What's a boy to do?

"Wolfgang" wrote in message
...

"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.

No, it is correct. Dr Math explains it better than I do at this ungodly hour
:-)
http://mathforum.org/dr.math/faq/faq.boy.girl.html

--riverman

"rw" wrote in message
m...
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.

And you'd be willing to bet dicklet a million dollars on that?

Wolfgang
the boy just WILL NOT learn!

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


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.


And you'd be willing to bet dicklet a million dollars on that?

I'm not interested in debating elementary combinatorics with
mathematical ignoramuses.

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

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


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.


And you'd be willing to bet dicklet a million dollars on that?

I'm not interested in debating elementary combinatorics with mathematical
ignoramuses.

Well, given that you didn't understand my question, I'd say that's probably
a very wise decision.......it must be humiliating enough to lose debates
with everyone else.

Wolfgang
have i mentioned recently that the boy just WILL NOT learn?

In article ,
says...
"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.


riverman is right!

I haven't weighed in on the original question until now, because I
wasn't sure that I really understood it (I know, I know, that never
stopped anybody before...) What I didn't understand was the role of
dart B in the problem, but the answer came to me in the shower this
morning (proving that those Fedex commercials are right also.)

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.

If we only throw two darts, and ask what is the probability that
the second dart is farther than the first, the correct analogy
would be, "Who has more money, a randomly chosen person or another
randomly chosen person?" In this case, the probability that the first
randomly chosen person has more money tha the second is indeed 50%.

But first we throw two darts, and ask what is the probability that a
third dart is farther from the center than the the first, given that the
second dart is farther from the center than the first. The money
analogy would be, "Who has more money, one random person, or the poorer
of two other randomly chosen people?"

Lets make some assumptions: Let a, b, and c be the distance of darts A,
B, and C, respectively from the center of the target, and let's suppose
that the radius of the target is 1. Then the sample space for the
experiment is the set { (a,b,c) | 0 = a,b,c = 1 }, which geometrically
is the unit cube. Furthermore, let's assume that every point on this
cube is equally likely to be chosen (more on this below.)

Now, the question is: Find P( c a | b a ); that is, find
P( c a AND b a )
------------------------
P( b a ) .
Looking at the denominator first, the requirement that b a
divides the sample space in half; the region is the triangular
wedge shape produced by splitting the unit cube with a vertical plane
passing through the points (0,0,0), (0,0,1), (1,1,0) and (1,1,1). This
wedge has volume 1/2.
Now, adding the additional requirement that c a splits this region
again, by passing a plane through the points (0,0,0), (0,1,0), (1,0,1),
and (1,1,1). (This is much easier to grasp if you sketch a graph of
this...) The remaining solid is a pyramid with the unit sqare for a
base and height 1, so the volume of the pyramid is (1/3)*1*1 = 1.
Therefore,
P( c a AND b a ) 1/3
------------------------ = ------- = 2/3.
P( b a ) 1/2


To make this more general, suppose the radius of the dartboard is R,
where R can be finite or infinite, and p(a,b,c) is a probability density
function over the sample space { (a,b,c) | 0 = a,b,c = R }. Then

/R /R /R
| | | p(a,b,c) dc db da
P( c a AND b a ) /0 /a /a
------------------------ = -------------------------------------
P( b a ) /R /R /R
| | | p(a,b,c) dc db da
/0 /0 /a

Now, I don't have a proof* that this always evaluates to 2/3 for any
choice of probability density function p(a,b,c), but I did some
experimenting with Maple, and for every pdf I tried, over finite and
infinite sample spaces, this always evaluated to 2/3.

Kevin

*yet, anyway. Thanks for ruining my weekend...


--
reply to: kevin dot vang at minotstateu dot edu

In article ,
says...

The math in the above post looks MUCH better if you view it
with a non-proportional font (say, Courier).

Kevin

--
reply to: kevin dot vang at minotstateu dot edu

In article drF2h.6625$mX4.1479@trndny03, lid
says...

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

Life isn't perfectg

I will certainly investigate the possibility for next semester,
though.

Kevin

--
reply to: kevin dot vang at minotstateu dot edu

"riverman" wrote in message ...

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.

No, it is correct. Dr Math explains it better than I do at this ungodly
hour :-)
http://mathforum.org/dr.math/faq/faq.boy.girl.html

Dr. Math says:

"In a two-child family, there are four and only four possible combinations
of children. We will label boys B and girls G; in each case the first letter
represents the oldest child: {BB, BG, GB, GG}. When we know that one child
is a boy, there cannot be two girls, so the sample space shrinks to: {BB,
BG, GB}. Two of the possibilities in this new sample space include girls:
{BG, GB}and since there are two combinations out of three that include
girls, the probability that the second child is a girl is 2/3." ( I changed
the spacing and added a couple of periods in the interest of saving space
and maintaining clarity in the process....I don't believe I have changed any
of the meaning in the process.)

I'd say your explanation is no worse than Dr. Math's. They appear identical
in all essentials.

Still smells funny to me. In the absence of any positional element (birth
order, size, where they might happen to be standing relative to one another
in a photo, etc.) how does {a boy and a girl} on the one hand differ from {a
girl and a boy} on the other? To be sure, {BG} LOOKS different than {GB}but
except with regard to the relative positions of the labels for {B}oy and
{G}irl they are identical entities. Why does one entity get counted as two
possibilities?

Wolfgang

On Fri, 3 Nov 2006 09:19:28 -0600, "Wolfgang" wrote:


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?

Since you've asked so nicely, it's only gonna cost you a little bit to
find out and you won't lose much. Here's the deal, one-time offer, not
negotiable. You agree to send a check for $25USD to St. Jude's. Don't
like St. Jude's? Tough ****. My offer, my choice. If you agree, then
I'll outline the rest, with the wager being that if you can't prove me
wrong, you send a check for $100USD (total) to St. Jude's and if you can
prove me wrong, you get whatever satisfaction you might get. Don't like
any of the above? Again, tough ****, etc.

Are you in or out?

R

Wolfgang wrote:
To be sure, {BG} LOOKS different than {GB}but
except with regard to the relative positions of the labels for {B}oy and
{G}irl they are identical entities. Why does one entity get counted as two
possibilities?

That's what bothers me as well. To be sure, the mathematics is correct
as modeled, but I think this debate arises from attempting to word a
precise, non-mathematical, real-world question to fit a desired
abstract (and, I would argue, a deliberatlely oblique) answer. I don't
care how many kids the guy has, the chances of any one of them being a
B or G is still the same as a coin flip (genetics notwithstanding).

It's clear to me that Dr. Math & Myron are both correct in their
analysis of the mathematics; but I take issue that the question as
asked accurately describes the model subsequently analyzed.

Joe F.