What's a boy to do?

On Thu, 02 Nov 2006 13:37:57 GMT, "Joe McIntosh"
wrote:


"Charlie Choc" wrote in message -- joe
responds

glad to see you returned --don"t we get a trip report about your western
summer? even if no fishing is included--most of the messages here do not.

I took a lot of pictures and helped build some houses, went to a pow-wow and
rodeo, and generally had a ball, although 8 weeks is plenty long enough to sleep
in what amounts to a tent with wheels.

As for the fishing part, I floated the Bighorn and Yellowstone with 'Bouncer'
and we had a fine time and caught some nice fish. I also fished in Yellowstone,
but water was low and the streams were crowded - especially the Firehole and in
Lamar Valley. The park seemed more crowded the 1st 2 weeks of September than it
has been in July, and that combined with the rutting elk and bison make for some
dangerous encounters. Several people were hospitalized while I was there, mostly
by virtue of getting too close to bull elk. My best 'catching' was probably in
Grand Teton and just outside it on the Gros Ventre and the Snake. Not as many
fishermen down there but the place was thick with photographers, and you had to
be careful and stay away from the rutting moose by the rivers.
--
Charlie...
http://www.chocphoto.com

"Scott Seidman" wrote in message
. 1.4...
"riverman" wrote in :

Thus, all the possible positions of A and C equal all the possible
positions of C and A...


But the question as you pose it has nothing to do with all the possible
positions of dart A-- it has to do with one specific position of dart A!
Let's say that A is 5cm away. Then you are looking for p(C5), which has
a value that depends only on the distribution of dart C.

Whatever high probability exists if A is close is
countered by the low probability if A is far.

True before dart A is thrown, but not after dart A is thrown. Now, you
have a real honest to goodness value for dart A.


Yes, but the possbilities for the value for A are limitless. You cannot
calculate all the different arrangements. Remember, the original question
did not specify where A landed, only that it did.


Thus, all the possible positions of A and C equal all the possible
positions
of C and A....its a combinatoric problem. Specifically because I
DON'T give the position of dart A.

This doesn't mean that you can just ignore the fact that Dart A is stuck
at a precise location in the dartboard, and it's why there isn't enough
info to offer a p-value.


No its not. Its stuck in any of an infinite number of locations. Each one
has a different effect on the probability of B being closer.


If its any comfort, I'm not making this problem up.

Then the person who did got it wrong.


Well, here's the results of one person who ran 1 million trials each, using
uniform, normal, gaussian and random distributions.
http://www.gatago.com/rec/puzzles/25594126.html

He came up with 2/3 every time. I just have an aversion to Monte Carlo
methods, but go ahead and try it yourself and let us know what the result
is.

--riverman

"Stan Gula" wrote in message
news:Fik2h.10075$gf5.7278@trndny01...
...In practice, I would prefer to find an approximation to the solution
through Monte Carlo simulation (with real darts, not a computer model),
accompanied by large quantities of fermented malt. In which case I'd
hazard (hah!) a guess that the probability approaches 1 as the amount of
malt consumed approaches unconsciousness.

All well and good as a purely intellectual exercise but in real world
situations one cannot afford a cavalier disregard (such as you have
demonstrated here) for the effects of important variables like (in this
instance) time. If the probability's approach to 1 is proportional to the
approach of looming unconsciousness (a proposition I am not prepared to
contest) it may nevertheless prove to be chimerical, depending, obviously,
on whether or not (and, if so, when) the blessed union with the cosmic
consciousness occurs. This, in turn, is a function of rate of consumption
over time.*

Wolfgang
*precise analyses do, of course, also depend on mitigating factors like body
mass, whether the malt is consumed in a raw or distilled form, etc., but
these details need not concern us here as they can (and, indeed, must) be
determined empirically in each experimental run.

On 1 Nov 2006 16:27:51 -0800, "riverman" wrote:


Calif Bill wrote:


You could have an almost infinite amount of darts the exact same distance
from the center. The only limiting number is how big the circle is from the
center and how big of diameter is the dart. There are an infinite number of
points equidistant from the center point. And it depends on neither the
precision or accuracy of the measurement. And in your measurement of the
distance it would be more accuracy and not precision. Precision only gives
more numbers after the decimal point.

Bill:
I'm not sure where to start, but there are a lot of little details in
your assertations that are erroneous. There's some truth also, so don't
lose hope :-)

First of all, yes the definition of a circle states there are an
infinite number of points in a plane that are equidistant from a given
point, but the liklihood of getting even two darts to land on that
circle is slim. (Just how slim is discussed in the second paragraph
below.) We don't even have to agree on how slim for now, but the the
more darts you want to have land on that circle, the less likely it is
to happen, and it approaches zero as the number of darts gets larger
and larger. Although the phrase 'almost infinite' is actually
meaningless, I assume you mean we are looking at numbers that are
growing huge beyond comprehension, so the liklihood of it happening is
shrinking tiny beyond comprehension.

Secondly, it IS a matter of precision, not accuracy. We don't care what
the actual distance from the center is, what we do care about is
whether or not two darts have the same measurement from the center,
even if that measurement is wrong. If we use an inaccurate tool, then
we might get a wrong amount (a broken ruler might show each dart to be
10.55 cm from the center, while they are both actually much less that
that). That's 'inaccurate', but if the numbers match, then we can still
assert that they are the same distance. If we use a ruler with really
fat indicator lines, we might get both measuring 10.55 cm, however if
we used a vernier caliper, calibrated or not, we might get one of them
measuring 10.550000000000001 cm and the other measuring
10.550000000000002 cm. Those are measures of high PRECISION, and my
assertation is that, no matter how the darts land, we can always use
more precise measuring devices until we find where the numbers vary.
And they always will, even if we have to go to electron microscope
levels. Just as no two snowflakes are alike, no two darts can land the
same distance from the center.

Now, I appreciate that some people might have an ingrained prejudice
against math because it doesn't always conform to their intuition (and
this might be you, or it might not). But when faced with something that
doesn't seem to 'fit' what we want to believe, there are two choices:
find out the rules of math and learn to analyze things according to
those rules, including learning the constraints and limitations and the
meaning of those, or else continue to assert that what we believe is
right because it 'feels right' to us, and use poorly structured
arguments or misnomers to claim that nothing has any validity, so we
can't possibly be wrong. That way lies madness.

--riverman

Whether or not _you_ can measure to _your_ satisfaction that two points
on a 2-dimensional plane are _absolutely_ the same distance from an
initially-chosen point (in this case, a "target"_), those two points
certainly exist. The random selection of a second point (the landing of
Dart A) "x" distance from the first point (the "target") creates a
radius from which a circumference may be scribed. The second dart (Dart
B) and its landing point have no relevance and can be ignored. A third
dart is thrown (Dart C). 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. Congratulations, you've talked your way
out of a fairly decent, secure job...yep, you're a Democrat...ah, well,
perhaps there's a job on Kerry's staff for ya...

On the practical side, it seems rather curious that you can measure to
your own satisfaction that it isn't the same distance, yet you cannot
measure to your satisfaction that it is the same distance.

HTH,
R

wrote in message
...


Whether or not _you_ can measure to _your_ satisfaction that two points
on a 2-dimensional plane are _absolutely_ the same distance from an
initially-chosen point (in this case, a "target"_), those two points
certainly exist. The random selection of a second point (the landing of
Dart A) "x" distance from the first point (the "target") creates a
radius from which a circumference may be scribed. The second dart (Dart
B) and its landing point have no relevance and can be ignored. A third
dart is thrown (Dart C). 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. Congratulations, you've talked your way
out of a fairly decent, secure job...yep, you're a Democrat...ah, well,
perhaps there's a job on Kerry's staff for ya...

On the practical side, it seems rather curious that you can measure to
your own satisfaction that it isn't the same distance, yet you cannot
measure to your satisfaction that it is the same distance.

Thus forcing one to conclude that forty bucks worth does not constitute an
"expensive education" so much as it does outright robbery.

Wolfgang
on the other hand, given what we get here for free, forty bucks worth of
entertainment would almost certainly kill any mere mortal.

On 1 Nov 2006 16:46:23 -0800, "Wolfgang" wrote:

SNI-I-I-I-IP

I will simply confine myself

Well, no, you didn't do either, but perhaps you should...

to making a proposition open to
anyone. Give me three darts and a prediction of where they will land
relative to one another in terms of distance from the center of the
target, and I will prove you wrong EVERY time.

Gee, it seems like this might be an attempt at a sucker bet...OK. I
accept. And I'd offer that you couldn't even do it ONE time... and that
you couldn't do it even if given a 3-dimensional "dartboard"...but don't
pee all over yourself, here's another hint: the taxpayers of Olathe,
Kansas are probably very glad you can't do it even that one time...why,
heck, one might say that's the essence of an industry...

HTH,
R
....I feel generous, here's another hint: ya better go back to sucker-bet
development school - with the "bet" above, it doesn't matter how, when,
or if you throw them...

Wolfgang

wrote in message
...
On 1 Nov 2006 16:27:51 -0800, "riverman" wrote:

Whether or not _you_ can measure to _your_ satisfaction that two points
on a 2-dimensional plane are _absolutely_ the same distance from an
initially-chosen point (in this case, a "target"_), those two points
certainly exist.

Yes, they do.

The random selection of a second point (the landing of
Dart A) "x" distance from the first point (the "target") creates a
radius from which a circumference may be scribed. The second dart (Dart
B) and its landing point have no relevance and can be ignored.

Not necessarily, it depends on what is being asked. "Conditional
probabilities" do exist. But in the case of what you are discussing (the
existance of arcs), I concur; we can ignore the second dart for now.

A third
dart is thrown (Dart C). 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.

Yes, that's correct also. There is a statement in calculus that asserts that
no matter what two numbers you choose on the number line, there is always
another number between them. No matter how close to the circumference you
get, you can always get closer. But you cannot get there unless you, well,
get there.

With
not being able to select a second point on the circumference, arcs, in
such a world, don't exist.

No one said you cannot select a second point. What is being said is that the
probability of another dart hitting that point, or any other point on that
circle, is zero. Thats because the point is infinitely small. The
probability of hitting something infinitely small is infinitely
small....zero, in fact.

If arcs don't exist,

....and everything after this antecedant is YOUR proposition, since I know
they do exist.

geometry, trig, etc.
begins to break down. In the failure cascade of interrelated bits , it
takes all math down with it. Congratulations, you've talked your way
out of a fairly decent, secure job...yep, you're a Democrat...ah, well,
perhaps there's a job on Kerry's staff for ya...


Not me. Say hello to John for us, willya? :-)

The way you are posing your interpretation of this is risky. I ask you; How
big is a point? If you answer 'infinitely small', then I ask; 'How is it
possible to construct anything out of points, then?" Certainly the
one-dimensionality of points and the existance of 2- and 3-dimensional
objects constructed of points are not exclusive. Math exists, life goes on,
and you get to remain a Republican.

--riverman

wrote in message
...
On 1 Nov 2006 16:46:23 -0800, "Wolfgang" wrote:

SNI-I-I-I-IP

I will simply confine myself

Well, no, you didn't do either, but perhaps you should...

to making a proposition open to
anyone. Give me three darts and a prediction of where they will land
relative to one another in terms of distance from the center of the
target, and I will prove you wrong EVERY time.

Gee, it seems like this might be an attempt at a sucker bet...OK. I
accept. And I'd offer that you couldn't even do it ONE time... and that
you couldn't do it even if given a 3-dimensional "dartboard"...but don't
pee all over yourself, here's another hint: the taxpayers of Olathe,
Kansas are probably very glad you can't do it even that one time...why,
heck, one might say that's the essence of an industry...

HTH,
R
...I feel generous, here's another hint: ya better go back to sucker-bet
development school - with the "bet" above, it doesn't matter how, when,
or if you throw them...

The beauty of saying nothing is that you can never be proved wrong and that
you never have to retract a statement, ainna? One can only suppose that
someone suggested this strategy to you and that you stick to it without a
hint as to its efficacy out of sheer dogged inability to think of anything
else to do. Well, that and the fact that so many play so gently with you.


Wolfgang
who, it must be admitted, has always been a bit rougher with his toys than
the other kids.

Scott Seidman wrote:
...
You could ask your question in a different way, to get the answer you
want, which is "you are going to throw three darts at a target. What is
the probability that the third dart will miss by more than the first
dart?" This is a VERY different question, but the answer is the one you
are describing.

Exactly correct. Good luck trying to convince the roffian gaggle. ;-)

--
Ken Fortenberry

"Jonathan Cook" wrote in message
...
Scott Seidman wrote:

You could ask your question in a different way, to get the answer you
want, which is "you are going to throw three darts at a target. What is
the probability that the third dart will miss by more than the first
dart?" This is a VERY different question, but the answer is the one you
are describing.

Yep.

I would agree that the probability above is 1/3 if you added the conditional
'given that the second dart missed by more than the first'. Otherwise, I'm
inclined (but not convinced) that the probability above is 1/2.

Since the outcome of the second dart is not taken into account, your
question is identical to 'you throw a dart, watch some TV and eat lunch,
then throw another. Whats the probability that the second is closer than the
first?' Where I am unsure is because of the existance of the middle dart...I
have to think that through more.

But I am certain that if you add that conditional, then you have my
question, with a probability of 1/3.

--riverman