has anyone else noticed that the probability of surrenders
in general (and dicklet's in particular) being gracious tends to decrease
over time?