What's a boy to do?
 

Jonathan Cook wrote:
Jonathan Cook wrote:

"you threw" versus "you throw"...

I suppose I should have wrote "you threw" versus "you are
going to throw".

Actually, it doesn't matter.

Are we in agreement that the statement and my statement are supposed to
be identical except for tense? If so, then its definitely not relevant.

The probability of some outcome does not change before or after the
event occurs. Stating 'you are going to throw three darts' presupposes
a range of possible outcomes. Stating 'you threw three darts' without
specifying what any of the specific results were does not eliminate
any of the presupposed outcomes. Stating 'you have thrown two darts,
and now you are about to throw the third' likewise, in the absence of
giving any specifics about the results of the previous two tosses, does
not eliminate any of the presupposed outcomes.

Scott is creating his own unsolvable problem by asserting that there is
an honest-to-goodness value for the position of dart A, and then
asserting that he has to know what it is to solve the problem.

You have to consider ALL the continuous values for dart A, which happen
to be all the values of dart B (minus the singularity where they occupy
the same point). As such, this is a completely symmetrical problem, and
all results are equally likely over the array of possible outcomes.

There is a similar problem using discrete variables rather than
continuous. A man says he has two kids, and mentions that one is a
girl. What is the probability that the other is also a girl? The answer
is NOT 50%, since some information has been disclosed. But if he says
"I have two kids, and here is one of them", the the probability that
the other is a girl IS 50%. All events are symmetric.

These problems; the Monty Hall one, the three darts one, the two kids
ones, and several more like them, are well-understood and often used to
illustrate what I mentioned early on: that its easy (especially when
dealing with infinities) to create indeterminate forms, or when dealing
with discrete outcomes, to try to solve problems using methods that
lead to dead ends. However, there are often methods that reduce the
problems to much more solvable forms, without prejudice. In fact, most
of probability is just that; finding a simpler, accurate way to model a
problem.

The key to solving the three darts problem is to see that each event is
independant and has the same distribution (regardless of what the
distribution is). As such, the darts are interchangable without
prejudice, and all those continuous random variables and infinite
possible arrangements simplify to a counting problem. Tense has nothing
to do with it, because absolutely no infomation is disclosed if a dart
or two or three have already been thrown. If information IS disclosed,
then you can no longer assume that three events are symmetrical, as the
P(dartB = dartA) becomes zero.

Try modeling it with a Monte Carlo program. You'll get 2/3.

--riverman

What's a boy to do?
 

"Kevin Vang" wrote:
snip
I don't have any nekkid cheerleaders...

Life isn't perfectg

, but I stand at one wall
and tell the class I plan to walk half of the distance to the other
wall, then walk half of the remaining difference, then walk half the
remaining difference, and so on. I ask "Will I make it to the other
wall?" I ask for a show of hands, and most students raise their hands
for "No". Then I say, "Watch me," and walk directly across the room
until I bump into the far wall. Lively discussion then ensues...

See, the problem of modeling a theoretical problem with real world finite
sized objects. When I was in school the last time, they had just built a 30
story library tower with a roughly square cross section and notched corners.
You could walk into one of these notches and have walls of brick about 10
feet wide and 30 stories tall in front of you. I had a little demonstration
for friends inspired by Zeno's Paradox (and as you might expect, this was
inspired by, and usually demonstrated after considerable beer had been
disappeared). I would start them off about 100 yards away from the corner
and have them walk toward the notch, fixing their vision at the top of the
building. The 'theory' was that the apparent height of the building would
increase as they got closer and the rate of the increase would appear to
accelerate as they got closer. If you could get right into the corner you
would see the effect of the building 'growing' very fast. It was hilarious
that almost everybody would walk smack into the brick wall, transfixed by
the illusion. Hilarious to me, that is. Unfortunately, being a State
building, the bricks started falling off the facade, and they closed access
to the building.

What's a boy to do?
 

"riverman" wrote in message
There is a similar problem using discrete variables rather than
continuous. A man says he has two kids, and mentions that one is a
girl. What is the probability that the other is also a girl? The answer
is NOT 50%, since some information has been disclosed. But if he says
"I have two kids, and here is one of them", the the probability that
the other is a girl IS 50%. All events are symmetric.

I'm having a problem with the wording of the questions, I suppose. In your
example above, I believe that once the first girl is revealed, the question
more accurately becomes, "What is the probability that both children are
girls?" I agree that's not .50.

I'll toss out the Rosencrantz and Guilderstern example. If I toss a coin,
the probability of heads is .50. If I toss it again, what is the
probability of heads? It's .50. How is that different from the man with
two children?

Joe F.

What's a boy to do?
 

On 2 Nov 2006 15:13:08 -0800, "Wolfgang" wrote:


wrote:
On Thu, 2 Nov 2006 14:26:25 -0600, "Wolfgang" wrote:


wrote in message
.. .
On Thu, 2 Nov 2006 13:00:17 -0600, "Wolfgang" wrote:


Again, I accept your proposal...wanna bet on the outcome?

Sure. You bring the dartboard......and the absinthe. I got Oprah and
Emeril on DVD.

Hang on a minute....first things first. How much would you like to
lose? 2500USD? 1000EURO? 1000GBP?

Sure. What time shall I expect you?

Yeah, that's what I thought...you puss out before you're forced to welsh
on the bet...run back under the porch, lil' pup...

O.k., let me see if I've got this straight. You want people to believe
that you were seriously offering to make a bet about something or other
that is a complete mystery to you and that I had taken you seriously
and accepted the bet and that somehow or other somebody was going to do
something or other to settle the bet in front of reliable witnesses to
everyone's satisfaction and that the results would then be duly
published here and that I would be appropriately contrite and confess
that I really DID know all along that you are
God......and.....something or other more. That about it?

Nope. While your claimed hesitance to take advantage of someone who
didn't understand what they were doing would be oh-so-noble, it's BS. If
one wagers, one risks losing. All you needed to do is put your money
where your mouth is. What other people believe or don't believe is up
to them. You made a proposition (which I note you snipped from above):

"I will simply confine myself 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."

and I twice made it clear that I accepted it I didn't qualify or
otherwise modify it. I then offered to wager and gave an idea of the
stakes in which I was willing to risk. If those stakes were too much for
you, you could have offered lesser with no shame. If you swung as big a
dick as you'd like to think, you would have simply accepted or declined
like a man, put the ball in my court, and have done as I did - not worry
about what "people" might or might not believe. Instead, you pussed. In
fact, I'd offer that I was wagering on that, too...

Like I said - run back under the porch, lil' pup,
R

What's a boy to do?
 

"rb608" wrote in message
news:WWF2h.7479$OK3.540@trndny09...
"riverman" wrote in message
There is a similar problem using discrete variables rather than
continuous. A man says he has two kids, and mentions that one is a
girl. What is the probability that the other is also a girl? The answer
is NOT 50%, since some information has been disclosed. But if he says
"I have two kids, and here is one of them", the the probability that
the other is a girl IS 50%. All events are symmetric.

I'm having a problem with the wording of the questions, I suppose. In
your example above, I believe that once the first girl is revealed, the
question more accurately becomes, "What is the probability that both
children are girls?" I agree that's not .50.

I'll toss out the Rosencrantz and Guilderstern example. If I toss a coin,
the probability of heads is .50. If I toss it again, what is the
probability of heads? It's .50. How is that different from the man with
two children?

The main difference is that one is a conditional probability, the other is
not.

The possible sample space for someone having two children is BB BG GB GG.
Since man gave us the condition that 'at least one is a girl', you know that
BB is not a possibility. Thus, the remaining possible sample space is BG, GB
or GG (these all meet his condition), and the only sucessful outcome is GG,
so the probability that "the other child is also a girl, given that at least
one is a girl" is 1/3.

The Rosencrantz/Guilderstern example differs. If you ask "what's the
probability that the second coin toss is H" the answer is .50, regardless of
what took place before---its not conditional. If you ask "whats the
probability of getting two heads, given that the FIRST toss is heads", then
we have to limit our possible outcomes to HT or HH (these are the only
outcomes that meet the condition), and the only successful outcome is HH,
with a probability of 1/2. That would be similar to the man identifying the
girl as his eldest, or his youngest. That serves to remove one of the BG/GB
options.

But if you ask "whats the probability of tossing a coin twice and getting
two Heads, given that you get at least one Heads", then the two problems are
identical. The possible outcomes are HH, HT and TH, and the successful
outcome is HH.

--riverman

What's a boy to do?
 

wrote in message
...
On 2 Nov 2006 15:13:08 -0800, "Wolfgang" wrote:


wrote:
On Thu, 2 Nov 2006 14:26:25 -0600, "Wolfgang" wrote:


wrote in message
.. .
On Thu, 2 Nov 2006 13:00:17 -0600, "Wolfgang"
wrote:


Again, I accept your proposal...wanna bet on the outcome?

Sure. You bring the dartboard......and the absinthe. I got Oprah
and
Emeril on DVD.

Hang on a minute....first things first. How much would you like to
lose? 2500USD? 1000EURO? 1000GBP?

Sure. What time shall I expect you?

Yeah, that's what I thought...you puss out before you're forced to welsh
on the bet...run back under the porch, lil' pup...

O.k., let me see if I've got this straight. You want people to believe
that you were seriously offering to make a bet about something or other
that is a complete mystery to you and that I had taken you seriously
and accepted the bet and that somehow or other somebody was going to do
something or other to settle the bet in front of reliable witnesses to
everyone's satisfaction and that the results would then be duly
published here and that I would be appropriately contrite and confess
that I really DID know all along that you are
God......and.....something or other more. That about it?

Nope. While your claimed hesitance to take advantage of someone who
didn't understand what they were doing would be oh-so-noble, it's BS. If
one wagers, one risks losing. All you needed to do is put your money
where your mouth is. What other people believe or don't believe is up
to them. You made a proposition (which I note you snipped from above):

"I will simply confine myself 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."

and I twice made it clear that I accepted it I didn't qualify or
otherwise modify it. I then offered to wager and gave an idea of the
stakes in which I was willing to risk. If those stakes were too much for
you, you could have offered lesser with no shame. If you swung as big a
dick as you'd like to think, you would have simply accepted or declined
like a man, put the ball in my court, and have done as I did - not worry
about what "people" might or might not believe. Instead, you pussed. In
fact, I'd offer that I was wagering on that, too...

Like I said - run back under the porch, lil' pup,

So, you're saying that you seriously DO expect people to believe that you
were seriously offering to make a bet about something or other that is a
complete mystery to you and that I had taken you seriously and accepted the
bet and that somehow or other somebody was going to do something or other to
settle the bet in front of reliable witnesses to everyone's satisfaction and
that the results would then be duly published here and that I would be
appropriately contrite and confess that I really DID know all along that you
are God......and.....something or other more?!!

You don't get any brighter as the week gets older, do you? :)

O.k., let's pretend (just for the moment) that you could find your way to
the real world for a brief visit. How would you propose that your little
bet be settled?

Wolfgang

What's a boy to do?
 

"riverman" wrote in message ...

"rb608" wrote in message
news:WWF2h.7479$OK3.540@trndny09...
"riverman" wrote in message
There is a similar problem using discrete variables rather than
continuous. A man says he has two kids, and mentions that one is a
girl. What is the probability that the other is also a girl? The answer
is NOT 50%, since some information has been disclosed. But if he says
"I have two kids, and here is one of them", the the probability that
the other is a girl IS 50%. All events are symmetric.

I'm having a problem with the wording of the questions, I suppose. In
your example above, I believe that once the first girl is revealed, the
question more accurately becomes, "What is the probability that both
children are girls?" I agree that's not .50.

I'll toss out the Rosencrantz and Guilderstern example. If I toss a
coin, the probability of heads is .50. If I toss it again, what is the
probability of heads? It's .50. How is that different from the man with
two children?

The main difference is that one is a conditional probability, the other is
not.

The possible sample space for someone having two children is BB BG GB GG.
Since man gave us the condition that 'at least one is a girl', you know
that BB is not a possibility. Thus, the remaining possible sample space is
BG, GB or GG (these all meet his condition), and the only sucessful
outcome is GG, so the probability that "the other child is also a girl,
given that at least one is a girl" is 1/3....

O.k., I will stress, once again, that I'm no mathematician, but this has a
funny smell to it. Looks to me like the math is unassailable as long as BG
is something different than GB......but it manifestly isn't. BG and GB are,
in fact, exactly the same thing and thus we a remaining sample space of
TWO, not three possibilities......unless we posit that birth order enters
the equation, in which case, the whole thing falls apart and we are dealing
with an entirely different problem.

Wolfgang

What's a boy to do?
 

Wolfgang wrote:

O.k., I will stress, once again, that I'm no mathematician,

There's no need to point that out. It's self evident.

--
Cut "to the chase" for my email address.

What's a boy to do?
 

"rw" wrote in message
m...
Wolfgang wrote:

O.k., I will stress, once again, that I'm no mathematician,

There's no need to point that out. It's self evident.

You don't learn.

What is the correct solution to Myron's problem?

Wolfgang

What's a boy to do?
 

Wolfgang wrote:
"rw" wrote in message
m...

Wolfgang wrote:

O.k., I will stress, once again, that I'm no mathematician,

There's no need to point that out. It's self evident.


You don't learn.

What is the correct solution to Myron's problem?

Myron gave the correct solution.

If we were told that the OLDEST child is a girl, the probability of the
other child being a girl is 1/2. But we weren't told that.

--
Cut "to the chase" for my email address.