What's a boy to do?
 

Wolfgang wrote:

Still smells funny to me.

Alert the mathematics world! This simple and uncontroversial
combinatorial reasoning "smells funny" to ROFF's Wolfgang.

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?

It's simply because they are two different people. Duh.

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?

Knowledge of birth order changes the combinatorics. The question as put
by Myron, devoid of birth-order information, is perfectly clear and
straightforward. The answer is a probability of 1/3 that the other child
is a girl.

As I explained before, if we were told that the OLDEST child was a girl,
the probability of the other child being a girl would be 1/2. But we
weren't told that.

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

What's a boy to do?
 

rw wrote:
Wolfgang wrote:

Still smells funny to me.

Alert the mathematics world! This simple and uncontroversial
combinatorial reasoning "smells funny" to ROFF's Wolfgang.

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?

It's simply because they are two different people. Duh.

Hm......

Bob and Sally are two different people than Sally and Bob? Do you
suppose that Bob and Sally know that? Then again, what will Sally and
Bob think when they find out?

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?

Knowledge of birth order changes the combinatorics.

Really? Well.......gosh. So?

The question as put
by Myron, devoid of birth-order information, is perfectly clear and
straightforward.

To be sure.

The answer is a probability of 1/3 that the other child
is a girl.

Ah! But, the question is.....the answer to what?

As I explained before, if we were told that the OLDEST child was a girl,
the probability of the other child being a girl would be 1/2.

Yes, you DID say that. However, limited sopace and time prohibit a
complete list of all the dumb and useless **** you've said here over
the years. With that in mind, what, exactly, is the purpose of setting
yet another bad precedent?

But we weren't told that.

No ****? Why wasn't I informed?

Wolfgang
who has seen vegetative matter that learns faster.

What's a boy to do?
 

Wolfgang wrote:
rw wrote:
Wolfgang wrote:

Still smells funny to me.

Alert the mathematics world! This simple and uncontroversial
combinatorial reasoning "smells funny" to ROFF's Wolfgang.

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?

It's simply because they are two different people. Duh.

Hm......

Bob and Sally are two different people than Sally and Bob? Do you
suppose that Bob and Sally know that? Then again, what will Sally and
Bob think when they find out?

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?

Knowledge of birth order changes the combinatorics.

Really? Well.......gosh. So?

The question as put
by Myron, devoid of birth-order information, is perfectly clear and
straightforward.

To be sure.

The answer is a probability of 1/3 that the other child
is a girl.

Ah! But, the question is.....the answer to what?

As I explained before, if we were told that the OLDEST child was a girl,
the probability of the other child being a girl would be 1/2.

Yes, you DID say that. However, limited sopace and time prohibit a
complete list of all the dumb and useless **** you've said here over
the years. With that in mind, what, exactly, is the purpose of setting
yet another bad precedent?

But we weren't told that.

No ****? Why wasn't I informed?

Wolfgang
who has seen vegetative matter that learns faster.

Heeeeyyyyy, wait just a doggone minute!

How come is it that if Bob and Sally are two different people from
Sally and Bob, Ethel and Elizabeth ain't two different people from
Elizabeth and Ethel?

O.k., let's see now......ya got yer {BB}, yer {Bb}, yer {bB}, yer {Bg},
yer {bG}, yer {BG}, yer {GB}, yer {Gb}, yer {gB}, yer {Gg}, yer {GG},
and yer {gG}.

Hm.......o.k......yeah......that's two.

Wolfgang
who be go ta hell if he can figure out where the xs and ys went to.
:(

What's a boy to do?
 

"rw" wrote in message
Knowledge of birth order changes the combinatorics.

Nonsense. The probabilities of a coin flip do not depend on the date of the
coin.

Joe F.

What's a boy to do?
 

"rb608" wrote in message
Nonsense. The probabilities of a coin flip do not depend on the date of
the coin.

I'm going to amend my response to acknowledge that I know it isn't nonsense
in the general case. In mathematical models of probability, considering or
ignoring order do make a difference. I'll stand by my second sentence,
though.

Joe F.

What's a boy to do?
 

"Kevin Vang" wrote in message
t...

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?"

I think you mean "who has LESS money..."


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



Very interesting geometric proof. I wouldn't have thought of solving it with
volumes. I approached it with areas, but also got stuck trying to figure out
what to do with the second dart.

I started with a target of radius 1. I made a graph where the x axis went
from 0-1 for the distance from the target of the first dart, and the Y axis
went 0-1 for Probability of hitting inside that distance. Since the area
inside that distance is a quadratic, I graphed P=sqrt(dist), and found the
area below the curve using a simple definite integral, and got 2/3. Then I
generalized for a target of infinite size by integrating sqrt(x) from 0 to
infinity, and got 2/3 again. But I couldn't figure out how the second dart
came into play, so I didn't post this solution.

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


Yow, that's an impressive triple integral. Never did much with PDFs myself.

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.


It gets better. Try different distributions, including those with random
pockets of P=0, and not even centered at the bullseye. In retrospect, its
not hard to generalize that, as long as the conditions are constant for all
three darts, it doesn't matter WHAT the distribution or even the target
point is, the probability is 2/3.

--riverman

What's a boy to do?
 

rb608 wrote:
"rw" wrote in message

Knowledge of birth order changes the combinatorics.


Nonsense. The probabilities of a coin flip do not depend on the date of the
coin.

I'll explain why it's not nonsense.

As Myron correctly pointed out, given only the information "at least of
the two children is a girl" the sample space is BG, GB, GG. (BB is
excluded.) Only GG meets the condition that the other child is a girl,
so the probability is 1/3.

Given the information "the oldest of the two children is a girl" and
using the first letter of a pair to denote the oldest child, the sample
space is GB, GG. Only GG meets the condition that the other child is a
girl, so the probability is 1/2.

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

What's a boy to do?
 

"Jonathan Cook" wrote in message
...
riverman wrote:

It gets better. Try different distributions, including those with random
pockets of P=0, and not even centered at the bullseye. In retrospect, its
not hard to generalize that, as long as the conditions are constant for
all
three darts, it doesn't matter WHAT the distribution or even the target
point is, the probability is 2/3.

Sure, for the problem that you are solving, that is
different than the one you originally posed. In email
I said you were assuming a particular distribution,
because I didn't think you were throwing darts
blindfolded. All of Kevin's analysis, and yours, is
correct(*), for a _different_ problem.

I am not assuming any 'particular' distribution, only a constant one. It
doesn't even have to be evenly distributed; the thrower can be the best
dartsman in the universe, aiming for the bullseye with all his skill. All
that will do is cluster the three darts closer to the center. In fact, give
him a target 1mm in diameter, and his final arrangement might not look much
different than one from a blindfolded man throwing at a target 10 meters in
diameter.

The type of distribution doesn't matter, as long as it is the same for all
three darts. Other than claiming that the thrower might improve between
throws, the bulk of your email was primarily just claiming that the model
was flawed, with no real specifics of how. You can go ahead and post the
email if you want; I don't mind.

The protest that tense plays a role is irrelevant, as long as no information
about the location of any of the darts is given. I believe we agree that the
answer to the question "if I toss a coin three times, the probability of
getting HHH is 1/8." What would you say is the answer to; "I have tossed a
coin twice, and now I'm about to toss it a third time. What is the
probability of getting HHH?"

--riverman

-riverman

What's a boy to do?
 

"Wolfgang" wrote in message
ups.com...

O.k., let's see now......ya got yer {BB}, yer {Bb}, yer {bB}, yer {Bg},
yer {bG}, yer {BG}, yer {GB}, yer {Gb}, yer {gB}, yer {Gg}, yer {GG},
and yer {gG}.

Hm.......o.k......yeah......that's two.

Wolfgang
who be go ta hell if he can figure out where the xs and ys went to
:(


Not that it makes any difference in the solution, but as a technical matter
you should not include BB, GG, BG, or GB as possibilities in your list
above, even if you are using age as a differentiation factor. Twins are not
born at *exactly* the same time.

Bob Weinberger

What's a boy to do?
 

Wolfgang wrote:
rw wrote:

Wolfgang wrote:

Still smells funny to me.

Alert the mathematics world! This simple and uncontroversial
combinatorial reasoning "smells funny" to ROFF's Wolfgang.


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?

It's simply because they are two different people. Duh.


Hm......

Bob and Sally are two different people than Sally and Bob? Do you
suppose that Bob and Sally know that? Then again, what will Sally and
Bob think when they find out?

Now that you've demonstrated that you're incapable of understanding
elementary mathematical logic, **** off.

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

What's a boy to do?
 

Bob Weinberger wrote:
"Wolfgang" wrote in message
ups.com...

O.k., let's see now......ya got yer {BB}, yer {Bb}, yer {bB}, yer {Bg},
yer {bG}, yer {BG}, yer {GB}, yer {Gb}, yer {gB}, yer {Gg}, yer {GG},
and yer {gG}.

Hm.......o.k......yeah......that's two.

Wolfgang
who be go ta hell if he can figure out where the xs and ys went to
:(


Not that it makes any difference in the solution, but as a technical matter
you should not include BB, GG, BG, or GB as possibilities in your list
above, even if you are using age as a differentiation factor. Twins are not
born at *exactly* the same time.

Noted and filed for future reference. Thanks, Bob. :)

Wolfgang

What's a boy to do?
 

rw wrote:
Wolfgang wrote:
rw wrote:

Wolfgang wrote:

Still smells funny to me.

Alert the mathematics world! This simple and uncontroversial
combinatorial reasoning "smells funny" to ROFF's Wolfgang.


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?

It's simply because they are two different people. Duh.


Hm......

Bob and Sally are two different people than Sally and Bob? Do you
suppose that Bob and Sally know that? Then again, what will Sally and
Bob think when they find out?

Now that you've demonstrated that you're incapable of understanding
elementary mathematical logic,

Well, maybe it's not so much incapacity on my part as it is poor
instruction. At least this can be tested experimentally. Why don't we
see if you can explain who are two different people from whom and then
we'll take a poll......see if anybody else can understand it.

**** off.

No, not just yet, I think. I worry about who would remind you that you
are a slow learner if I were to go away. Tell you what......see if you
can find a sufficiently qualified volunteer, and we'll talk about
it.....o.k.?

Wolfgang

What's a boy to do?
 

On Sun, 5 Nov 2006 00:53:30 +0800, "riverman" wrote:

The type of distribution doesn't matter, as long as it is the same for all
three darts.

That's the problem I have with your example - I think the distribution changes
based on the prior throw. Change your dart throwing to fly casting to a rising
fish and maybe then I'll believe you're totally random. g
--
Charlie...
http://www.chocphoto.com

What's a boy to do?
 

"Charlie Choc" wrote in message
...
On Sun, 5 Nov 2006 00:53:30 +0800, "riverman" wrote:

The type of distribution doesn't matter, as long as it is the same for all
three darts.

That's the problem I have with your example - I think the distribution
changes
based on the prior throw. Change your dart throwing to fly casting to a
rising
fish and maybe then I'll believe you're totally random. g


LOL, but I see the problem. Not 'random'. 'Consistent'.

--riverman

What's a boy to do?
 

On Sun, 5 Nov 2006 09:58:16 +0800, "riverman" wrote:


"Charlie Choc" wrote in message
.. .
On Sun, 5 Nov 2006 00:53:30 +0800, "riverman" wrote:

The type of distribution doesn't matter, as long as it is the same for all
three darts.

That's the problem I have with your example - I think the distribution
changes
based on the prior throw. Change your dart throwing to fly casting to a
rising
fish and maybe then I'll believe you're totally random. g


LOL, but I see the problem. Not 'random'. 'Consistent'.

I said I think the distribution changes based on the prior throw (or cast for
that matter) - IOW not consistent (and for this problem, random or consistent
gives the same result). There may be a consistent distribution for 1st throws,
2nd throws, etc - but I believe they will be different distributions.
--
Charlie...
http://www.chocphoto.com

What's a boy to do?
 

"Charlie Choc" wrote in message
...
On Sun, 5 Nov 2006 09:58:16 +0800, "riverman" wrote:


"Charlie Choc" wrote in message
. ..
On Sun, 5 Nov 2006 00:53:30 +0800, "riverman" wrote:

The type of distribution doesn't matter, as long as it is the same for
all
three darts.

That's the problem I have with your example - I think the distribution
changes
based on the prior throw. Change your dart throwing to fly casting to a
rising
fish and maybe then I'll believe you're totally random. g


LOL, but I see the problem. Not 'random'. 'Consistent'.

I said I think the distribution changes based on the prior throw (or cast
for
that matter) - IOW not consistent (and for this problem, random or
consistent
gives the same result). There may be a consistent distribution for 1st
throws,
2nd throws, etc - but I believe they will be different distributions.

Well, I won't disagree that this is a mathematical model, while IRL they
would probably be able to adjust their throw a bit based on their previous
toss. But unless they improve a LOT, the results (within a small degree of
error) would hold. Maybe a few percentage points off, or fractions of a
percentage, so dealing with it as a mathematical model is not a vain
exercise. Throwing darts still has enough of a random component that even
the experts are not hitting perfect games consistently. Possibly it would be
more salable if I said "after warming up" at the beginning.

Anyway, this has been a long discussion, and I'm pretty much fried out on
it. Those with the mathematical insight know that this is acually the Monte
Hall problem in another form, which makes it fascinating that people were
more likely to adjust their misconceptions of the MH puzzle when faced with
some physical demonstration or layman's description, while this problem had
many folks stubbornly refusing to change their point of view, and unwilling
to accept a layman analysis. Interesting.

--riverman

What's a boy to do?
 

"Jonathan Cook" wrote in message
...
riverman wrote:

This is EOT for me. The problem, as stated, does not contain
enough information to solve.

You mean 'the problem, as you chose to state it, does not contain enough
information to solve.' The coin toss example was merely to show that, if no
information was disclosed, then tense is irrelevant.

I'm okay with EOT, and you can even have the last word. It won't change the
solution g

--riverman

the definitive answer
 

Wolfgang wrote:
RE Obviously, the five dollars must be under one of the other two.

The way you stated this, you removed what at one time looked like
a one in three chance. But, the way you stated this, you took away
one of the three choices, and the one you took away was known
to be false.

So there are now two choices left, one of which is guaranteed
to be correct. And you have no evidence to indicate one choice
over the other.

The current 50-50 condition is unrelated to a previous condition,
when three chances were involved. And it doesn't matter how many times
to you do it (if you follow the sequence of events you specified).

If you restate the problem, and say you now remove one of three
choices, leaving two that might be false, or two choices that
contain at most one true, then it is a different problem.

Jesus, forget mathematicians. When you need answers to difficult
problems,
always ask a sliver digger (a carpenter).

the definitive answer
 

salmobytes wrote:
Wolfgang wrote:
RE Obviously, the five dollars must be under one of the other two.

The way you stated this, you removed what at one time looked like
a one in three chance. But, the way you stated this, you took away
one of the three choices, and the one you took away was known
to be false.

So there are now two choices left, one of which is guaranteed
to be correct. And you have no evidence to indicate one choice
over the other.

The current 50-50 condition is unrelated to a previous condition,
when three chances were involved. And it doesn't matter how many times
to you do it (if you follow the sequence of events you specified).

If you restate the problem, and say you now remove one of three
choices, leaving two that might be false, or two choices that
contain at most one true, then it is a different problem.

Jesus, forget mathematicians. When you need answers to difficult
problems,
always ask a sliver digger (a carpenter).

I won't speak for anyone else, but when I need the answer to a
difficult question (or even what may turn out to be a not so difficult
question, for that matter) I think I'll ask someone who can at least
make his or her position on a previous question (not to mention an
explication thereof) clear.

Thanks, anyway.

Wolfgang
still, the venture was not entirely without profit.......we have at
least learned something about the origins of old expression, "to a man
with a hammer......."

the definitive answer
 

you said you took a false choice away
That means the remaining choice is 50/50
no matter what.

Any previous condition is like Wolfgang:
totally irrelevant :-)

the definitive answer
 

salmobytes wrote:
you said you took a false choice away

I said nothing of the sort. I said that I revealed one of the losing
tickets. This is not the same thing at all. Firstly, I took nothing
away. Secondly, there were no false choices. The only choices to be
made were the player's initial pick of what he or she hoped to be the
winner, mine to reveal one of the losers, and the player's to switch or
stand pat. Only one of those three is subject to a value judgment that
makes any sort of sense in context. The player can make the WRONG
second choice, but there can be nothing "false" about it or any of the
others.

That means the remaining choice is 50/50
no matter what.

Most people learn very early in life to get used to the fact that they
will not always understand what is said to them. Those of us who are
lucky also learn, eventually, that this is not really as much of a
problem as it first appear, because often enough we DO understand what
has been said to us.....even when those who said it clearly do not.

This is one of those instances.

You are absolutely right; you have no idea of what the correct solution
to the problem is (despite it's having been put forward here in various
guises several times), and thus your chances of correctly guessing what
to do are 50/50 since there are only two possibilities to choose from.

Unfortunately, for you (well, not JUST you......as we have seen here
over the past few days.....as well as the past few years), you guessed
the right answer to the wrong question. The question, as clearly
stated in the original exposition of the problem, concerned what you
SHOULD DO.....NOT your chances of guessing correctly what you should
do.

So, after several days, you have a fuzzy notion (at best) of what the
question was, no idea of what the correct answer is, and only a dim
(and wrong) notion of what you said above.

Here's what you should do. Whenever tempted to say something, you
should first determine whether or not you really have something to say.
Then, if you meet the first condition in the affirmative, you should
take some time to try to figure out whether there is any chance of
determining what it is you have to say. If that yields positive
results, you should then ask yourself whether it was worth the trouble
and would be worth the extra trouble of trying to phrase it in a way
that might be comprehensible and of interest to a reader. If so, then
you should do just that, as opposed to tapping on the keyboard in a
random fashion.

Any previous condition is like Wolfgang:
totally irrelevant :-)

And yet, here you are.

Wolfgang
p.s. for the benefit of anyone who might still be interested in all of
this, i believe it was Scott who stated early on in this thread that
the MH problem hinged on the fact that the exposure of a losing ticket
was not random. i agreed. on further reflection, it turns out that
this is not true. as i pointed out to Jon a bit later, as long as a
losing ticket comes up in the exposure, it doesn't make any difference
whether that ticket was chosen deliberately or at random. this
shouldn't be difficult for anyone to grasp. it's also not the
interesting part. what IS interesting.....because it is SO obvious in
retrospect.....is that turning up the WINNING ticket (whether by chance
or deliberately) doesn't make any differece either. remember, you, the
player, have been given the option of sticking with your original pick
or switching. the question is "what should you do." the exposure of
the winner simply makes the choice a no brainer. :)

What's a boy to do?
 

Wolfgang said
"the five dollars must be under one of the other two"
(after removing one choice)

Wolfgang also said:
" so what SHOULD you do now?"

Sandy says:
"it doesn't matter."

You know one choice is true and one choice is false.
It's one of the two.
But you have no evidence which is which.
And any previous choice (regarding three possibilities)
is irrelevant.

Your insulting, passive aggressive attempts to
set me off have done the opposite. I'm laughing.

What's a boy to do?
 

salmobytes wrote:
Wolfgang said
"the five dollars must be under one of the other two"
(after removing one choice)

Wolfgang also said:
" so what SHOULD you do now?"

Sandy says:
"it doesn't matter."

Yeah, we know that. And Sandy is STILL wrong. Ask stevie. :)

You know one choice is true and one choice is false.

No, there is nothing "true" or "false" about any of the choices in this
problem. The player's first choice might as well be random (and, in a
very real sense, IS), my choice, as has now been demonstrated, could
also have been random for all the difference it makes to the player, or
it could just as well be deliberate (as appears to be implicit in the
original statement of the problem), and the player's second choice can
be a good one (the correct choice.....to switch) or a bad one (the
wrong one.....to stick), but "true" and false" are meaningless in the
context of the problem.

It's one of the two.

True.

But you have no evidence which is which.

False. You have probablilities. Switching doubles the odds of winning
(except, as outlined in my previous post, in cases in which the winner
is exposed in the modified version of the game. I'm content to leave
calculation of the odds in that case as an exercise for you.).

And any previous choice (regarding three possibilities)
is irrelevant.

That looks suspiciously like it might mean something.

Your insulting, passive aggressive attempts to
set me off have done the opposite.

You may rest assured that there is nothing passive about anything I do
here.

I'm laughing.

Sure, you are.

Wolfgang

What's a boy to do?
 

On Nov 5, 4:01 pm, "Wolfgang" wrote:
salmobytes wrote:
Wolfgang said
"the five dollars must be under one of the other two"
(after removing one choice)

Wolfgang also said:
" so what SHOULD you do now?"

Go back and read my answer.

The answer to that question is, as I said, : "flip a coin".

cheers

oz.......with a number of sucker bar bets. Any takers?

What's a boy to do?
 

MajorOz wrote:
On Nov 5, 4:01 pm, "Wolfgang" wrote:
salmobytes wrote:
Wolfgang said
"the five dollars must be under one of the other two"
(after removing one choice)

Wolfgang also said:
" so what SHOULD you do now?"

Go back and read my answer.

No. I find more than enough fresh dumb **** to read in here......no
need to recycle.*

The answer to that question is, as I said, : "flip a coin".

Flip whatever you like. Nobody cares. Trust me on this one.

cheers

oz.......with a number of sucker bar bets. Any takers?

Heh, heh, heh. See all those dorsal fins circling you?

Wolfgang
*yet.

What's a boy to do?
 

"Wolfgang" wrote in message
oups.com...

MajorOz wrote:
On Nov 5, 4:01 pm, "Wolfgang" wrote:
salmobytes wrote:
Wolfgang said
"the five dollars must be under one of the other two"
(after removing one choice)

Wolfgang also said:
" so what SHOULD you do now?"

Go back and read my answer.

No. I find more than enough fresh dumb **** to read in here......no
need to recycle.*

The answer to that question is, as I said, : "flip a coin".

Flip whatever you like. Nobody cares. Trust me on this one.

cheers

oz.......with a number of sucker bar bets. Any takers?

Heh, heh, heh. See all those dorsal fins circling you?

Wolfgang
*yet.
cut his ****ing throat and take the ****in
money..........................

What's a boy to do?
 

Faaart wrote in message ...

"Wolfgang" wrote in message
oups.com...

MajorOz wrote:
On Nov 5, 4:01 pm, "Wolfgang" wrote:
salmobytes wrote:
Wolfgang said
"the five dollars must be under one of the other two"
(after removing one choice)

Wolfgang also said:
" so what SHOULD you do now?"

Go back and read my answer.

No. I find more than enough fresh dumb **** to read in here......no
need to recycle.*

The answer to that question is, as I said, : "flip a coin".

Flip whatever you like. Nobody cares. Trust me on this one.

cheers

oz.......with a number of sucker bar bets. Any takers?

Heh, heh, heh. See all those dorsal fins circling you?

Wolfgang
*yet.
cut his ****ing throat and take the ****in
money..........................

Nah......I don't need the money. Besides, a live fool is cheap
entertainment.......a dead one is just litter.

Wolfgang

What's a boy to do?
 

On Tue, 7 Nov 2006 07:27:18 -0600, "Wolfgang"
wrote:

Nah......I don't need the money. Besides, a live fool is cheap
entertainment.......a dead one is just litter.

Wolfgang


Good for compost, though.
--

r.bc: vixen
Speaker to squirrels, willow watcher, etc..
Often taunted by trout. Almost entirely harmless. Really.

http://www.visi.com/~cyli