What's a boy to do?

"Kevin Vang" wrote:
Two darts could conceivable land equidistant from the center; however,
the probability of that happening is 0. Explaining why will require a
bit of less than elementary probability theory, with integral calculus
as a prerequisite. We can go there, if you are up to it...

Kevin

I was wondering when you would pop up. We could let it go with an "It's
intuitively obvious and not important in solving the problem at hand". In
practice, we're talking about physical objects which are not made up of
infinitesimal particles, but a finite set of rather large (relatively)
molecules of stuff, so the set of positions where a dart could penetrate is
a very small subset (albeit a very large set) of the set of points on a
circle through the location of any dart. 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.

--
Stan
from the applied side.

"riverman" wrote in news:1162440298.459352.48600
@e3g2000cwe.googlegroups.com:

The combinatoric arrangement merely gives all
the arrangements.


Yes, and has next to nothing to do with the probability of any outcomes.

In five years, there are two possible outcomes to the experiment "Who has
more money, me, or Donald Trump". I could have more money, or he can have
more money. By your logic, I therefore have a 50% chance of having more
money than The Donald. Good to know.

--
Scott
Reverse name to reply

"Stan Gula" wrote in message
news:Fik2h.10075$gf5.7278@trndny01...
"Kevin Vang" wrote:
Two darts could conceivable land equidistant from the center; however,
the probability of that happening is 0. Explaining why will require a
bit of less than elementary probability theory, with integral calculus
as a prerequisite. We can go there, if you are up to it...

Kevin

I was wondering when you would pop up. We could let it go with an "It's
intuitively obvious and not important in solving the problem at hand". In
practice, we're talking about physical objects which are not made up of
infinitesimal particles, but a finite set of rather large (relatively)
molecules of stuff, so the set of positions where a dart could penetrate
is a very small subset (albeit a very large set) of the set of points on a
circle through the location of any dart. 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.

--
Stan
from the applied side.

I concur. But I was also about to propose to Jon (who IS a computer geek) to
write a computer simulation and run it a few hundred million times.

--riverman

On 2 Nov 2006 12:37:19 GMT, Scott Seidman wrote:

"riverman" wrote in news:1162440298.459352.48600
:

The combinatoric arrangement merely gives all
the arrangements.


Yes, and has next to nothing to do with the probability of any outcomes.

In five years, there are two possible outcomes to the experiment "Who has
more money, me, or Donald Trump". I could have more money, or he can have
more money. By your logic, I therefore have a 50% chance of having more
money than The Donald. Good to know.

You might as well run for president, too. You have a 50% chance of being
elected.
--
Charlie...
http://www.chocphoto.com

"Scott Seidman" wrote in message
. 1.4...
"riverman" wrote in news:1162440298.459352.48600
@e3g2000cwe.googlegroups.com:

The combinatoric arrangement merely gives all
the arrangements.


Yes, and has next to nothing to do with the probability of any outcomes.

In five years, there are two possible outcomes to the experiment "Who has
more money, me, or Donald Trump". I could have more money, or he can have
more money. By your logic, I therefore have a 50% chance of having more
money than The Donald. Good to know.


That's a seriously lame strawman argument.

Speaking of straws....put three straws in your hand...one short, one medium,
one long. What's the probability of drawing the shortest one last?

And what's the probability of drawing the longest one first, and the
shortest one after that?

--riverman

"riverman" wrote in :

And what's the probability of drawing the longest one first, and the
shortest one after that?

--riverman



Yes, that problem is a matter of combinatorials, but that isn't the
problem you posed.

You toss three darts at a target. Dart A misses the target, then Dart
B
misses by even more. What is the probability that Dart C will miss by
more than Dart A?

I just need to know the distribution of dart C, and the location of Dart
A. I don't even care if darts A,B, and C are independent. It is not a
combinatorial problem. If dart A is 1mm from the target, the probability
is very good that dart C will miss by more. If dart A is a mile from the
target, the probability is very poor.

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.


--
Scott
Reverse name to reply

Scott Seidman wrote in
. 1.4:

If dart A is 1mm from the target, the probability
is very good that dart C will miss by more. If dart A is a mile from
the target, the probability is very poor.


To express this better, the location of the dart is a continuous random
variable, but you aren't treating it that way.


--
Scott
Reverse name to reply

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

And what's the probability of drawing the longest one first, and the
shortest one after that?

--riverman



Yes, that problem is a matter of combinatorials, but that isn't the
problem you posed.

You toss three darts at a target. Dart A misses the target, then Dart
B
misses by even more. What is the probability that Dart C will miss by
more than Dart A?

I just need to know the distribution of dart C, and the location of Dart
A. I don't even care if darts A,B, and C are independent. It is not a
combinatorial problem. If dart A is 1mm from the target, the probability
is very good that dart C will miss by more. If dart A is a mile from the
target, the probability is very poor.


You're getting close. Considering that dart A and dart C have the same
distribution, then either dart can occupy whatever spot the other dart
occupies (minus the situation where they occupy the same point). As you
said, the probability changes according to the location of dart A. Whatever
high probability exists if A is close is countered by the low probability if
A is far.

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.

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

--riverman

"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.

spent a couple of day in Graham co last week with Jeff----then Dene and I
spent a week bumming around snowbird area--one big rain so tried to improve
my nymphing ability with slight success, but did dig out one big rainbow in
big pool just upstream from snowbird cabin. They call them steelhead up that
way as they come into streams from lakes to lay eggs.

Went to eye doctor yesterday and after he said I was still 20/20 I
explained that tennis balls I once returned now go flying by, and Jeff had
to tie on small flies for me. Dr said perhaps I need to clean my glasses and
gave me a bill for $121

Kayaking in backwater salt most every day--not catching much but just
cannot say inside and do homework when Carolina in Oct is smiling

Indian Joe

"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.


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.

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

Then the person who did got it wrong.

--
Scott
Reverse name to reply