The probability of boys
Dennis Rakestraw
I agree with the algebra teacher and Marilyn on this one. Actually this
puzzle has been around for many years and it has probably survived
because the correct answer is counter-intuitive. I am trying to find the
earliest known publication of the problem. Here his one from 1980, (Life
Science Library, Mathematics, Time-Life Books, Alexandria, Virginia):
If a man says, "of my two children at least one is a boy," what is the
probability that both are boys? The answer, oddly, is not "50-50." If
the sex of the first-born is unknown, there are three possible sequences
of children:
BOY, THEN BOY
BOY, THEN GIRL
GIRL, THEN BOY
Of these three possible arrangements, only one includes two boys: The
probability that the other child is a boy is thus one in three, or 1/3.
Had we known ;the man's first child was a boy, only the upper two figures
above would be possible; and the probability of his having two boys
would then be one out of two or "50-50."
Dennis Rakestraw
drake...@hotmail.com
-------------------==== Posted via Deja News ====-----------------------
http://www.dejanews.com/ Search, Read, Post to Usenet
Jim Balter
Dennis Rakestraw wrote:
>
> I agree with the algebra teacher and Marilyn on this one. Actually this
> puzzle has been around for many years and it has probably survived
> because the correct answer is counter-intuitive.
Like Monty Hall and the 3 doors, it's counter-intuitive for those
without good intuitions about conditional probabilities, which turns
out to be almost everyone who hasn't spent a lot of time with them,
including most mathematicians. The fact that mathematicians expect
to have good intuitions about things mathematical causes many of them
to react violently when told that they have the wrong answer to these
seemingly simple probability questions. They expect to have good
intuitions largely because they have a faulty, often metaphysical,
notion of what intuition is and, in the case of probability, they
often have a faulty notion of what *it* is.
> I am trying to find the
> earliest known publication of the problem. Here his one from 1980, (Life
> Science Library, Mathematics, Time-Life Books, Alexandria, Virginia):
>
> If a man says, "of my two children at least one is a boy," what is the
> probability that both are boys? The answer, oddly, is not "50-50."
It isn't odd at all to anyone who recognizes that the answer can't be
the same as if we knew which one was the boy. In fact, once one
realizes that probability theory is all about information, one
immediately looks to see what information has been provided.
> If
> the sex of the first-born is unknown, there are three possible sequences
> of children:
>
> BOY, THEN BOY
>
> BOY, THEN GIRL
>
> GIRL, THEN BOY
>
> Of these three possible arrangements, only one includes two boys: The
> probability that the other child is a boy is thus one in three, or 1/3.
> Had we known ;the man's first child was a boy, only the upper two figures
> above would be possible; and the probability of his having two boys
> would then be one out of two or "50-50."
Yes, probabilities are a function of what is known. Another strike
against Platonism, esentialism, and absolutism.
--
<J Q B>
No SPAM
Dennis Rakestraw <drake...@hotmail.com> wrote in article
<882203102....@dejanews.com>...
[stuff deleted]
> If a man says, "of my two children at least one is a boy," what is the
> probability that both are boys? The answer, oddly, is not "50-50."
[more stuff deleted]
A finer example of sexism in mathematics will ne'er be found.
If men who have one boy and one girl are as likely to start,
"of my two children at least one is a girl" as they are to start,
"of my two children at least one is a boy", then the answer is
indeed 50-50.
--
Standard disclaimers apply.
David Ullrich
If we start thinking about how likely people are to say
what under what conditions then it's no longer a strictly
mathematical question. One needs to rephrase (or at least interpret)
the question
If a man says, "of my two children at least one is a boy," what is the
probability that both are boys?
as
Given that a man has two children, at least one is a boy, what is the
probability that both are boys?
or the question can't be answered.
--
David Ullrich
sig.txt not found
William L. Bahn
No SPAM <nos...@apgea.army.mil> wrote in article
<01bd0a37$f1ea8700$b8f2...@orpc07.apgea.army.mil>...
>
>
> Dennis Rakestraw <drake...@hotmail.com> wrote in article
> <882203102....@dejanews.com>...
> [stuff deleted]
>
> > If a man says, "of my two children at least one is a boy," what is the
> > probability that both are boys? The answer, oddly, is not "50-50."
> [more stuff deleted]
>
> A finer example of sexism in mathematics will ne'er be found.
So if a man states that of his two children at least one of them is a boy,
you consider it sexist to conclude that there is no possibility that both
of his children are girls? That's all that is involved here.
You probably also believe that the reason wall receptacles are female and
plugs are male (as a rule) is purely because these sexist male architects,
engineers and electricians were expressing their beliefs that females
should be firmly bound to one location while males should be free to roam
around and mate with any female they choose (I have actually found people
that latched on to that notion when I proposed it).
It never ceases to amaze me the lengths people will go to in order to claim
sexism or racism.
How about this:
If a person says "Of my two children, at least one of them is Gender A,
what is the probability that both are of Gender A?"
The answer is that the odds are 2/3 that both children are of the same
gender as mentioned by the person asking the question.
In a nutshell, the key is that the information that the person has given
you has specifically excluded the possibility that both children are of
Gender B.
Now please explain again how this is sexist?
> If men who have one boy and one girl are as likely to start,
> "of my two children at least one is a girl" as they are to start,
> "of my two children at least one is a boy", then the answer is
> indeed 50-50.
You are so intent of being able to make your claim of sexism, that you
refuse to even look at what you are saying. Consider the following: Is a
man who has two boys equally likely to start: "of my two children at least
one is a girl" as they are to start, "of my two children at least one is a
boy"?
Jim Balter
No SPAM wrote:
>
> Dennis Rakestraw <drake...@hotmail.com> wrote in article
> <882203102....@dejanews.com>...
> [stuff deleted]
>
> > If a man says, "of my two children at least one is a boy," what is the
> > probability that both are boys? The answer, oddly, is not "50-50."
> [more stuff deleted]
>
> A finer example of sexism in mathematics will ne'er be found.
>
> "of my two children at least one is a girl" as they are to start,
> "of my two children at least one is a boy", then the answer is
> indeed 50-50.
Bzzzt! Wrong. If the man (or woman) says "at least one boy",
then the chances are 1/3 that the other is a boy.
If the man (or woman) says "at least one girl",
then the chances are 1/3 that the other is a girl.
That's where the parity comes in.
If you want to avoid sexism, then just phrase the problem
differently every other time.
--
<J Q B>
Jim Balter
David Ullrich wrote:
>
> No SPAM wrote:
> >
> > Dennis Rakestraw <drake...@hotmail.com> wrote in article
> > <882203102....@dejanews.com>...
> > [stuff deleted]
> >
> > > If a man says, "of my two children at least one is a boy," what is the
> > > probability that both are boys? The answer, oddly, is not "50-50."
> > [more stuff deleted]
> >
> > A finer example of sexism in mathematics will ne'er be found.
> >
> > If men who have one boy and one girl are as likely to start,
> > "of my two children at least one is a girl" as they are to start,
> > "of my two children at least one is a boy", then the answer is
> > indeed 50-50.
>
> what under what conditions then it's no longer a strictly
> mathematical question. One needs to rephrase (or at least interpret)
> the question
>
> probability that both are boys?
>
>
> Given that a man has two children, at least one is a boy, what is the
> probability that both are boys?
>
Why not? Contrary to NO SPAM's claim, it doesn't matter how often
the man says "at least one boy" when he has both a boy and a girl.
*When* he says "at least one boy", the chances are 1/3 that the
other is a boy. Remember, it's a *conditional* probability.
It would matter, of course, if we knew he only made the statement when
he had a boy and a girl, or if we knew he only made it when he had two
boys, or if we knew he was lying or something. Under *those* sorts of
conditions, then indeed we have other information that would affect
the probability. But no amount of information about a counterfactual
such as how often he says "I have at least one girl" matters in the
least.
--
<J Q B>
x
No SPAM
William L. Bahn <ba...@pcisys.net> wrote in article
<01bd0a72$90cbb260$0400a8c0@BAHN>...
[much stuff deleted]
> How about this:
>
> If a person says "Of my two children, at least one of them is Gender A,
> what is the probability that both are of Gender A?
[more stuff deleted]
The probability is 1/2 because Gender A is not specific so that both the
Brown
and Smith families (see below) get to play this version of the game. If
you think
this is wrong, please explain why.
Family Children
Brown Boy, Boy
Johnson Boy, Girl
Jones Girl, Boy
Smith Girl, Girl
----------------------------------------------------------------------------
-----------------------------
Some people got the point of my first post and some were too upset by the
charge of
sexism to see it. I will try to clarify (and hopefully they will try to
calm down and understand).
Boy version: Of my two children, at least one is a boy, what is the
probability that both are boys?
Girl version: Of my two children, at least one is a girl, what is the
probability that both are girls?
It was stated that the answer to the boy version is 1/3. That is correct
under the usual assumptions.
But what are the consequences of those assumptions? Smith, the poor
mathematician with only daughters, never gets to play the game. Suppose
Smith does want to play this game; he can only
play the girl version. So the answer to the girl version is 1 because only
Smith plays the girl version.
If we assume that Johnson and Jones play both versions equally often, then
the answer to both
versions is 1/2.
For anyone having trouble with the conditioning of the probabilities,
assume that you're at a party
and are equally like to encounter Brown, Johnson, Jones, or Smith. No
matter which one you meet,
they will play the game with you (i.e., ask the question). But Brown can
only play the boy version and
Smith can only play the girl version. What do we assume about Johnson and
Jones? That they
always play the boy version? or that they play each version half the time?
--
Standard Disclaimers Apply.
Xcott Craver
William L. Bahn <ba...@pcisys.net> wrote:
[rant rant rant]
[rant rant rant]
[rant rant rant]
>> If men who have one boy and one girl are as likely to start,
>> "of my two children at least one is a girl" as they are to start,
>> "of my two children at least one is a boy", then the answer is
>> indeed 50-50.
>
>You are so intent of being able to make your claim of sexism, that you
>refuse to even look at what you are saying. Consider the following: Is a
>man who has two boys equally likely to start: "of my two children at least
>one is a girl" as they are to start, "of my two children at least one is a
>boy"?
Actually, he's correct here. If someone with a boy and a girl is
as likely to say "of my two children at least one is a girl," as to say,
"of my two children at least one is a boy," then the answer indeed becomes
50-50.
The answer is usually 2/3 because of the people who say
"of my two children at least one is a boy," 1/3 of them have 2 boys,
and 2/3 have one boy and one girl. If half of those 2/3 people
say "girl" instead of "boy," the sizes of the boy-boy and boy-girl
populations become equal, reducing the probability to 50%.
The fact that someone with two boys will never say "blah blah
blah at least one is a girl" keeps the remaining 1/3 of the population
the same whilst the other is chopped in half.
-Caj
>
>
>
>
>
William L. Bahn
William L. Bahn <ba...@pcisys.net> wrote in article
<01bd0a72$90cbb260$0400a8c0@BAHN>...
> No SPAM <nos...@apgea.army.mil> wrote in article
<snip>
> The answer is that the odds are 2/3 that both children are of the same
> gender as mentioned by the person asking the question.
<snip>
Brain Fart !!!
That should read 1/3 that both children are of the same gender (an increase
from 1/4). I just got to typing too quickly and failed to take a step back
and ask if the result I was reciting (without thinking about it) from
having worked the problem so many times in the past made sense (which,
obviously, it doesn't). No SPAM or anyone else can do a Deja News search to
verify that I have worked the problem numerous times and am not being
revisionist here.
Happy Holidays to all.
Xcott Craver
Jim Balter <j...@sandpiper.com> wrote:
>Why not? Contrary to NO SPAM's claim, it doesn't matter how often
>the man says "at least one boy" when he has both a boy and a girl.
>*When* he says "at least one boy", the chances are 1/3 that the
>other is a boy. Remember, it's a *conditional* probability.
Jim, look at it this way: you poll 1200 2-kid families, to find
that:
300 have 2 boys
300 have 2 girls
600 have one of each.
Now suppose (as NO SPAM said,) that dads with one of each are
equally likely to say "at least one girl" as "at least one boy."
Dads with only boys will always say "at least one boy," and dads
with only girls will say "at least one girl."
Our distribution is now:
300 have 2 boys, and say "at least one boy."
300 have 2 girls, and say "at least one girl."
300 have one of each, and say "at least one boy."
300 have one of each, and say "at least one girl."
Of the dads who say "at least one boy," 50% have both boys
and 50% have one of each. So, _if_ the dad says "at least one boy"
in this situation, we have a 50% chance the other is a boy.
This is caused by the set of 600 dads being reduced to half
its original size by the extra rule. This change affects the
conditional probability.
><J Q B>
-Caj
Darrell Ryan
No SPAM wrote:
>
> William L. Bahn <ba...@pcisys.net> wrote in article
> <01bd0a72$90cbb260$0400a8c0@BAHN>...
>
> > How about this:
> >
> > If a person says "Of my two children, at least one of them is Gender A,
> > what is the probability that both are of Gender A?
> [more stuff deleted]
>
> The probability is 1/2 because Gender A is not specific
Why is Gender A not specific? Gender A is *specifically* Gender A
(whatever that gender is). All that was done was rename male/female
to Gender A/Gender B (not necessarily respectively). We do not need
to know if Gender A is male or if it is female. All we need to know
is Gender A is Gender A and the other is Gender B and these are the
only two possible genders.
> so that both the
> Brown
> and Smith families (see below) get to play this version of the game. If
> you think
> this is wrong, please explain why.
>
> Family Children
> Brown Boy, Boy
> Johnson Boy, Girl
> Jones Girl, Boy
> Smith Girl, Girl
Maybe I missed something, but why are we talking about these different
families? Where do you get boys and girls? We're now talking about
Gender A and Gender B, with only *one* family and we know this *one*
family has at least one child (out of two) that is of Gender A.
We are told that one of the children is Gender A. Here's the sample
space of the three possible scenarios for the gender of the children:
scenario #1:
child#1 A
child#2 A
scenario#2:
child#1 A
child#2 B
scenario#3:
child#1 B
child#2 A
Clearly only one out of three scenarios results in both children being
of Gender A, so the probability is 1/3.
--
Darrell Ryan
Happy Holidays!
Visit my Xmas enhanced website at http://www.cafes.net/darrell/
Darrell Ryan
Xcott Craver wrote:
>
>
> Actually, he's correct here. If someone with a boy and a girl is
> as likely to say "of my two children at least one is a girl," as to say,
> "of my two children at least one is a boy," then the answer indeed becomes
> 50-50.
What the man says doesn't matter, as long as he is telling the truth.
If I had 99 boys and 1 girl, I may choose to say that at least one of
my children is a boy and be completely correct and this in no way,
shape, or form, changes the probability of the gender of another
child.
Yeah, if I have two kids, one of each gender, I could choose to say
that one of them is a girl the 1st time. Then I could choose to say
that one of them is a boy the 2nd time. Or I could choose to say one
of them is a boy 75% of the time. Or I could choose to say one of
them is a girl 75% of the time. That may not be likely, but it
ultimately doesn't matter what I say, as long as it's the truth. Just
because I may actually have one boy and one girl, that in no way shape
or form changes the *mathematical probability* of 1/3 to someone that
*doesn't know* I have one of each.
The only *relevent* information is I have two children, and at least
one of them is a boy. There are only three possible scenarios:
scenario #1
1st child is a boy
2nd child is a boy
scenario #2
1st child is a boy
2nd child is a girl
scenario #3
1st child is a girl
2nd child is a boy
In only one scenario, I have both boys, so the probability that I have
both boys is 1/3.
>
> The answer is usually 2/3 because of the people who say
> "of my two children at least one is a boy," 1/3 of them have 2 boys,
Yes, that's the question and 1/3 is the correct answer. That was the
question, right? If I tell you that one of my two children is a boy,
then there is a 1/3 probability that the other is also a boy, right?
Sean Case
In article <882203102....@dejanews.com>,
drake...@hotmail.com (Dennis Rakestraw) wrote:
>If a man says, "of my two children at least one is a boy," what is the
>probability that both are boys?
[proposed answer deleted]
Let's begin at the beginning. Who is this man in the problem, and why
is he telling us about his children? It's not the sort of remark one
makes off-hand. Does he mean he's not sure about the sex of the other
one? Does he disapprove of his child's lifestyle, or is he just
forgetful?
Normally, if one says something like "one of my children is a boy,"
there is something of an implication that the other child is a girl -
or at least something other than a boy. People who say "one of my
children is a boy, and so is the other" are harshly dealt with by
society. But in the problem statement, we have the remarkable phrase
"at least one." Why is this?
Well, the person posing the problem would really like to get the answer
1/3 - although they'd be quite happy to get 5/16 or 1/e or anything
other than the "obvious" answer. To get the surprise, we need to make
the two cases "boy-girl" and "girl-boy" indistinguishable. Now, children,
as you may have noticed, are not unlike sub-atomic particles. They can
only swap places like that when we aren't looking. Once you can tell
the difference between the children, the magic is lost. If we knew the
elder child to be male, then the sex of the younger would be 50-50 - but
the same applies if we know the sex of the taller child, the nearer
child, or even the more easterly child. Somehow, the problem must convey
the information that we have either one or two boys, without giving us
any other information. Let's try some examples:
Scenario 1: On visiting the house, we see a boy, but do not know whether
it is the older or the younger. Some older puzzle books have this version.
As pointed out above, it doesn't help. The probability that the "unseen"
child is a boy stays resolutely stuck at 1/2.
Scenario 2: On visiting the house, we see a boy's bicycle. (We choose to
ignore, for the purposes of the exercise, the fact that many people ride
bicycles of "inappropriate gender.") Assuming the bicycle to belong to an
individual, however, we are left knowing the sex of the bicycle owner,
which leaves us an even split on the sex of the other child.
Scenario 3: While interviewing the father, we ask whether both his
children are girls, and insist that he confine himself to "yes" or "no."
He answers in the negative, and we at last have a 1/3 probability that
both children are boys.
Scenario 4: In an interview, the father is asked to pick a true statement
from the following list:
a. At least one of my children is a boy.
b. At least one of my children is a girl.
As before, there are twice as many "mixed" families as "boy/boy" families,
but the fathers in the former group have a choice of either statement. If
half of them pick each statement, then the probability that both children
are boys, given that statement a was picked, is 1/2.
Let's just compare the last two. The subject makes the same statement in
each one. What is different is what the subject _didn't_ say, and I'm not
the first person ever to point out the importance of what is not said.
(Again with the cheap analogies to quantum mechanics!)
In each of the two cases, I have artificially constrained the other things
that the person might have chosen to say. This is purely to make the
calculation of probability possible. Note that we got the 1/3 answer by
restricting the options to either "it is true that at least one child is
a boy" or "it is false that..." I don't regard this as a natural situation
in terms of conversational English, but it has a certain appeal from the
standpoint of mathematics. This is how some people manage to convince
themselves that it is the "correct" interpretation. Personally, I find
option 4 more appealing, although it, too, is artificial.
To take yet another point of view, the hidden assumption that makes the
answer come out as 1/3 is that a father with one boy and one girl is just
as likely to come out with that blasted sentence as a father with two boys.
I think that this assumption is unjustified. I don't know what the relative
probabilities are, but I don't think it reasonable to assume that there
is no difference.
So, just as in "that game show problem," the whole question hinges on some
assumptions about what the people in the puzzle are up to. I have rarely
seen a convincing statement of the problem that genuinely forces the answer
1/3, but I think replacing the children with tossed coins would be a good
start.
Feel free to insert references to Shannon, Bayes, and so on, in appropriate
places above. It's getting late.
Sean Case
Dave Seaman
In article <34983933...@cafes.net>,
Darrell Ryan <dar...@cafes.net> wrote:
>Xcott Craver wrote:
>>
>>
>> Actually, he's correct here. If someone with a boy and a girl is
>> as likely to say "of my two children at least one is a girl," as to say,
>> "of my two children at least one is a boy," then the answer indeed becomes
>> 50-50.
>
>
>
>What the man says doesn't matter, as long as he is telling the truth.
>If I had 99 boys and 1 girl, I may choose to say that at least one of
>my children is a boy and be completely correct and this in no way,
>shape, or form, changes the probability of the gender of another
>child.
The only argument you made here is that the man's statement does not
change the probability. That happens to be true, but it is irrelevant
to the issue. To see why, it's necessary to apply the definition of
conditional probability.
Let H (the hypothesis) be the event that the man has exactly two
children, including at least one boy. Let A be the event that both
children are boys. Then the conditional probability is
P{A|H} = P{AH} / P{H}
= P{A} / P{H} (because A implies H, or AH = A)
= (1/4) / (3/4)
= 1/3. (1)
Now, let H' (the alternate hypothesis) be the event that the man SAYS
he has two children, including at least one boy. Let's also suppose
the man is truthful. That is, H' implies H, or HH' = H'. Then the
conditional probability is
P{A|H'} = P{AH'} / P{H'} (2)
but we now have a problem in simplifying this unless we have additional
information. For example, we can't write P{AH'} = P{A}, analogous to
above, because we were not told that A implies H' (we are not
guaranteed that the man will tell us he has at least one boy, even if
it's true). We are still assuming that the man tells the truth if he
makes a statement (HH' = H'), but that is no help in simplifying the
conditional probability (2).
In particular, if we suppose he is guaranteed to tell us either that he
has at least one boy, or that he has at least one girl, and that he
chooses at random if he has one of each, then we get P{AH'} = P{A}
(because event A means he has two boys, and our assumption in this
paragraph is that in that case he must therefore tell us that at least
one child is a boy). Under this assumption, we get
P{A|H'} = P{A} / P{H'}
= (1/4) / (1/2)
= 1/2. (3)
On the other hand, if we assume he is guaranteed to tell us he has at
least one boy if it happens to be true, then we get
P{A|H'} = P{A|H} (under the assumption H' = H)
= 1/3, by (1).
Therefore, we see that expression (2) is ambiguous. Under different
assumptions we get different values for P{A|H'}.
Notice that in all cases, the unconditional probability P{A} is always
1/4. Nobody's statements will change this in the least. Comparing
your arguments above, we see that everything you said about the
absolute probability P{A} is true, but your conclusion about the
conditional probability P{A|H'} is incorrect.
--
Dave Seaman dse...@purdue.edu
++++ stop the execution of Mumia Abu-Jamal ++++
++++ if you agree copy these lines to your sig ++++
++++ see http://www.xs4all.nl/~tank/spg-l/sigaction.htm ++++
David A Karr
Darrell Ryan <dar...@cafes.net> wrote:
>> > If a person says "Of my two children, at least one of them is Gender A,
>> > what is the probability that both are of Gender A?
>>
>> The probability is 1/2 because Gender A is not specific
>
>Why is Gender A not specific? Gender A is *specifically* Gender A
>(whatever that gender is).
Of my two children, at least one of them is Gender A.
What is the probability that *you* are Gender A?
If you don't know which gender (male or female) Gender A is, then I
haven't told you anything about my children in the statement above,
other than the fact there are two of them.
To put it another way, you meet a man of whom you have no prior
knowledge, and he says to you, "Of my two children, at least one of
them is Gender A." Now, assuming he isn't lying, one of the following
is true:
He has two boys, and Gender A is male.
He has a boy and a girl, and Gender A is male.
He has a boy and a girl, and Gender A is female.
He has two girls, and Gender A is female.
What's your assessment of the relative probabilities of those four
possibilities? Why?
David A. Karr
Darrell Ryan
Sean Case wrote:
>
> In article <882203102....@dejanews.com>,
> drake...@hotmail.com (Dennis Rakestraw) wrote:
>
> >If a man says, "of my two children at least one is a boy," what is the
> >probability that both are boys?
>
> [proposed answer deleted]
>
> Let's begin at the beginning. Who is this man in the problem, and why
> is he telling us about his children? It's not the sort of remark one
> makes off-hand. Does he mean he's not sure about the sex of the other
> one? Does he disapprove of his child's lifestyle, or is he just
> forgetful?
Of course, he's sure about the sex of both children! The point is the
*we* are not sure, and are asked to find the probability that both are
boys, given that one is a boy for sure. Forgetful? Disapprovement of
lifestyle? These are of no mathematical consequence. Mathematical
probability is not based on such biases or forgetfullness.
>
> Normally, if one says something like "one of my children is a boy,"
> there is something of an implication that the other child is a girl -
> or at least something other than a boy. People who say "one of my
> children is a boy, and so is the other" are harshly dealt with by
> society.
Again, there seems to be some confusion between proper socialogical
behavior and *mathematical* probability.
> But in the problem statement, we have the remarkable phrase
> "at least one." Why is this?
It means what it says literally. There are two kids, and at least one
is a boy. That means the other could be a boy or a girl. We don't
know, and are asked to find the *probability* that both are boys. Why
he said what he said is of no *mathematical* relevence at all. If
some wish to pick apart the reasons and the liklihood that he said "at
least one is a boy", that's fine, but understand that these issues are
not mathematical and do not change the mathematical probability of 1/3
that both are boys.
>
> Well, the person posing the problem would really like to get the answer
> 1/3 - although they'd be quite happy to get 5/16 or 1/e or anything
> other than the "obvious" answer. To get the surprise, we need to make
> the two cases "boy-girl" and "girl-boy" indistinguishable. Now, children,
> as you may have noticed, are not unlike sub-atomic particles. They can
> only swap places like that when we aren't looking. Once you can tell
> the difference between the children, the magic is lost. If we knew the
> elder child to be male, then the sex of the younger would be 50-50 - but
> the same applies if we know the sex of the taller child, the nearer
> child, or even the more easterly child. Somehow, the problem must convey
> the information that we have either one or two boys, without giving us
> any other information. Let's try some examples:
>
> Scenario 1: On visiting the house, we see a boy, but do not know whether
> it is the older or the younger. Some older puzzle books have this version.
> As pointed out above, it doesn't help. The probability that the "unseen"
> child is a boy stays resolutely stuck at 1/2.
>
> Scenario 2: On visiting the house, we see a boy's bicycle. (We choose to
> ignore, for the purposes of the exercise, the fact that many people ride
> bicycles of "inappropriate gender.") Assuming the bicycle to belong to an
> individual, however, we are left knowing the sex of the bicycle owner,
> which leaves us an even split on the sex of the other child.
Absolutely irrelevant! You are making a non-mathematical assumtion
based on something that is not even part of the problem.
>
> Scenario 3: While interviewing the father, we ask whether both his
> children are girls, and insist that he confine himself to "yes" or "no."
> He answers in the negative, and we at last have a 1/3 probability that
> both children are boys.
We are not allowed to ask such a question. We are are *told* that one
is a boy.
>
> Scenario 4: In an interview, the father is asked to pick a true statement
> from the following list:
>
> a. At least one of my children is a boy.
> b. At least one of my children is a girl.
Again, this is not allowed! We are simply told that one is a boy,
that's all.
>
> As before, there are twice as many "mixed" families as "boy/boy" families,
> but the fathers in the former group have a choice of either statement. If
> half of them pick each statement, then the probability that both children
> are boys, given that statement a was picked, is 1/2.
It may very well be that there exists twice as many mixed families as
boy-boy families. But understand what this is saying. If there are
twice as many mixed families as boy-boy families, what does that say
about the ratio of boy-boy families to mixed families? If two out of
three families are mixed, then 1 out of three are boy-boy, i.e. 1/3
will be boy-boy.
> To take yet another point of view, the hidden assumption that makes the
> answer come out as 1/3 is that a father with one boy and one girl is just
> as likely to come out with that blasted sentence as a father with two boys.
Irrelevant. If I have two kids, one boy one girl, I may for some
reason choose to say "at least one of my two children is a boy" 99
times out of 100. That does not affect the *mathematical probability*
that both are boys!
> So, just as in "that game show problem," the whole question hinges on some
> assumptions about what the people in the puzzle are up to.
The question does not hinge on any such assumptions. Such assumptions
are *not* mathematical, and have absolutely no bearing whatsoever on
the mathematical probability of both being boys!
> I have rarely
> seen a convincing statement of the problem that genuinely forces the answer
> 1/3, but I think replacing the children with tossed coins would be a good
> start.
OK, for some reason, people seem to be more likely to separate
themselves from any biases if we are dealing with objects instead of
people. Why are you not conviced the answer is 1/3? Does the same
problem done with coins (which have no bias nor make any
non-mathematical assumptions) somehow justify that the broblem about
the genders is somehow flawed, just because that real people may have
biases and make non-mathematical assumptions? Remember, our human
faults in no way, shape, or form, affect the true *mathematical*
probability. Am I the only one who sees this?
I flipped a coin exactly two times. Here are all the possible
scenarios:
Case #1
1st flip came up heads
2nd flip came up heads
Case #2
1st flip came up heads
2nd flip came up tails
Case #3
1st flip came up tails
2nd flip came up heads
Case #4
1st flip came up tails
2nd flip came up tails
I am also telling you that *at least* one of the two flips came up
heads. That eliminates Case#4 altogether. Out of the three cases
left, only one (case#1) has heads for both flips. The probability
that both came up heads given that at least one came up heads is
therefore 1/3. For reason unbeknownst to me, replacing "1st flip"
with "child#1", "2nd flip" with "child#2", "heads" with "male", and
"tails" with "female" all somehow changes the answer!
Dr D F Holt
In article <34983933...@cafes.net>,
Darrell Ryan <dar...@cafes.net> writes:
>Xcott Craver wrote:
>>
>>
>> Actually, he's correct here. If someone with a boy and a girl is
>> as likely to say "of my two children at least one is a girl," as to say,
>> "of my two children at least one is a boy," then the answer indeed becomes
>> 50-50.
>
>
>
>What the man says doesn't matter, as long as he is telling the truth.
Yes it does.
[stuff deleted]
>
>Yes, that's the question and 1/3 is the correct answer. That was the
>question, right? If I tell you that one of my two children is a boy,
>then there is a 1/3 probability that the other is also a boy, right?
>--
No, wrong. Why not just work it out mathematically, rather than treating
it like a topic for debate?
The situation we are considering is the following.
The man has two children - we all agree on that.
As usual, we assume that all children are equally likely to be male or female.
Let BB = probability that both children are boys. Clearly BB = 1/4.
Now the man makes a true statement of the form
"At least one my children is a XXXXX",
If he has two boys, he says "boy", if he has two girls he says "girl", and
if he has one of each, he says "boy" with probability p, and "girl" with
probability 1-p.
Overall, the probability 1B that he says
"At least one my children is a boy" is 1/4 + p/2.
So, if we are now given that he does in fact say
"At least one my children is a boy",
then the conditional probability CBB that he has two boys is BB/1B = 1/(1+2p),
which certainly depends on p.
For example, if p=1/2, then CBB = 1/2 (as Xcott Craver correctly said).
In the extreme cases, if p=0 then CBB=1 and if p=1 then CBB = 1/3 (which
is the standard over-simplistic answer to the problem).
Derek Holt.
Darrell Ryan
David A Karr wrote:
>
> Darrell Ryan <dar...@cafes.net> wrote:
> >> > If a person says "Of my two children, at least one of them is Gender A,
> >> > what is the probability that both are of Gender A?
> >>
> >> The probability is 1/2 because Gender A is not specific
> >
> >Why is Gender A not specific? Gender A is *specifically* Gender A
> >(whatever that gender is).
>
> Of my two children, at least one of them is Gender A.
> What is the probability that *you* are Gender A?
>
> If you don't know which gender (male or female) Gender A is, then I
> haven't told you anything about my children in the statement above,
> other than the fact there are two of them.
????????????
If you know that there are only two *possible* genders, namely gender
A and Gender B and that *at least* one is of Gender A, then...
Go back to calling them male and female if you wish. Same
difference. Only the names have changed.
>
> To put it another way, you meet a man of whom you have no prior
> knowledge, and he says to you, "Of my two children, at least one of
> them is Gender A." Now, assuming he isn't lying, one of the following
> is true:
>
> He has two boys, and Gender A is male.
> He has a boy and a girl, and Gender A is male.
> He has a boy and a girl, and Gender A is female.
> He has two girls, and Gender A is female.
You need to pick one or the other. Use Gender A/Gender B or
male/female but not both.
>
> What's your assessment of the relative probabilities of those four
> possibilities? Why?
The way you stated the possibilities is erronious for the reasons
given. You are somewhow concluding that Gender A can take on more
than one value (male or female). Rather, it is a constant value. It
is either male or female but not both!
>
> David A. Karr
John Rickard
Darrell Ryan <dar...@cafes.net> wrote:
: > >If a man says, "of my two children at least one is a boy," what is the
: > >probability that both are boys?
: Of course, he's sure about the sex of both children! The point is the
: *we* are not sure, and are asked to find the probability that both are
: boys, given that one is a boy for sure.
No, we're not. We're asked for the probability that both are boys
given that the man said "of my two children at least one is a boy".
If we assume that the man will say that if and only if he has two
children at least one of which is a boy, then these come to the same
thing; however, to many of us that seems a very unreasonable
assumption.
: I flipped a coin exactly two times.
[...]
: I am also telling you that *at least* one of the two flips came up
: heads.
[...]
: The probability that both came up heads given that at least one came
: up heads is therefore 1/3.
Agreed. It doesn't follow that that is the probability that both came
up heads given that you told us that at least one came up heads.
Consider this variant:
I flip a fair coin twice and tell you "At least one of the flips came
up ...", but the last word is drowned out by the noise of a passing
motorcycle. (Assume that I confirm, when you ask, that only one word
was lost and that it was either "heads" or "tails".) What is the
probability that the flips both came up the same way?
Now you ask me what I said, and I reply "At least one of the flips
came up tails". What is the probability now?
--
John Rickard <j...@atml.co.uk>
Xcott Craver
Darrell Ryan <dar...@cafes.net> wrote:
>
>> The answer is usually 2/3 because of the people who say
>> "of my two children at least one is a boy," 1/3 of them have 2 boys,
>
>question, right? If I tell you that one of my two children is a boy,
>then there is a 1/3 probability that the other is also a boy, right?
Well, put it this way: If you poll 4000 (or some large # of)
people with 2 kids, and ask, "is at least one of your 2 kids a boy?"
Then of the people who (truthfully) answer "yes," 1/3 of them will
have two boys. This is because you get the following expected
breakdown:
o 1000 have 2 boys (and answer "yes")
o 2000 have 1 boy (and answer "yes")
o 1000 have 0 boys (and do not answer "yes")
So if you take the population of people who agree that
one of their kids is a boy, the fraction of people with 2 boys is
(1000/(1000+2000)) = 1/3.
HOWEVER: Suppose you poll another 4000 2-kid parents, and this
time say, "Complete the following sentence with either `boy' or `girl':
`Of my two children, at least one is a [blank].'" Parents with 2
boys will always say "boy" and parents with 2 girls will always say "girl."
If we assume that parents with one of each choose either answer with
equal probability, we get the following expected breakdown:
o 1000 have 2 boys (and answer "boy")
o 1000 have 1 boy (and answer "boy")
o 1000 have 1 boy (and answer "girl")
o 1000 have 0 boys (and answer "girl")
So 2000 people tell you that of their two children, at least
one is a boy. Half of these people have two boys. The fraction is
now (1000/(1000+1000)) = 1/2. So is the conditional probability.
See, the reason you get 1/3 in the first case is because of
the relative (expected) sizes of both groups: one is twice as large as
the other. If you decide to assume that a percentage of one group doesn't
say,"of my 2 kids, at least one is a boy," then you're changing the
relative sizes of the groups, and hence the conditional probability.
-Xcott
Rainer Thonnes
In article <01bd0a72$90cbb260$0400a8c0@BAHN>,
"William L. Bahn" <ba...@pcisys.net> writes:
>
> It never ceases to amaze me the lengths people will go to in order to claim
> sexism or racism.
>
>
>
> The answer is that the odds are 2/3 that both children are of the same
> gender as mentioned by the person asking the question.
>
> Gender B.
Another key would be the circumstances in which the statement is made.
The probability can be either 1/3 or 1/2.
It would be 1/3 if a random person (chosen from a population of parents-of-two)
were specifically asked "One or both of your 2 children is a boy, true or
false?" and answered "true".
But suppose the father was nervously pacing up and down the corridor
in the maternity ward, awaiting details of his second child, only just
born. Clearly the only possibility excluded here is that the first child
is a girl. Therefore in this case the probability is 1/2.
--
I don't know whether there really is a ukol.com.
In any case I'm not there, I'm at dcs.ed.ac.uk.
Darrell Ryan
> >Xcott Craver wrote:
> >>
> >>
> >> Actually, he's correct here. If someone with a boy and a girl is
> >> as likely to say "of my two children at least one is a girl," as to say,
> >> "of my two children at least one is a boy," then the answer indeed becomes
> >> 50-50.
> >
> >
> >
> >What the man says doesn't matter, as long as he is telling the truth.
?????????
The fact that he actually *told* us pretty much guarantees that he
told us!
>
> Therefore, we see that expression (2) is ambiguous. Under different
> assumptions we get different values for P{A|H'}.
This is the problem! We are to not make any assumtions! We are told
that he has two kids, at least one being a boy. This is "given"
information, i.e. take it as truth without throwing in any monkey
wrenches.
>
> Notice that in all cases, the unconditional probability P{A} is always
> 1/4. Nobody's statements will change this in the least.
Given no additional information, that is true (see below). But we are
given the additional information that at at least one is a boy (also
see below). It is not mathematical to question whether or not this is
true or speculate as to why it was said. It is "given" information.
> Comparing
> your arguments above, we see that everything you said about the
> absolute probability P{A} is true, but your conclusion about the
> conditional probability P{A|H'} is incorrect.
Then my textbook's answer is also incorrect! This very problem is in
the book, worked out in complete detail. I usually don't quote
exactly from texts, but I will make exception on this occasion in hope
that (if nothing else) I have shown that at least one other person
agrees with me... even if the only other person is the author of the
text :-)
From Finite Mathematics, 3rd ed. Smith, Karl J.
Brooks/Cole
ISBN 0-534-16878-7
pages 299-300
***begin quote
Conditional Probability
The probability of an event depends on what information is known about
the event. For example, suppose you know that a family has two
children and you are interested in the probability that both children
are boys. This probability depends on additional information, as
illustrated by Example 1.
EXAMPLE 1
What is the probability that a family with two children has two boys
if:
a. You have no additional information.
b. You know there is at least one boy
c. You know that the youngest child is a boy.
SOLUTION
Let D = {2 boys]
E = {at least 1 boy}
F = {youngest is a boy}
a. Consider the sample space:
B G
/ \ / \
B G B G
Sample space BB BG GB GG <--- Four possibilities
^^
success
Thus, P(D) = 1/4
b. The sample space from part a is altered because of the additional
information:
Sample space BB BG GB XGGX <--- Three
possibilities
^^
success
This [meaning the GG possibility] is crossed out because we have
additional information that the family has at least one boy. This is
called altering the sample space.
We write P(D|E) to mean the proability of D given the addional
information that E has occured. Thus P(D|E) = 1/3.
***end of quote***
Darrell Ryan
> >Xcott Craver wrote:
> >>
> >>
> >> Actually, he's correct here. If someone with a boy and a girl is
> >> as likely to say "of my two children at least one is a girl," as to say,
> >> "of my two children at least one is a boy," then the answer indeed becomes
> >> 50-50.
> >
> >
> >
> >What the man says doesn't matter, as long as he is telling the truth.
>
>
> [stuff deleted]
> >
> >question, right? If I tell you that one of my two children is a boy,
> >then there is a 1/3 probability that the other is also a boy, right?
>
> No, wrong. Why not just work it out mathematically, rather than treating
> it like a topic for debate?
I have worked it out mathematically. The answer is 1/3.
>
> The situation we are considering is the following.
> The man has two children - we all agree on that.
> As usual, we assume that all children are equally likely to be male or female.
Agreed.
>
> Let BB = probability that both children are boys. Clearly BB = 1/4.
Agreed. There are 4 possibilities:
BB
BG
GB
GG
Only one of these possibilites yields both boys, so that probability
is 1/4.
>
> Now the man makes a true statement of the form
> "At least one my children is a XXXXX",
> If he has two boys, he says "boy", if he has two girls he says "girl", and
> if he has one of each, he says "boy" with probability p, and "girl" with
> probability 1-p.
????????????????????
Our job is not to consider what he actually has and use this
information to calculate any probability of him actually making the
statement. This is where the confusion lies. All he said was at
least one is a boy. Simply take the above sample space and throw out
anything that does not have at least one "B" in it. We are left with
only *three* possibilities, and only one of those remaining
possibilities has two Bs. Why is this not easily understood?
It is not our job to consider the probability of him making any one
specific statement. It is "given" information. You are complicating
the simple. It is simply our job to determine the *mathematical*
probability that both are boys and I have done just that.
>
> Overall, the probability 1B that he says
> "At least one my children is a boy" is 1/4 + p/2.
>
> So, if we are now given that he does in fact say
> "At least one my children is a boy",
...we were given that from the start. We were *told* that he has two
kids, with at least one being a boy. why this is so hard to
understand is unbeknownst to me.
Darrell Ryan
Xcott Craver wrote:
>
> Darrell Ryan <dar...@cafes.net> wrote:
> >
> >> The answer is usually 2/3 because of the people who say
> >> "of my two children at least one is a boy," 1/3 of them have 2 boys,
> >
> >question, right? If I tell you that one of my two children is a boy,
> >then there is a 1/3 probability that the other is also a boy, right?
>
> people with 2 kids, and ask, "is at least one of your 2 kids a boy?"
> Then of the people who (truthfully) answer "yes," 1/3 of them will
> have two boys.
...which really answers the original question, doesn't it? If in a
large number of fair trials, you get a result of 1/3, then the
PROBABILITY of any ONE man who has two kids and at least one is a boy
must also be 1/3. The only difference between the original problem
and what you suggest above is we are not asking the question. The man
is giving us the answer before the question is asked.
<...>
Darrell Ryan
Rainer Thonnes wrote:
> Another key would be the circumstances in which the statement is made.
>
> The probability can be either 1/3 or 1/2.
>
> It would be 1/3 if a random person (chosen from a population of parents-of-two)
> were specifically asked "One or both of your 2 children is a boy, true or
> false?" and answered "true".
This is the same as our scenario, only we are not asking the question
of a random person. The random person is telling us this information
before we ask it.
>
> But suppose the father was nervously pacing up and down the corridor
<...>
Why are we supposing any such thing? We are simply told by a person
that he has two kids and at least one is a boy. This is *random*
person because we are not given any evidence that suggests this is
*not* a random person.
David A Karr
Darrell Ryan <dar...@cafes.net> wrote:
>Dave Seaman wrote:
>> P{A|H'} = P{AH'} / P{H'} (2)
>>
>> but we now have a problem in simplifying this unless we have additional
>> information. For example, we can't write P{AH'} = P{A}, analogous to
>> above, because we were not told that A implies H' (we are not
>> guaranteed that the man will tell us he has at least one boy, even if
>> it's true).
>
>told us!
The fact that you wrote this pretty much guarantees that you didn't
understand any of Dave Seaman's math. Which is too bad, because
it was perfectly good math.
>Then my textbook's answer is also incorrect! This very problem is in
>the book, worked out in complete detail.
No, it isn't. At any rate it is nowhere in the part you quoted, of
which the relevant portion was:
>What is the probability that a family with two children has two boys
>if:
>
>a. You have no additional information.
>b. You know there is at least one boy
>c. You know that the youngest child is a boy.
The book solves problems (a) and (b) correctly (and presumably (c) as
well, but you omitted that part). Unfortunately, the problem that we
are trying to work on is
d. The father of the children has recently said, "At least one of my
children is a boy."
If you cannot see that (b) and (d) are different, it's not worth
continuing this conversation. If you can see the difference, then I
suggest you
1. study the concepts of prior probabilities and Bayes' rule, and then
2. reread Dave Seaman's analysis.
If the analysis still doesn't make sense, go back to step 1, and
repeat until you understand the math.
David A. Karr
Dave Seaman
In article <349985C0...@cafes.net>,
>> he has two children, including at least one boy. Let's also suppose
>> the man is truthful. That is, H' implies H, or HH' = H'. Then the
>> conditional probability is
>>
>>
>> but we now have a problem in simplifying this unless we have additional
>> information. For example, we can't write P{AH'} = P{A}, analogous to
>> above, because we were not told that A implies H' (we are not
>> guaranteed that the man will tell us he has at least one boy, even if
>> it's true).
>
>
>The fact that he actually *told* us pretty much guarantees that he
>told us!
If you are claiming that P{H'} = 1, then you are quite incorrect. That
would imply that either some fathers lie, or else the probability of
having two girls is zero, both of which are contrary to the assumptions
made. The fact that H' holds on a single trial does not prove P{H'} =
1, just as tossing a coin one time and having it come up heads does not
prove it is a two-headed coin.
The point of saying that "he told us", i.e. event H' has occurred, is
not to say that P{H'} = 1, but merely to say that we are to evaluate
the conditional probability P{A|H'}.
If you are claiming that H and H' are the same event, then you are
making an assumption not explicitly stated (yet). In fact, I made that
provisional assumption later on in order to show that the assumption
makes a difference in the result.
>> Therefore, we see that expression (2) is ambiguous. Under different
>> assumptions we get different values for P{A|H'}.
>
>This is the problem! We are to not make any assumtions! We are told
>that he has two kids, at least one being a boy. This is "given"
>information, i.e. take it as truth without throwing in any monkey
>wrenches.
And therefore, if it is possible to arrive at conflicting answers by
making assumptions that are consistent with what is given, it means the
problem is not well-posed and does not have a definite answer.
If you disagree, then let's see you derive a different answer for
P{A|H} = 1/3 by making different assumptions. You can't. Remember,
you can only make assumptions that are consistent with what is given.
That is where P{A|H'} differs from P{A|H}. The latter is a well-posed
problem, but the former is not.
>> Notice that in all cases, the unconditional probability P{A} is always
>> 1/4. Nobody's statements will change this in the least.
>
>Given no additional information, that is true (see below). But we are
>given the additional information that at at least one is a boy (also
>see below). It is not mathematical to question whether or not this is
>true or speculate as to why it was said. It is "given" information.
Then, according to your argument, the conditional probability P{A|Y}
should also be 1/3, where Y is the event that the YOUNGER child is a
boy. After all, Y implies H, and therefore your argument says that H
is a "given", and therefore the answer must be P{A|O} = P{A|H} = 1/3.
Wrong. In fact, P{A|Y} = 1/2, and your argument fails by not making
use of all the information that was provided. It's true that H is a
"given", but it's not the ONLY given.
By the same token, we are given H', a stronger hypothesis than H, but
your argument ignores the difference and assumes that P{A|H'} = P{A|H}
= 1/3. Again, wrong. Again, for the very same reason. You have
failed to make use of all the information that was provided.
>> Comparing
>> your arguments above, we see that everything you said about the
>> absolute probability P{A} is true, but your conclusion about the
>> conditional probability P{A|H'} is incorrect.
>
>Then my textbook's answer is also incorrect! This very problem is in
>the book, worked out in complete detail. I usually don't quote
>exactly from texts, but I will make exception on this occasion in hope
>that (if nothing else) I have shown that at least one other person
>agrees with me... even if the only other person is the author of the
>text :-)
No, your textbook is correct, as far as it goes.
>What is the probability that a family with two children has two boys
>if:
>
>a. You have no additional information.
>b. You know there is at least one boy
>c. You know that the youngest child is a boy.
Case (a) is the unconditional probability P{A}. Case (b) asks for the
conditional probability P{A|H}, and case (c) asks for P{A|Y}. Nowhere
is the case P{A|H'} even considered.
Case (b) really should say "You know there is at least one boy AND
NOTHING ELSE", since otherwise there would be a conflict between cases
(b) and (c). If you know the younger child is a boy, then you know
that at least one child is a boy, and therefore it would appear that
cases (b) and (c) should both apply simultaneously, giving conflicting
answers of 1/3 and 1/2. Add the "AND NOTHING ELSE" clause to (b), and
the apparent conflict disappears, because cases (b) and (c) are now
disjoint. P{A|Y} does not equal P{A|H}, even though Y implies H. The
problem is that H does not imply Y.
You have made the very same error in confusing case (b) with the
unstated case (d), "he TELLS you at least one child is a boy." Nowhere
is it stated that he had to tell you that, even if it happened to be
true. He could have told you instead that he had at least one girl, or
that he had two boys, or that he is going bowling tonight, or anything
else that came into his head, or nothing at all. The fact that he SAID
he has at least one boy violates the "AND NOTHING ELSE" clause, and
therefore the textbook's analysis for case (b) does not apply to case
(d). P{A|H'} need not equal P{A|H}, even though H' implies H. The
problem is that H does not imply H'.
Darrell Ryan
You miss my point. I suggest making *no* assumptions. Just use the
given information and *only* the given information.
>
> >> Notice that in all cases, the unconditional probability P{A} is always
> >> 1/4. Nobody's statements will change this in the least.
> >
> >Given no additional information, that is true (see below). But we are
> >given the additional information that at at least one is a boy (also
> >see below). It is not mathematical to question whether or not this is
> >true or speculate as to why it was said. It is "given" information.
>
> Then, according to your argument, the conditional probability P{A|Y}
> should also be 1/3, where Y is the event that the YOUNGER child is a
> boy. After all, Y implies H, and therefore your argument says that H
> is a "given", and therefore the answer must be P{A|O} = P{A|H} = 1/3.
> Wrong. In fact, P{A|Y} = 1/2, and your argument fails by not making
> use of all the information that was provided. It's true that H is a
> "given", but it's not the ONLY given.
If the problem was "I have two kids, and the younger one is a boy.
What's the probability that the older is also a boy?" I would approach
the problem essentially the same thing by 1st writing out the sample
space:
Case#1
older child B
younger child B
Case#2
older child G
younger child B
...and it's quite easy to see that this probability is 1/2 as you
say. I never said it wasn't and don't understand why you feel I
implied different. This is a different problem with a different
*altered* sample space.
???
Are you saying that unless we are specifically told, we are to assume
we have more information than the given information? You are implying
that because it didn't specifically say "NOTHING ELSE" then that
somehow means we knew more than he told us. Sorry, but I don't see
how.
No conflict. Part (b) tells us that one is a boy. Part (c) tells us
WHICH ONE is a boy. There is additional information included in part
(c) that is not included in part b. that does not mean they
"conflict." When considering part (c), the sample space is yet again
altered in addition to what was already altered for part (b).
Part d implies part b (in addition to providing extra information).
The altered sample space from part (b) is
BB BG GB
What is the extra information? The younger child is a boy, so we
throw away the last possibility of GB (with the convention that the
first letter represents the younger of the two)
We have left:
BB BG
of which there is 1/2 probability that the other is a boy. No
conflict there.
> If you know the younger child is a boy, then you know
> that at least one child is a boy, and therefore it would appear that
> cases (b) and (c) should both apply simultaneously, giving conflicting
> answers of 1/3 and 1/2. Add the "AND NOTHING ELSE" clause to (b), and
> the apparent conflict disappears, because cases (b) and (c) are now
> disjoint. P{A|Y} does not equal P{A|H}, even though Y implies H. The
> problem is that H does not imply Y.
>
> You have made the very same error in confusing case (b) with the
> unstated case (d), "he TELLS you at least one child is a boy." Nowhere
> is it stated that he had to tell you that, even if it happened to be
> true.
I agree, but the fact is he DID tell us that. We "know" he has one
boy.
> He could have told you instead that he had at least one girl, or
> that he had two boys,
...and of course those would be different problems because we have
different information.
> or that he is going bowling tonight, or anything
> else that came into his head, or nothing at all. The fact that he SAID
> he has at least one boy violates the "AND NOTHING ELSE" clause,
Say what? The fact that he SAID he has at least one boy somehow
violates "AND NOTHING ELSE" ???
If it violates it, then please explain just what else we are supposed
to know besides what he has told us!
I'm not trying to be hard to deal with, as I'm sure you're not
either. I just feel it's pretty obvious that no such clause is
necessary. To show it is necessary means to show that we somehow are
supposed to know more than what we are told.
Darrell Ryan
William L. Bahn wrote:
>
> Darrell Ryan <dar...@cafes.net> wrote in article
> <3499BB3E...@cafes.net>...
> > You miss my point. I suggest making *no* assumptions. Just use the
> > given information and *only* the given information.
>
> Which you can't do. On what basis, using ONLY the given information, can
> you determine that the probability of a particular child being a boy is
> equal to that child being a girl. That is an assumption that you have to
> make and which is not derivable from only the given information. If I
> remember correctly, the odds of a child being a girl are actually slightly
> higher due to a wider envelope of chemical environments in the womb that a
> female fetus can tolerate - I think it has to do more with environments
> that will kill the fetus as the transformation from female to male is
> occurring. At least this is what I recall from a science special I saw on
> PBS years ago and it may be an explanation that hasn't held up - but the
> demonstrated higher birth rate for girls than boys in the long term is a
> fact in most countries.
>
> We nearly always have to make assumptions of one kind or another. The big
> issue is how reasonable it is to make those particular assumptions. In this
> case, the assumption that the probabilities of a particular child being a
> boy or a girl are equal is quite reasonable.
Yes, I too agree that the "real world" probability of any one child
being a boy is not really 1/2. Ask any labor and delivery nurse.
You are taking me way too literally when I say not to assume
anything. For the purposes of this discussion, it was agreed that
this probably is 1/2. In other words, the sample space is:
B G
...and we are not considering any further "conditions" like the ones
you stated above. You raise an interesting and very valid point about
the real world feasibility of this result, but remember that those
conditions you pointed out are not being considered in this
discussion. If they are to be considered, we need to agree on an
actual figure of what that probability is. If we can't agree on an
actual figure, then the safest thing to say is the *mathematical*
probability of a boy is 1/2, but certain other non-mathematical
real-world factors may influence this result over the long haul.
That's why I and others have said the probability of having 2 boys
(given no other information) is 1/4 and it is so because each of the 4
possible ways to have 2 kids
BB BG GB GG
are assumed to be *equally likely* to occur. If it's not agreed that
they are each equally likely to occur, al bets are off.
>
> When the man says, "At least one of my children is a boy." He is giving us
> information. But we USE that information in order to exclude some possible
> states by asking under what circumstances the person could or could not
> make that statement (which we assume is true - another assumption that we
> are making with no justification based ONLY on the given information). But
> it is completely reasonable to look not only at what the information given
> both explicitly states and indirectly requires, but what information is
> implied.
What other information is implied? Seems pretty vague to me.
>
> And that is were the focus of this debate seems to be centered - on how
> reasonable it is to use information implied by the statement when different
> assumptions about what that information is are possible and lead to
> different conclusions.
In other words, the probability may be different from one person to
another, depending on what information each person has. My
point---the only information given was that he has two kids, and at
least one is a boy. If you want to consider some information that you
may be aware of, that's fine go ahead. But if that's the case, you
have *different* information than what the problem gives us. You are
assuming facts not in evidence.
> And is disregarding the information even an option?
Only if you want to consider something other than *mathematical*
probability.
>
> In this case we must make an assumption regarding the likelihood of the man
> making that particular statement from among other possible statements of
> similar vein that he could have made.
Why "must" we make such an assumption? All I'm simply saying is to
use the information literally as it is given. As you know,
probability is different depending on how much information we have. I
choose to simply take the information that is given, and not make any
non-mathematical assumptions about how the information was obtained.
> People have claimed (reasonably,
> IMHO) that a man that has one boy and one girl could just as easily of said
> that he has "at least one girl" instead of "at least one boy." But is this
> a reasonable assumption and are there any additional layers of assumptions
> being made? People have argued that the probability of him saying one or
> the other is 50/50. But why? Because you have two ways of relaying
> information. Or because there is one boy and one girl?
The *mathematical* probability of him saying one over the other is
50/50. Of course, being the imperfect human beings that we are, there
may be other *non*-mathematical factors that my influence this. What
I am suggesting is to simply assume the statement he makes to be a
true statement, and not consider any other information that we may
have (with the exception that a 50/50 chance of a boy in any one
birth).
>
> If we were talking about a men with five children - of the ones that have
> three boys and two girls, is it reasonable to assume that 60% of them would
> be expected to state that they have at least one boy and the other 40% to
> state that they have at least one girl? I would feel uncomfortable with
> making this assumption.
But in reality, we are a biased people like it or not. I don't have
any evidence to support any specific figures, but I can see that it
may very well be more likely for more than 50% of them to say "I have
a boy" simply because they have more boys than girls. But here's my
point that I think everyone is missing----as long as they are telling
the truth, we *must* consider all the possibilities having equally
likely chances of occuring. Is it reasonable for me to believe that
more of these people will say "I have a boy" ? Yes, it is
reasonable. Is it mathematical? No. Is it information that is
"given" in the problem? No.
Mathematical probability makes no "reasonable" assumptions. It makes
*mathemematical* assumptions. All this talk about the lilkihood of
the man saying what he said is not mathematically founded. Rather, it
is founded on our human, "reasonable" assumptions, and these
assumptions can vary from person to person.
<...>
Darrell Ryan
Xcott Craver wrote:
>
> Darrell Ryan <dar...@cafes.net> wrote:
> >b. The sample space from part a is altered because of the additional
> >information:
> >
> >Sample space BB BG GB XGGX <--- Three
> >possibilities
> > ^^
> > success
>
> Right. Now, NO-SPAM's variation: Suppose that a parent
> with one of each kid says, "at least one of my kids is a boy" 50% of
> the time, and "blah blah yak yak girl" the other 50%.
I don't know where this information was obtained, but it was not
obtained from the problem I read! What i read was something to the
effect of "I have 2 kids, at least one is a boy." Where does "a
parent with one of each kid" come from?
<...>
> The other two possibilities are crossed out because we have
> the additional information that *HALF* of each set (BG and GB) would
> never have told you that they had at least one boy.
Again, where did this information come from? Are we talking about the
same problem here?
Darrell Ryan
> >information to calculate any probability of him actually making the
> >statement.
>
> I'm not saying that the answer to the (well-known) problen is not 1/3,
> or that someone who answers 1/2 is justified in drawing some extra
> assumption about the unstated probability of saying "girl" instead of
> "boy". What I am saying is that, if someone _makes_ that assumption,
> that those who have a choice say "boy" p of the time and "girl" q of
> the time, it will make a difference in the final answer.
I'll buy that. But why are we making that assumtion? *Must* we make
such an assumption?
Do we agree that if you _don't_ make that assumption the probability
is 1/3? If so, what are we debating?
>
> There's the rub. Yes, you ARE given from the start that
> the fella says, "of my two kids, at least one is a boy." Nevertheless,
> if different kinds of fellas are more or less likely to say this,
> it effects the probability. This is what conditional probability is
> all about.
Yes! I agree. That is what conditional probability is all about.
But in our problem, all we are given is there are two kids, one being
a boy. This other information you are referring to is a matter of
opinion and not included in the given information.
>
> A simpler example: suppose men are just as likely as women
> to watch soap-operas. If some randomly chosen American says, "I sure
> do love them soap-operas," you can conclude, knowing nothing else,
> a 50% chance that this someone is a man. Dig?
You bet.
>
> [I am assuming a population 50% male. So sue me.]
>
> Okay, now suppose that of the men who watch soap-operas, only
> 1/10 of them will actually admit it (boy, am I a sexist brute.)
> Sure, you were *TOLD* that this someone told you "I sure do love them
> soap-operas." It's given to you from the start. However, this extra
> detail changes the size of the population of men who would state this,
> while the size of the corresponding female population remains unchanged.
> The probability of male-ness becomes 1/11. Agree? Disagree?
I'll agree to accept you answer of 1/11 without checking it. it's the
reasoning I'm more concerned with. In this case you have the
additional information that only 1/10 will actually admit it. Please,
please, PLEASE someone tell me where in the original problem is any
additional information given other than "I have two kids, at least one
is a boy." It just isn't there. Facts (opinions, rather) are being
assumed that are NOT in evidence. I'm thru shouting now. Sorry :-)
Darrell Ryan
John Rickard wrote:
>
> Darrell Ryan <dar...@cafes.net> wrote:
> : > >If a man says, "of my two children at least one is a boy," what is the
> : > >probability that both are boys?
> : Of course, he's sure about the sex of both children! The point is the
> : *we* are not sure, and are asked to find the probability that both are
> : boys, given that one is a boy for sure.
>
> given that the man said "of my two children at least one is a boy".
> If we assume that the man will say that if and only if he has two
> children at least one of which is a boy,
in other words, if we assume he is not lying.
> then these come to the same
> thing; however, to many of us that seems a very unreasonable
> assumption.
yes, to many that apparantly is unreasonable. the "why" part is what
doesn't convince me!
>
> : I flipped a coin exactly two times.
> [...]
> : I am also telling you that *at least* one of the two flips came up
> : heads.
> [...]
> : The probability that both came up heads given that at least one came
> : up heads is therefore 1/3.
>
> Agreed. It doesn't follow that that is the probability that both came
> up heads given that you told us that at least one came up heads.
So do you agree or not? You say "agreed" then you immediately say "it
doesn't follow..."
>
> Consider this variant:
>
> I flip a fair coin twice and tell you "At least one of the flips came
> up ...", but the last word is drowned out by the noise of a passing
> motorcycle. (Assume that I confirm, when you ask, that only one word
> was lost and that it was either "heads" or "tails".) What is the
> probability that the flips both came up the same way?
There are 4 ways to flip a coin twice:
Way#1
flip one h
flip two h
Way#2
flip one h
flip two t
Way#3
flip one t
flip two h
Way#4
flip one t
flip two t
Without knowing what one of the flips was (cause I didn't hear you), I
have to first consider the possibility that you said heads, then after
that consider the possibility that you said tails. Of course, it is
assumed you were equally likely to have said either one. If you said
heads, then Way#4 is out of the picture, leaving one of the three
remaining ways yielding the desired result. Probability - 1/3.
If you said tails, then Way#1 is out of the picture and Way#4 is back
in. Of the three remaining ways, only one gives the desired result.
Probability - 1/3.
So the answer to the question is 1/3, because no matter which one you
actually said, the probability is 1/3 regardless. All I need to know
is that you actually said one or the other.
>
> Now you ask me what I said, and I reply "At least one of the flips
> came up tails". What is the probability now?
Already answered from the above scenario. What is your point? In the
discussion at hand, we actually hear the man say "one is a boy." That
being the case, what is the relevence of the 1st scenario you bring
up, the one where we don't "hear" what was said? What makes scenario
#2, the one where you tell me tails, any different from the original
problem of the man with the boys?
Xcott Craver
Darrell Ryan <dar...@cafes.net> wrote:
>
>Our job is not to consider what he actually has and use this
>information to calculate any probability of him actually making the
>statement.
Well, it is our job if it makes a difference in the final answer.
I'm not saying that the answer to the (well-known) problen is not 1/3,
or that someone who answers 1/2 is justified in drawing some extra
assumption about the unstated probability of saying "girl" instead of
"boy". What I am saying is that, if someone _makes_ that assumption,
that those who have a choice say "boy" p of the time and "girl" q of
the time, it will make a difference in the final answer.
>This is where the confusion lies.
HELL yeah.
>> So, if we are now given that he does in fact say
>> "At least one my children is a boy",
>
>...we were given that from the start. We were *told* that he has two
>kids, with at least one being a boy. why this is so hard to
>understand is unbeknownst to me.
There's the rub. Yes, you ARE given from the start that
the fella says, "of my two kids, at least one is a boy." Nevertheless,
if different kinds of fellas are more or less likely to say this,
it effects the probability. This is what conditional probability is
all about.
A simpler example: suppose men are just as likely as women
to watch soap-operas. If some randomly chosen American says, "I sure
do love them soap-operas," you can conclude, knowing nothing else,
a 50% chance that this someone is a man. Dig?
[I am assuming a population 50% male. So sue me.]
Okay, now suppose that of the men who watch soap-operas, only
1/10 of them will actually admit it (boy, am I a sexist brute.)
Sure, you were *TOLD* that this someone told you "I sure do love them
soap-operas." It's given to you from the start. However, this extra
detail changes the size of the population of men who would state this,
while the size of the corresponding female population remains unchanged.
The probability of male-ness becomes 1/11. Agree? Disagree?
>Darrell Ryan
-Caj
Xcott Craver
No, your textbook is simply talking about a different problem.
than is NO-SPAM. Allow me to illustrate:
>***begin quote
>
>b. You know there is at least one boy
>b. The sample space from part a is altered because of the additional
>information:
>
>Sample space BB BG GB XGGX <--- Three
>possibilities
> ^^
> success
Right. Now, NO-SPAM's variation: Suppose that a parent
with one of each kid says, "at least one of my kids is a boy" 50% of
the time, and "blah blah yak yak girl" the other 50%.
The sample space is altered because of the additional information.
Half of the middle two sets are parents who would not have told you
that "at least one of my kids is a boy," just like all the parents
who both have girls. As such, they are excluded from the sample space:
1/4 1/4 1/4 1/4
BB BG GB XGGX <--- Three out of Four
| / \ / \ |
BB BG XBGX GB XGBX XGGX
1/4 1/8 1/8 1/8 1/8 1/4
The other two possibilities are crossed out because we have
the additional information that *HALF* of each set (BG and GB) would
never have told you that they had at least one boy.
P(D|E) = 1/4 / (1/4+1/8+1/8) = 1/2.
>***end of quote***
-Xcott
William L. Bahn
Darrell Ryan <dar...@cafes.net> wrote in article
<3499BB3E...@cafes.net>...
> Dave Seaman wrote:
> >
> > In article <349985C0...@cafes.net>,
> > Darrell Ryan <dar...@cafes.net> wrote:
> > >Dave Seaman wrote:
> > >> Now, let H' (the alternate hypothesis) be the event that the man
SAYS
> > >> he has two children, including at least one boy. Let's also suppose
> > >> the man is truthful. That is, H' implies H, or HH' = H'. Then the
> > >> conditional probability is
> > >>
> > >> P{A|H'} = P{AH'} / P{H'} (2)
> > >>
> > >> but we now have a problem in simplifying this unless we have
additional
> > >> information. For example, we can't write P{AH'} = P{A}, analogous
to
> > >> above, because we were not told that A implies H' (we are not
> > >> guaranteed that the man will tell us he has at least one boy, even
if
> > >> it's true).
> > >
<snip>
> > >
> > >This is the problem! We are to not make any assumtions! We are told
> > >that he has two kids, at least one being a boy. This is "given"
> > >information, i.e. take it as truth without throwing in any monkey
> > >wrenches.
> >
> > And therefore, if it is possible to arrive at conflicting answers by
> > making assumptions that are consistent with what is given, it means the
> > problem is not well-posed and does not have a definite answer.
> >
> > If you disagree, then let's see you derive a different answer for
> > P{A|H} = 1/3 by making different assumptions. You can't.
>
> You miss my point. I suggest making *no* assumptions. Just use the
> given information and *only* the given information.
Which you can't do. On what basis, using ONLY the given information, can
you determine that the probability of a particular child being a boy is
equal to that child being a girl. That is an assumption that you have to
make and which is not derivable from only the given information. If I
remember correctly, the odds of a child being a girl are actually slightly
higher due to a wider envelope of chemical environments in the womb that a
female fetus can tolerate - I think it has to do more with environments
that will kill the fetus as the transformation from female to male is
occurring. At least this is what I recall from a science special I saw on
PBS years ago and it may be an explanation that hasn't held up - but the
demonstrated higher birth rate for girls than boys in the long term is a
fact in most countries.
We nearly always have to make assumptions of one kind or another. The big
issue is how reasonable it is to make those particular assumptions. In this
case, the assumption that the probabilities of a particular child being a
boy or a girl are equal is quite reasonable.
When the man says, "At least one of my children is a boy." He is giving us
information. But we USE that information in order to exclude some possible
states by asking under what circumstances the person could or could not
make that statement (which we assume is true - another assumption that we
are making with no justification based ONLY on the given information). But
it is completely reasonable to look not only at what the information given
both explicitly states and indirectly requires, but what information is
implied.
And that is were the focus of this debate seems to be centered - on how
reasonable it is to use information implied by the statement when different
assumptions about what that information is are possible and lead to
different conclusions. And is disregarding the information even an option?
If we choose to disregard the information altogether, can not the claim be
made that we are really just making whatever assumption about the implied
information that doesn't happen to change the outcome? As the RUSH song
says, "You can choose not to decide, but you still have made a choice."
In this case we must make an assumption regarding the likelihood of the man
making that particular statement from among other possible statements of
similar vein that he could have made. People have claimed (reasonably,
IMHO) that a man that has one boy and one girl could just as easily of said
that he has "at least one girl" instead of "at least one boy." But is this
a reasonable assumption and are there any additional layers of assumptions
being made? People have argued that the probability of him saying one or
the other is 50/50. But why? Because you have two ways of relaying
information. Or because there is one boy and one girl?
If we were talking about a men with five children - of the ones that have
three boys and two girls, is it reasonable to assume that 60% of them would
be expected to state that they have at least one boy and the other 40% to
state that they have at least one girl? I would feel uncomfortable with
making this assumption.
And if we are going to consider implied information based upon what WAS
said versed what COULD have been said, don't we need to be a lot more
careful to categorize what all of the possibilities are? After all, if we
are going to claim implied information because the parent of one boy and
one girl could also have said that they have at least one girl. Then don't
we also have to claim implied information because the parent of two boys
could also have said that they have no girls or that they have all boys?
And this means that we also have to define what possible statements would
be considered "of similar vein."
Is the following a good enough list? This is the list most people seem to
be working from.
BB: I have at least one boy.
BG: I have at least one boy.
I have at least one girl.
GB: I have at least one boy.
I have at least one girl.
GG: I have at least one girl.
But what about this list.
BB: I have no girls.
I have at least one boy.
I have all boys.
My youngest is a boy.
My oldest is a boy.
BG: I have at least one boy.
I have at least one girl.
My youngest is a boy.
My oldest is a boy.
My youngest is a girl.
My oldest is a girl.
GB: I have at least one boy.
I have at least one girl.
My youngest is a boy.
My oldest is a boy.
My youngest is a girl.
My oldest is a girl.
GG: I have no boys.
I have at least one girl.
I have all girls.
My youngest is a girl.
My oldest is a boy.
I can think of situations in which any of these statements would be a
reasonable thing for a parent to say - but I have no idea what the relative
probabilities are and I would think that those probabilities are highly
context sensitive. I would doubt that the parent would say "at least" very
often because that type of exactitude is rare in common usage. But the
phrase "I have a son," spoken by a person that has one or more sons is not
uncommon - again, there are contexts where it is perfectly reasonable to
assume that it means "at least one" and others where it is only reasonable
to assume it means "exactly one".
Given the list of potential statements above, and assuming that each is
equally likely for a given parent to choose relative to the allowed
alternatives, I would come to the conclusion that the probability that a
parent that states that they have at least one boy actually has two boys to
be 5/17 or 0.294.
Another way to approach this is to assume that there are so many different
ways that each parent could state this kind of information, that in the
limit you assume that each is equally likely. That leads directly back to
the 1/3 probability of both children being boys. (So this is where I'll
hitch my horse for now.)
But even before we get into all of this, we need to ask if the fact that a
parent is presenting this information is fair game to be analyzed at all.
Isn't THE most reasonable assumption to make that the person posing the
problem wants us to determine the conditional probability based on the
knowledge that at least one of the two children is a boy and the story
about a man making that particular statement is just a schtick to relay the
information as opposed to simply spelling it out. Just like the old "Two
trains leave a station at the same time ...." story problems. It wasn't
reasonable (unless the course material was related to this) to ask about
whether the clocks at the two stations were set to same master clock or if
one train might be considered to "have left" the moment it's wheel's start
turning and the other only after it's caboose has cleared station property.
Making the claim that "you can't assume anything" is a cop-out. In
virtually any problem someone could come up with a reason to say, "There is
no answer based on the information provided." The challenge is to decide
what assumption MUST be made (because they affect the conclusion) and
whether there is a sufficiently reasonable way to make that assumption.
That should include a statement to the point that "This is under the only
reasonable assumption that ....." or "This is under the particular
assumption that ..." And of course, sometimes the statement. "This problem
has no reasonable answer because the impact of unstated assumptions is too
strong." is sometimes itself the most reasonable course to take.
Just my $1.98
John Rickard
Darrell Ryan <dar...@cafes.net> wrote:
: John Rickard wrote:
[...]
: > If we assume that the man will say that if and only if he has two
: > children at least one of which is a boy,
:
: in other words, if we assume he is not lying.
No. If one has a boy and a girl, it's not lying to say "of my two
children at least one is a girl", or to say nothing at all.
: > : The probability that both came up heads given that at least one came
: > : up heads is therefore 1/3.
: >
: > Agreed. It doesn't follow that that is the probability that both came
: > up heads given that you told us that at least one came up heads.
:
: So do you agree or not? You say "agreed" then you immediately say "it
: doesn't follow..."
I agree that the probability given that at least one came up heads is
1/3; I don't agree that the probability given that *you told us* that
at least one came up heads is 1/3.
: > I flip a fair coin twice and tell you "At least one of the flips came
: > up ...", but the last word is drowned out by the noise of a passing
: > motorcycle. (Assume that I confirm, when you ask, that only one word
: > was lost and that it was either "heads" or "tails".) What is the
: > probability that the flips both came up the same way?
[...]
: So the answer to the question is 1/3, because no matter which one you
: actually said, the probability is 1/3 regardless. All I need to know
: is that you actually said one or the other.
OK, this is consistent, depending on the assumptions you make. (For
example, it is correct under the assumption that I decide beforehand
what I might say, and then either say it or not depending on whether
it turns out to be true.)
But change the scenario slightly: before I flip, I tell you that after
I flip I will say either "At least one of the flips came up heads" or
"At least one of the flips came up tails" (but not both). Then
everything proceeds as before. Do you agree that in this case the
probability is 1/2?
--
John Rickard <j...@atml.co.uk>
QSCGZ
> If a man says, "of my two children at least one is a boy," what is the
> probability that both are boys?
The question is not clearly enough formulated.
It is important, where and when and why that statement is made.
If you choose all these statements done in 1997 in USA ,
(other countries e.g. India might be different)
exclude lies and families with >2 children,
and then do a statistics, I guess, you get a value for BB between
1/3 and 1/2 , more close to 1/3 .
The point is : how likely is a BB , BG ,GB parent to make
that statement . I suppose, the poser of the question assumes
that they are equally likely , but that's not absolutely clear.
-------------------------------------------------------
If you choose all those B-statements given at a
boys-school-class-parents-meeting (e.g. stated to some childs
that were invited too, and that didn't know, that it's
a boys-school) , then the statistics would be close to 1/2 ,
as BB parents are about twice as likely to be there as BG or GB
parents.
------------------------------------
if all 2 child families _have to_ do a B or G statement,
then 1/2 of the B statements are from BB families.
-----------------------
> Scenario 2: On visiting the house, we see a boy's bicycle. (We choose to
> ignore, for the purposes of the exercise, the fact that many people ride
> bicycles of "inappropriate gender.") Assuming the bicycle to belong to an
> individual, however, we are left knowing the sex of the bicycle owner,
> which leaves us an even split on the sex of the other child.
I don't agree.I would estimate, that in BB families the children more
often have to share _one_ bicycle than in GB or BG families.
So, I guess the answer is about ~ 0.4 here .
Dave Seaman
In article <3499BB3E...@cafes.net>,
Darrell Ryan <dar...@cafes.net> wrote:
>Dave Seaman wrote:
>>
>> In article <349985C0...@cafes.net>,
>> Darrell Ryan <dar...@cafes.net> wrote:
>> >Dave Seaman wrote:
>> >> P{A|H'} = P{AH'} / P{H'} (2)
>> >> Therefore, we see that expression (2) is ambiguous. Under different
>> >> assumptions we get different values for P{A|H'}.
>> >
>> >This is the problem! We are to not make any assumtions! We are told
>> >that he has two kids, at least one being a boy. This is "given"
>> >information, i.e. take it as truth without throwing in any monkey
>> >wrenches.
No, the given information is that the man *says* he has two kids, at
least one being a boy. In other words, the hypothesis H'.
You keep assuming that H' = H, but nowhere in the problem statement
does it say that. Without making assumptions that are not contained in
the problem, it is not possible to arrive at an answer. The problem is
ambiguous.
>You miss my point. I suggest making *no* assumptions. Just use the
>given information and *only* the given information.
Then why do you keep assuming that H' = H? Are you saying that only
*you* are allowed to make assumptions here, but nobody else?
Let's make things a bit more explicit. There are four possible
distributions of the sexes for two children: BB, BG, BG, and GG. I
hope you won't mind if we make some additional assumptions, even though they
are not stated in the problem. The assumptions I propose are:
1. The man does in fact have exactly two children.
(The problem doesn't really tell us this -- only that he SAYS
he has exactly two children).
2. P{BB} = P{BG} = P{GB} = P{GG} = 1/4.
(Notice that I am not talking about conditional
probabilities here, and therefore the fact that he SAYS he
has at least one boy does not make the a priori probability
P{GG} = 0. We'll get to the conditional probabilities in a
minute.)
Applying Bayes' Theorem, we find that the conditional probability of
the man having two sons on the hypothesis that he says he has at least
one son, is
P{H'|BB} P{BB}
P{BB|H'} = -----------------------------------------------------------------
P{H'|BB} P{BB} + P{H'|BG} P{BG} + P{H'|GB} P{GB} + P{H'|GG} P{GG}
P{H'|BB}
= -----------------------------------------
P{H'|BB} + P{H'|BG} + P{H'|GB} + P{H'|GG}
where I have used assumption (1) in listing the four possibilities, and
assumption (2) to simplify the expression. Without further assumptions, no
further simplification is possible.
At this point, you would no doubt like to add the additional assumptions
3. P{H'|BB} = 1,
4. P{H'|BG} = 1,
5. P{H'|GB} = 1,
6. P{H'|GG} = 0.
In other words, assumptions 3-5 say that the man is certain to tell us
he has at least one son if the statement is true, and assumption 6 says
he is certain NOT to tell us that if the statement is false. Assumptions 3-5
together say that H implies H', while assumption 6 says H' implies H. Putting
it all together, assumptions 3-6 say that H = H', which is the assumption you
have been making all along without showing us where it is stated in the
problem.
With all the assumptions 1-6 operative, we may deduce that the answer
we seek is
P{BB|H'} = 1 / (1+1+1+0) = 1/3,
but without those assumptions we can't deduce an answer.
>> >> Notice that in all cases, the unconditional probability P{A} is always
>> >> 1/4. Nobody's statements will change this in the least.
>> >
>> >Given no additional information, that is true (see below). But we are
>> >given the additional information that at at least one is a boy (also
>> >see below). It is not mathematical to question whether or not this is
>> >true or speculate as to why it was said. It is "given" information.
You are confusing conditional probability with unconditional
probability here. You are talking about the former, while I was
talking about the latter. Looking back at the enumerated assumptions
above, assumption (2) is about unconditional probability, while
assumptions (3)-(6) and the final conclusion are all about conditional
probability. If you insist on confusing the two concepts, you are
going to make it very difficult to carry on a conversation.
>> Then, according to your argument, the conditional probability P{A|Y}
>> should also be 1/3, where Y is the event that the YOUNGER child is a
>> boy. After all, Y implies H, and therefore your argument says that H
>> is a "given", and therefore the answer must be P{A|O} = P{A|H} = 1/3.
^
|
typo - should be Y
>> Wrong. In fact, P{A|Y} = 1/2, and your argument fails by not making
>> use of all the information that was provided. It's true that H is a
>> "given", but it's not the ONLY given.
>...and it's quite easy to see that this probability is 1/2 as you
>say. I never said it wasn't and don't understand why you feel I
>implied different. This is a different problem with a different
>*altered* sample space.
Correct. I didn't say you were actually making that argument for the
P{A|Y} case, but only that you were making the equivalent incorrect
argument for the P{A|H'} case.
Suppose a hypothetical person X is making the following argument:
If the younger child is a boy, then it certainly follows that
at least one child is a boy. That is a "given" in the
problem. We have already agreed that the conditional
probability of two sons, given that there is at least one son,
is 1/3. Specifying that the younger child is a boy does not
make the hypothesis false. Therefore, the probability is still
1/3.
What would you say to convince person X that his argument is
incorrect? Write down your argument. Then change each occurrence of
hypothesis Y in your argument to H', and read what you wrote. The
resulting argument should convince you that your reasoning about
P{A|H'} (or P{BB|H'}, in the notation I used when applying Bayes'
Theorem) is incorrect.
Xcott Craver
Darrell Ryan <dar...@cafes.net> wrote:
>
>Again, where did this information come from? Are we talking about the
>same problem here?
No, we're talking about a variation on the problem, which was
pointed out by NO SPAM. I believe you were well aware of this, considering
your previous article:
=====================CUT=======================================
Xcott Craver wrote:
>
>
> Actually, he's correct here. If someone with a boy and a girl is
> as likely to say "of my two children at least one is a girl," as to say,
> "of my two children at least one is a boy," then the answer indeed becomes
> 50-50.
What the man says doesn't matter, as long as he is telling the truth.
If I had 99 boys and 1 girl, I may choose to say that at least one of
my children is a boy and be completely correct and this in no way,
shape, or form, changes the probability of the gender of another
child.
=====================CUT========================================
As this variation shows, what the man says *does* matter.
-Caj
Xcott Craver
William L. Bahn <ba...@pcisys.net> wrote:
> Darrell Ryan wrote:
>> You miss my point. I suggest making *no* assumptions. Just use the
>> given information and *only* the given information.
>Which you can't do.
I wouldn't consider this a debate over the fairness or
meaningfulness of the original problem, but rather a look at one
interesting (and clearly distinct) variation. It is impossible
for a puzzle to disclaim away every assumption that would change
its answer, and unrealistic to criticize the problem poser for
stating the one obvious answer as correct.
> On what basis, using ONLY the given information, can
>you determine that the probability of a particular child being a boy is
>equal to that child being a girl. That is an assumption that you have to
>make and which is not derivable from only the given information.
Similarly, a newly minted US Penny is biased towards one side
because of the surface features. Nevertheless, when a puzzle speaks
of flipping a fair coin, the obvious meaning is that there is an equal
probability of either side.
On what basis can one determine that the parents aren't lying,
that single-child families are more or less likely to have another child
if their first is a girl, or that people with two girls are mysteriouslyj
less likely to ever mention their children?
The answer: on the basis that this is a story problem, and the
original interpretation is pretty damn obvious. That's why the problem
has been printed in countless books for countless years by countless
expositors of mathematical games, and not one ever thought it necessary
to disclaim any one of these. Had Marilyn vos Savant not mentioned it,
it would likely last centuries more without any focus on the starting
assumptions.
College students who disagree are invited to write wrong answers
to simple questions on Prob&Stat exams, and try to convince their profs
to give them their points back on the basis of some undisclaimed
interpretation of the problem. "You didn't explicitly state that this fair
coin was not a fair *double-headed* coin, or that it was flipped by an
unfair person, so the probability could have been 0. Or 1/3. Or 3/4.
Or Pi/4. Or that it was Monty Hall, who really uses subtle psychological
manipulation (which magically changes the objective probabilities),
who flipped the coin, offering me $100 if he could flip it again."
-Caj
Darrell Ryan
John Rickard wrote:
>
> Darrell Ryan <dar...@cafes.net> wrote:
> : John Rickard wrote:
> [...]
> : > If we assume that the man will say that if and only if he has two
> : > children at least one of which is a boy,
> :
> : in other words, if we assume he is not lying.
>
> No. If one has a boy and a girl, it's not lying to say "of my two
> children at least one is a girl", or to say nothing at all.
...and I never said that it would be a lie to say that. As far as
"saying nothing at all," that's another story! Remember, you said if
and only if he has two children and at least one boy, then he WILL
tell us. A lie of omission is still a lie. Allow me to include a
little more of your quote than you included:
> No, we're not. We're asked for the probability that both are boys
> given that the man said "of my two children at least one is a boy".
> If we assume that the man will say that if and only if he has two
> children at least one of which is a boy,
...and I replied:
> : in other words, if we assume he is not lying.
Well, if the man said "of my two children at least one is a boy" and
we assume he'll say that if and only if that actually is the case,
then isn't it fairly obvious that he *won't* say it if it's not the
case? You said "if and only if" which means just what it says. If
it's true, he'll say it and ONLY if it's true will he say it. That
implies if it is NOT true, he will NOT say it (i.e. he won't lie to
us). You already assumed that he WOULDN't say it unless it was
actually so. All I did was state that the same thing can be said by
assuming he ain't lying! If he ain't lying, then he's telling the
truth, right? Saying nothing at all in this case would mean he is
really lying by ommission, if it's true that he actually DOES have at
least one boy because you said if that's the case he WILL tell us (if
and only if).
>
> I agree that the probability given that at least one came up heads is
> 1/3; I don't agree that the probability given that *you told us* that
> at least one came up heads is 1/3.
Then please explain whow that would effect the probability. How is
that any diferent from you actually witnessing one of the tosses? How
is that any different from a third party that witnessed one of the
tosses telling you the result? Isn't the only relevent infomation
here the information of knowing what one of the tosses was? Why does
it make one bit of difference where you get this information, as long
as it is accurate information? Remember, nothing in the problem gives
us any additional information other than one of the tosses came up
heads and the coin was tossed exactly two times. If you want to
consider some other factors not explicitely stated in the problem,
i.e. credability, liklihood of me saying what I said, etc. then GO
AHEAD! But if you do consider anything else, then you are working on
a DIFFERENT problem than I am because you are using additional
information that I am not using!
I'm not really a teacher, but for the sake of argument let's say I
am. Put it to you this way, If I posed the question on a test:
Given that a person tells you s(he) tosses a fair coin exactly two
times, and also tells you that at least one of the tosses comes up
heads: What is the probability that both tosses come up heads?
An answer of anything other than 1/3 is incorrect. I do not want to
hear the student bringing in additional information other than what
was given. If I wanted the student to consider any additional
information, I would have included the information!
I can just imagine it now, getting answers such as "Well, depending on
the air current in the room and the amount of force applied from the
act of tossing, and the liklihood that someone would bother to give me
this information in the 1st place, etc. etc. etc." They may know
exactly what they are talking about, but the point is this is a
DIFFERENT PROBLEM than the one posed. It is different because someone
is assuming additional conditions OTHER than the given conditions. As
you know, conditional probability depends on what information we
have. In order for any two people to agree on the answer to a
conditional probability problem, it *must* be agreed upon as to what
information we have!
>
> : > I flip a fair coin twice and tell you "At least one of the flips came
> : > up ...", but the last word is drowned out by the noise of a passing
> : > motorcycle. (Assume that I confirm, when you ask, that only one word
> : > was lost and that it was either "heads" or "tails".) What is the
> : > probability that the flips both came up the same way?
> [...]
> : So the answer to the question is 1/3, because no matter which one you
> : actually said, the probability is 1/3 regardless. All I need to know
> : is that you actually said one or the other.
>
> OK, this is consistent, depending on the assumptions you make. (For
> example, it is correct under the assumption that I decide beforehand
> what I might say, and then either say it or not depending on whether
> it turns out to be true.)
Yes, you can do that. But examine both cases. If it's true and you
choose to tell me, then I have a probability problem to calculate (as
was already done with a result of 1/3). If it is true and you choose
NOT to tell me, then I really don't have any probability problems to
calculate, now do I? Unless, of course, I can read your mind and
choose to do the problem.
>
> But change the scenario slightly: before I flip, I tell you that after
> I flip I will say either "At least one of the flips came up heads" or
> "At least one of the flips came up tails" (but not both). Then
> everything proceeds as before. Do you agree that in this case the
> probability is 1/2?
???
How many diferent scenarios are you gonna put me thru? If the point
ain't been made by now, I doubt my point will get across by doing one
more scenario. I'm trying my best to answer your questions, but you
are not answering mine! Again, I ask how the scenario of the coin
tosses is any different from that of the man with the two kids. You
have not answered that as of yet. Instead, you just keep throwing me
different "what if" scenarios.
Xcott Craver
Darrell Ryan <dar...@cafes.net> wrote:
>
>I'll buy that. But why are we making that assumtion? *Must* we make
>such an assumption?
We are making the assumption because I said, "If we make
this assumption, then blah blah yak yak." No, we don't have to make
such an assumption, and the original problem doesn't make it.
>Do we agree that if you _don't_ make that assumption the probability
>is 1/3? If so, what are we debating?
I agree. The only thing under debate locally is that it
would make a difference if we threw in this assumption (that it would
make a difference was denied by several people). Nobody's claiming
that this assumption is part of the original problem.
I do think that some people are trying to argue that
this difference renders the original problem meaningless, because
this assumption wasn't explicitly denied; but I think that's a bit
extreme, and only being argued because Marilyn vos Savant mentioned
the puzzle in Parade magazine. People often assume she is wrong when
the publishes a surprising answer (like 1/3), and when she is vindicated
the typical behavior is to mutter something about the problem not being
stated clearly, or having alternate interpretations.
>Darrell Ryan
-Xcott
William L. Bahn
From reading this post, it is pretty clear that you are confused about what
"if and only if" means. The statement "... the man will say that (namely,
that he has at least one boy) IF AND ONLY IF he has two children, at least
one of which is a boy," means that the man is PROHIBITIED from saying he
has at least one girl in the event that he has one of each. Because if he
has one of each, he MUST state that he has at least one boy in order to
satisfy the "if" portion of the clause. The "only if" portion is the part
that stipulates that the statement, if said, is true.
When you respond "in other words, if we assume he is not lying," you are
NOT restating the same thing. If the above assumption just said "ONLY IF"
then you could have rephrased it to say, "If the mans says that at least
one of his two children is a boy, then he is telling the truth." When you
throw in the "if and only if" condition, it becomes a two way requirement.
IF he says he has at least one boy, then he has at least one boy PLUS the
condition that IF he has at least one boy, he will say he has at least one
boy and therefore is not allowed to say another statement such as 'I have
at least one girl' when he has one boy and one girl.
At points in this post, you yourself indicate that you agree with this -
When you said:
You said "if and only if" which means just what it says. If
it's true, he'll say it and ONLY if it's true will he say it. That
You explicitly state the understanding that this "if and only if"
assumption demands that if it is true, that he will say it. But earlier you
claimed that the "if and only if" assumption was the same as the "not
lying" assumption and when John pointed out that saying that he had one
girl in a GB or BG situation was a violation of the "if and only if"
assumption but not the "not lying" assumption you agreed that it was not a
violation of the "not lying" assumption. However, your own words confirm
that it IS a violation of the "if and only if" assumption and if is
violates one and not the other, then the two ways of stating the assumption
are not equivalent.
Darrell Ryan <dar...@cafes.net> wrote in article
Dave Seaman
In article <67ednd$6a2$1...@gannett.math.niu.edu>,
>>is 1/3? If so, what are we debating?
>
> I agree.
I don't agree. If you don't make any assumptions, the problem as
stated has no answer. I listed six additional assumptions that make it
possible to conclude the probability is 1/3.
1. The man is telling the truth when he says he has exactly two kids,
2. P{BB} = P{BG} = P{GB} = P{GG} = 1/4,
3. P{H'|BB} = 1,
4. P{H'|BG} = 1,
5. P{H'|GB} = 1,
6. P{H'|GG} = 0,
where H' is the event that the man says at least one child is a boy.
More generally, if we drop assumptions 3-6, the correct answer is
P{H'|BB}
P{BB|H'} = -----------------------------------------
P{H'|BB} + P{H'|BG} + P{H'|GB} + P{H'|GG}
where P{H'|BB} is the conditional probability that the man says he has
at least one son, on the hypothesis that he actually has two sons.
Similarly for the other conditionals.
Darrell Ryan
> >> >This is the problem! We are to not make any assumtions! We are told
> >> >that he has two kids, at least one being a boy. This is "given"
> >> >information, i.e. take it as truth without throwing in any monkey
> >> >wrenches.
>
> No, the given information is that the man *says* he has two kids, at
> least one being a boy. In other words, the hypothesis H'.
What is the logical difference between "we are told that he has two
kids" and "the man says he has two kids."? In order to be told
something, someone has to say something, right?
>
> You keep assuming that H' = H, but nowhere in the problem statement
> does it say that. Without making assumptions that are not contained in
> the problem, it is not possible to arrive at an answer. The problem is
> ambiguous.
Forgive my laziness, but it has slipped my mind just exactly what you
mean by H and H'. I'll go back and look later today when I get more
time. At any rate, I never said H'=H. that must be a conclusion you
are making based on something else I may have said. Again, let me go
back and study just what was said and the context in which it was
said.
>
> >You miss my point. I suggest making *no* assumptions. Just use the
> >given information and *only* the given information.
>
> Then why do you keep assuming that H' = H? Are you saying that only
> *you* are allowed to make assumptions here, but nobody else?
I'm not saying that. You are taking me way too literally when said
make no assumtions.
>
> Let's make things a bit more explicit. There are four possible
> distributions of the sexes for two children: BB, BG, BG, and GG. I
> hope you won't mind if we make some additional assumptions, even though they
> are not stated in the problem. The assumptions I propose are:
>
> 1. The man does in fact have exactly two children.
> (The problem doesn't really tell us this -- only that he SAYS
> he has exactly two children).
Sure, it's reasonable to assume he's not lying.
>
> 2. P{BB} = P{BG} = P{GB} = P{GG} = 1/4.
> (Notice that I am not talking about conditional
> probabilities here, and therefore the fact that he SAYS he
> has at least one boy does not make the a priori probability
> P{GG} = 0. We'll get to the conditional probabilities in a
> minute.)
Agreed. This assumes each case is equally likely to occur. BB BG GB
GG is the original, unaltered sample space, i.e. the non-conditional
probability.
>
> Applying Bayes' Theorem, we find that the conditional probability of
> the man having two sons on the hypothesis that he says he has at least
> one son, is
OK, then H' must be the hypothesis that he says he has two kids and at
least one son. Sorry, I forgot that earlier. I guess H is the
hypothesis that he actually has two kids and at least one son.
>
> P{H'|BB} P{BB}
> P{BB|H'} = -----------------------------------------------------------------
> P{H'|BB} P{BB} + P{H'|BG} P{BG} + P{H'|GB} P{GB} + P{H'|GG} P{GG}
>
> P{H'|BB}
> = -----------------------------------------
> P{H'|BB} + P{H'|BG} + P{H'|GB} + P{H'|GG}
>
> where I have used assumption (1) in listing the four possibilities, and
> assumption (2) to simplify the expression. Without further assumptions, no
> further simplification is possible.
I'm sure your math is correct.
>
> At this point, you would no doubt like to add the additional assumptions
>
> 3. P{H'|BB} = 1,
> 4. P{H'|BG} = 1,
> 5. P{H'|GB} = 1,
> 6. P{H'|GG} = 0.
>
> In other words, assumptions 3-5 say that the man is certain to tell us
> he has at least one son if the statement is true, and assumption 6 says
> he is certain NOT to tell us that if the statement is false. Assumptions 3-5
> together say that H implies H', while assumption 6 says H' implies H. Putting
> it all together, assumptions 3-6 say that H = H', which is the assumption you
> have been making all along without showing us where it is stated in the
> problem.
If I understand correctly, you are saying that I believe just because
he has two kids, with one being a boy, then he *must* tell us this.
Therefore H implies H'. I do not believe that. Nothing said he must
tell us this. Of course, if he does tell us this (assuming he ain't a
liar), then H' implies H. That is, if he tells us he has two kids
with one being a boy, then he actually has two kids and one boy. I do
not understand why you believe that I feel H implies H'. Again, I'll
go back and look and see if I can determine what I said that gave you
that idea.
>
> With all the assumptions 1-6 operative, we may deduce that the answer
> we seek is
>
> P{BB|H'} = 1 / (1+1+1+0) = 1/3,
>
> but without those assumptions we can't deduce an answer.
I must admit, I am somewhat rusty with Baye, but I don't see how we
*have* to assume that the man *must* tell us this information in order
to conclude a conditional proability of 1/3. Whether he had to or
not, he did tell us. If he didn't tell us, then we don't even have a
problem to consider. Or at least, it would be a considerably
different problem.
>
> >> >> Notice that in all cases, the unconditional probability P{A} is always
> >> >> 1/4. Nobody's statements will change this in the least.
> >> >
> >> >Given no additional information, that is true (see below). But we are
> >> >given the additional information that at at least one is a boy (also
> >> >see below). It is not mathematical to question whether or not this is
> >> >true or speculate as to why it was said. It is "given" information.
>
> You are confusing conditional probability with unconditional
> probability here. You are talking about the former, while I was
> talking about the latter. Looking back at the enumerated assumptions
> above, assumption (2) is about unconditional probability, while
> assumptions (3)-(6) and the final conclusion are all about conditional
> probability. If you insist on confusing the two concepts, you are
> going to make it very difficult to carry on a conversation.
I don't think I'm confusing them at all. I started with the original
unaltered sample space (the unconditional probability) and if I was to
stop right there, the probability of a BB situation is 1/4. I then
considered the *condition* that at least one child is a boy, thus
eliminating part of the sample space. How that can be interpreted as
confusion between unconditional and conditional probability is beyond
me :-)
>
> >> Wrong. In fact, P{A|Y} = 1/2, and your argument fails by not making
> >> use of all the information that was provided. It's true that H is a
> >> "given", but it's not the ONLY given.
>
> >...and it's quite easy to see that this probability is 1/2 as you
> >say. I never said it wasn't and don't understand why you feel I
> >implied different. This is a different problem with a different
> >*altered* sample space.
>
> Correct. I didn't say you were actually making that argument for the
> P{A|Y} case, but only that you were making the equivalent incorrect
> argument for the P{A|H'} case.
Again, I'l have to go back and review earlier posts to see why you
feel this way.
>
> Suppose a hypothetical person X is making the following argument:
>
> If the younger child is a boy, then it certainly follows that
> at least one child is a boy. That is a "given" in the
> problem. We have already agreed that the conditional
> probability of two sons, given that there is at least one son,
> is 1/3. Specifying that the younger child is a boy does not
> make the hypothesis false. Therefore, the probability is still
> 1/3.
>
> What would you say to convince person X that his argument is
> incorrect? Write down your argument.
OK, I'll just write it here as I have done before. This ain't Baye,
but it is still perfectly legitimate. 1/3 is not the correct answer
to that problem. True, if the younger is a boy, then at least one is
a boy.
original unaltered nonconditional sample space:
BB BG GB GG
If at least one is a boy, then throw away GG:
BB BG GB
If we were to stop right there, then yes the proability of BB would be
1/3. But we can't stop there. We are given that the younger is a
boy. (I'm using the 1st letter in each pair to represent the
younger). So, throw away the GB:
BB BG
...and probability of BB is now 1/2. This is how I would explain it.
Specifying that the younger is a boy does not change the fact that at
least one is a boy. That's true. But that does not mean the
probality is 1/3 just because the first hypothesis is still valid and
it's probability is 1/3. The second statement that the younger is a
boy, although it does include the hypothesis that at least one is a
boy, does not in and of itself mean that we can just say "therefore
the probability is 1/3." The second statement implies the first, but
the first statement does not imply the second. Isn't this pretty easy
to see? The second statement implies the first, in addition to
implying something else. It's the "something else" that make the two
statements different and results in further altering of the sample
space.
I'm not going to plug in any H', because H' and H are not the same
thing, remember?
Darrell Ryan
> >such an assumption?
>
> We are making the assumption because I said, "If we make
> this assumption, then blah blah yak yak." No, we don't have to make
> such an assumption, and the original problem doesn't make it.
Thank you, that's what I thought. So the answer to the *original*
problem is 1/3.
>
> >Do we agree that if you _don't_ make that assumption the probability
> >is 1/3? If so, what are we debating?
>
> I agree. The only thing under debate locally is that it
> would make a difference if we threw in this assumption (that it would
> make a difference was denied by several people). Nobody's claiming
> that this assumption is part of the original problem.
...and all I'm questioning is *why* we are throwing in these
assumptions. I think I now know the answer to that.
>
> I do think that some people are trying to argue that
> this difference renders the original problem meaningless, because
> this assumption wasn't explicitly denied; but I think that's a bit
> extreme, and only being argued because Marilyn vos Savant mentioned
> the puzzle in Parade magazine. People often assume she is wrong when
> the publishes a surprising answer (like 1/3), and when she is vindicated
> the typical behavior is to mutter something about the problem not being
> stated clearly, or having alternate interpretations.
Thank you. So the answer of 1/3 is "vindicated." I feel better now.
Darrell Ryan
William L. Bahn wrote:
>
> From reading this post, it is pretty clear that you are confused about what
> "if and only if" means. The statement "... the man will say that (namely,
> that he has at least one boy) IF AND ONLY IF he has two children, at least
> one of which is a boy," means that the man is PROHIBITIED from saying he
> has at least one girl in the event that he has one of each.
Why is he probited? Nothing about "if and only if" implies he cannot
make another statement if he chooses. The "if and only if" clause is
applying to the statement about one being a boy.
> Because if he
> has one of each, he MUST state that he has at least one boy in order to
> satisfy the "if" portion of the clause.
Yes.
> The "only if" portion is the part
> that stipulates that the statement, if said, is true.
>
Yes.
> When you respond "in other words, if we assume he is not lying," you are
> NOT restating the same thing. If the above assumption just said "ONLY IF"
> then you could have rephrased it to say, "If the mans says that at least
> one of his two children is a boy, then he is telling the truth."
Yes.
> When you
> throw in the "if and only if" condition, it becomes a two way requirement.
> IF he says he has at least one boy, then he has at least one boy PLUS the
> condition that IF he has at least one boy, he will say he has at least one
> boy
Yes.
> and therefore is not allowed to say another statement such as 'I have
> at least one girl' when he has one boy and one girl.
?
Why is this not allowed? There was no "if then" clause associated
with any such statement about a girl! You argument would be valid if
we had something like the following:
We're asked for the probability that both are boys given that the man
said "of my two children at least one is a boy." We assume the man
will say that and NOTHING ELSE if and only if he has two children, and
at least one is a boy.
Nothing about the context in which the original if-then clause was
used suggests that the man was prohibited from saying he had one
girl. Below is a copy of that paragraph, I believe written by John
Rickard:
_____CUT_____
> No, we're not. We're asked for the probability that both are boys
> given that the man said "of my two children at least one is a boy".
> If we assume that the man will say that if and only if he has two
> children at least one of which is a boy,
_____CUT_____
Nothing in the above paragraph implies that he is prohibited from
making any such statement about a girl, a dog, the weather, or
anything else.
>
> At points in this post, you yourself indicate that you agree with this -
> When you said:
>
> You said "if and only if" which means just what it says. If
> it's true, he'll say it and ONLY if it's true will he say it. That
Yeah, I said that, but how does that somehow imply that I think he is
prohibited from saying anything about a girl? it just ain't there.
What is the "it" that I refer to? It's the statement that the man has
two kids, and at least one is a boy.
>
> You explicitly state the understanding that this "if and only if"
> assumption demands that if it is true, that he will say it. But earlier you
> claimed that the "if and only if" assumption was the same as the "not
> lying" assumption and when John pointed out that saying that he had one
> girl in a GB or BG situation was a violation of the "if and only if"
> assumption but not the "not lying" assumption you agreed that it was not a
> violation of the "not lying" assumption. However, your own words confirm
> that it IS a violation of the "if and only if" assumption and if is
> violates one and not the other, then the two ways of stating the assumption
> are not equivalent.
Here's what I said:
_____CUT_____
> > > No. If one has a boy and a girl, it's not lying to say "of my two
> > > children at least one is a girl", or to say nothing at all.
> >
> > ...and I never said that it would be a lie to say that. As far as
> > "saying nothing at all," that's another story! Remember, you said if
> > and only if he has two children and at least one boy, then he WILL
> > tell us. A lie of omission is still a lie. <snip>
_____CUT_____
Can't you see that I am pointing out two things:
1. It ain't a lie to say he has a girl, if he actually has one.
2. If he says NOTHING, but in reality he does have a boy, then the
if-then clause is violated, i.e. it is a lie of omission.
Nothing I said suggests that the man is prohibited from saying he has
a girl. Just as long as he also says he has a boy, if applicable by
the if-then clause. When I equate the situation to one of "not lying"
all I mean is he does not violate the if-then clause. I don't know
where you get the idea that somehow this is supposed to prevent him
from saying any true statements about any girls that he may have.
William L. Bahn
> Xcott Craver wrote:
> >
> > Darrell Ryan <dar...@cafes.net> wrote:
> > >
> > >I'll buy that. But why are we making that assumtion? *Must* we make
> > >such an assumption?
> >
> > We are making the assumption because I said, "If we make
> > this assumption, then blah blah yak yak." No, we don't have to make
> > such an assumption, and the original problem doesn't make it.
>
> Thank you, that's what I thought. So the answer to the *original*
> problem is 1/3.
The point remains that the only way to NOT make that assumption, is to make
another assumption in its place. Either way you MUST make an assumption.
Someone else could say, "Assuming that their is a probability p that the
child is neither boy or girl, but is a transsexual (I think that's the
correct term)." And everyone could debate whether that was a reasonable
assumption and most people would pretty strongly argue that the problem
should be worked using only boys and girls with equal probability of each.
But in denying the transsexual assumption, we have replaced it with another
assumption - that there is NO possibility of having a child that is
distinctly categorized as either a boy or a girl. There is not enough
information in the original problem statement to be able to say something
like, "the original problem doesn't make that assumption." The original
problem pretty much has to make one assumption or the other (or some other
alternative assumption). This means that the person working the problem has
to make SOME assumption, even if it is an unconscious one. Once again, it
all comes down to what is reasonable - which covers the gammit from
obviously reasonable to obviously unreasonable with a big, gray, debatable
area in between.
William L. Bahn
> William L. Bahn wrote:
> >
> > From reading this post, it is pretty clear that you are confused about
what
> > "if and only if" means. The statement "... the man will say that
(namely,
> > that he has at least one boy) IF AND ONLY IF he has two children, at
least
> > one of which is a boy," means that the man is PROHIBITIED from saying
he
> > has at least one girl in the event that he has one of each.
>
> Why is he probited? Nothing about "if and only if" implies he cannot
> make another statement if he chooses. The "if and only if" clause is
> applying to the statement about one being a boy.
Let's try this again:
Rule: The man will say, "I have two children and at least one of them is a
boy." if and only if the man has two children, at least one of which is a
boy.
A man having a boy and a girl makes the statement, "I have two children and
at least one of them is a girl."
He has violated the Rule because the rule states that IF he has two
children, at least one of which is a boy, then he WILL say "I have two
children and at least one of them is a boy."
The underlying assumption that I, and probably most other people have been
making, is that the man is going to make exactly one statement regarding
his children. This assumption stems from the fact that the discussion
involves the probability of the man saying one of two things - That he has
at least one boy or that he has at least one girl.
To try to clear this up:
If these are the rules:
RULE A: A man will walk into the room make one or more statements regarding
his children. From that (those) statement(s) you are to determine the
probabilities of the various possible combinations of gender for his
children.
RULE B: The man will say, "I have two children and at least one of them is
a boy." if and only if the man has two children, at least one of which is a
boy.
then:
If a man walks into the room and says, "I have at two children and at least
one of them is a girl." You know that both of his children are girls.
If these are the rules:
RULE A: A man will walk into the room make one or more statements regarding
his children. From that (those) statement(s) you are to determine the
probabilities of the various possible combinations of gender for his
children.
RULE B: The man will not lie.
then:
If a man walks into the room and says, "I have at two children and at least
one of them is a girl." You only know that both children are not boys. The
probabilities of the other possible combinations is a function of the
probabilities you assign to him making that particular statement under each
combination.
The nit you have been hanging your hat on is that the man with a boy and a
girl could still say "I have at least one girl." PROVIDED that this is IN
ADDITION to making the pronouncement that he has at least one boy.
Because if he says "at least one girl." Then he failed to say, "at least
one boy."
If he walks in an says the following statement when he has one boy and one
girl, "I have two children, at least one of which is a girl." AND NOTHING
ELSE. Then he has violated the "if and only if" rule but he HAS NOT
violated your "not lying" rule.
We don't have to restrict him to only making a single statement, the fact
remains that he CAN make that one single pronouncement and in doing so
violate one rule and not the other. Therefore the two rules ARE NOT
equivalent.
Why must "we assume he is not lying" force him to tell us that he has at
least one boy in a BG or GB situation?
Darrell Ryan
William L. Bahn wrote:
> The point remains that the only way to NOT make that assumption, is to make
> another assumption in its place. Either way you MUST make an assumption.
> Someone else could say, "Assuming that their is a probability p that the
> child is neither boy or girl, but is a transsexual (I think that's the
> correct term)." And everyone could debate whether that was a reasonable
> assumption and most people would pretty strongly argue that the problem
> should be worked using only boys and girls with equal probability of each.
> But in denying the transsexual assumption, we have replaced it with another
> assumption - that there is NO possibility of having a child that is
> distinctly categorized as either a boy or a girl.
I assume you mean "neither" a boy or a girl.
> There is not enough
> information in the original problem statement to be able to say something
> like, "the original problem doesn't make that assumption." The original
> problem pretty much has to make one assumption or the other (or some other
> alternative assumption). This means that the person working the problem has
> to make SOME assumption, even if it is an unconscious one.
I agree wholeheartedly. As a silly (yet legitimate) example, I assume
that the man actually wants me to answer his question.
> Once again, it
> all comes down to what is reasonable - which covers the gammit from
> obviously reasonable to obviously unreasonable with a big, gray, debatable
> area in between.
That sounds reasonable to me. I just disagree somewhat with some of
these assumptions and whether or not it is "reasonable" that they be
introduced.
If someone asked me what is 1+1, I would probably assume they are
using base 10 numbers and give an answer accordingly, although it is
technically not wrong of me to say "not enough information." It would
just be unreasonable of me to say that, IMO.
This has been an interesting discussion, but I feel I really have
nothing further to say that hasn't already been said. I want to thank
all those I have corresponded with. If nothing else, I have
determined that my Usenet access seems to be functioning properly :-)
Dr D F Holt
In article <34998AA8...@cafes.net>,
Darrell Ryan <dar...@cafes.net> writes:
>> >
>> >Yes, that's the question and 1/3 is the correct answer. That was the
>> >question, right? If I tell you that one of my two children is a boy,
>> >then there is a 1/3 probability that the other is also a boy, right?
>> >--
>
>It is not our job to consider the probability of him making any one
>specific statement. It is "given" information. You are complicating
>the simple. It is simply our job to determine the *mathematical*
>probability that both are boys and I have done just that.
I see that we are no longer in disagreement about the mathematical
arguments involved. But I do disagree with your belief that 1/3 is
the only reasonable answer to the original problem. I think that
there is simply not enough information given in the simple
statement "A man 2 children and at least one is a boy" to
arrive at an unambiguous solution. The point of the different
assumptions is just to show that with perfectly reasonable
assumptions, different solutions are possible, and I think that
some of these possibilities are just as reasonable as the assumptions you
are making that lead to an answer 1/3 (which basically amount to the fact
that the possibilties BB, BG and GB are all equally likely).
Let me give another example of a scenario which can lead to the
probability 1/2, which actually happened to me once.
Somebody told me that Mr X had two children, but he had no idea
whether they were boys or girls. A few days later, a girl came up to
me and said "Hi, you don't know me, but I'm X's daughter."
It occurred to me that I now knew that X had two children and that at
least one was a girl, but the probability that they were both girls was
clearly 1/2.
But I am not arguing that 1/2 is THE answer to the problem, just that
there is not enough information given for there to be a unique
correct answer.
Derek Holt.
QSCGZ
> Somebody told me that Mr X had two children, but he had no idea
> whether they were boys or girls. A few days later, a girl came up to
> me and said "Hi, you don't know me, but I'm X's daughter."
> It occurred to me that I now knew that X had two children and that at
> least one was a girl, but the probability that they were both girls was
> clearly 1/2.
even this scenario is not clearly formulated.
When and why were you told that Mr X has two children ?
Why came the girl up to you ? Why did she say that ?
And why did you remember that, when you did this posting ?
I mean , are GG girls just as likely to do that as BG or GB girls ?
And couldn't it even happen, that both children visit you
_together_ ? In this case you probably wouldn't have told us about
this scenario at all !
But I would guess, averaging all events, whenever this scenario
occurred in England the last years, that 1/2 would be a good
approximation to the statistics.
Christian Bau
In article <349ACB3B...@cafes.net>, Darrell Ryan <dar...@cafes.net>
wrote:
> Dave Seaman wrote:
> >
> > In article <3499BB3E...@cafes.net>,
> > Darrell Ryan <dar...@cafes.net> wrote:
>
> > >> >This is the problem! We are to not make any assumtions! We are told
> > >> >that he has two kids, at least one being a boy. This is "given"
> > >> >information, i.e. take it as truth without throwing in any monkey
> > >> >wrenches.
> >
> > No, the given information is that the man *says* he has two kids, at
> > least one being a boy. In other words, the hypothesis H'.
>
> What is the logical difference between "we are told that he has two
> kids" and "the man says he has two kids."? In order to be told
> something, someone has to say something, right?
If the father or mother has the choice to give either the information "at
least one boy" or the information "at least one girl" or to give no
information, then his or her choice gives you additional information.
You ask a man who has two children to choose a true statement from the
following list:
1. At least one of my children is a boy.
2. At least one of my children is a girl.
Fathers of two boys have to choose (1). Fathers of two girls have to
choose (b). Fathers of one girl and one boy choose (1) with a propability
p and (2) with a propability 1-p. Because all people are more or less
sexist in one direction or the other, the probability p is not equal to
1/2, but might be quite far away. If the probability of a child being a
boy or a girl is 1/2 (which it is not, but it is close to 1/2), then the
probability is
1/4 that the father has two boys and chooses (1)
p/2 that the father has one boy and one girl and chooses (1)
1/4 + (1-p)/2 = 3/4 - p/2 that the father chooses (2).
If you knew that mans character, and you knew this man absolutely hates
boys and would never admit to having a boy unless he was forced to, and if
this man said that he has at least one boy, then you would know that both
children must be boys (because if only one was a boy he would never have
chosen (1)).
Koen Samyn
I remember having the same discussion in high-school.
I've picked the possibilities which lead to the 1/3 solution
from an earlier posting.
scenario #1:
child#1 A
child#2 A
scenario#2:
child#1 A
child#2 B
scenario#3:
child#1 B
child#2 A
My remark:
We don't know if the child mentioned is the firstborn or not,
so in my humble opinion there are four possibilities:
scenario#1:
child#1:son mentioned
child#2:other son
scenario#2:
child#1:other son
child#2:son mentioned
scenario#3:
child#1:daughter
child#2:son mentioned
scenario#4:
child#1:son mentioned
child#2:daughter
I'm interested to hear valid counterarguments.
Don't tell me the order doesn't matter because then the
number of scenario's would be reduced to two.
To me it's clear :the probability of two sons is 50%
Koen Samyn
Dave Seaman
In article <349E5B...@se.bel.alcatel.be>,
Koen Samyn <sam...@se.bel.alcatel.be> wrote:
>I remember having the same discussion in high-school.
>I've picked the possibilities which lead to the 1/3 solution
>from an earlier posting.
[snip]
>We don't know if the child mentioned is the firstborn or not,
>so in my humble opinion there are four possibilities:
>
>scenario#1:
>child#1:son mentioned
>child#2:other son
>
>scenario#2:
>child#1:other son
>child#2:son mentioned
How do you know which son was mentioned? How do you now the father even had a
particular son in mind, when he said he has at least one son?
>I'm interested to hear valid counterarguments.
>Don't tell me the order doesn't matter because then the
>number of scenario's would be reduced to two.
>
>To me it's clear :the probability of two sons is 50%
The order does matter. There are four cases, one of which is ruled out
by the fact that the children can't both be girls. There are three
equally likely cases remaining, only one of which has two boys, and
therefore the conditional probability is
P{BB|H} = P{BB & H} / P{H}
= (1/4) / (3/4)
= 1/3.
Dennis Rakestraw
In article <67h678$t...@crocus.csv.warwick.ac.uk>,
ma...@csv.warwick.ac.uk (Dr D F Holt) wrote:
>
>
> Let me give another example of a scenario which can lead to the
> probability 1/2, which actually happened to me once.
> whether they were boys or girls. A few days later, a girl came up to
> me and said "Hi, you don't know me, but I'm X's daughter."
> It occurred to me that I now knew that X had two children and that at
> least one was a girl, but the probability that they were both girls was
> clearly 1/2.
>
This one also works back to the probability of 1/3 as there is only a 2/3
probability that a daughter will be the first X child encountered, if you
consider the original problem with a gender change. Of course you would
never ask the son if both he and is sister were girls, but nevertheless
the two of them would fall into the original sample space.
Thanks for keeping this thread alive, but I never did find an older
confirmed publication date for Marilyn's problem as stated, than the 1980
date in my original post. That's probably old enough to prove that
Marilyn had nothing to do with the original problem, which was my main
objective.
Dennis Rakestraw
drake...@hotmail.com
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Xcott Craver
Dave Seaman <a...@seaman.cc.purdue.edu> wrote:
>In article <67ednd$6a2$1...@gannett.math.niu.edu>,
>
>I don't agree. If you don't make any assumptions, the problem as
>stated has no answer. I listed six additional assumptions that make it
>possible to conclude the probability is 1/3.
But this is a slippery slope. Anyone can introduce an
unspecified assumption into any story problem which is not entirely
unrealistic and which would change the answer.
Personally, I would rather read and enjoy mathematical
puzzles, rather than declaring them all insoluble on the grounds
that they don't (and likely can't) disclaim every assumption.
Nor would I accept answers on student homework along these lines,
unless the problem's interpretation is really hard to get.
-Xcott
Xcott Craver
Dennis Rakestraw <drake...@hotmail.com> wrote:
>the two of them would fall into the original sample space.
>
>Thanks for keeping this thread alive, but I never did find an older
>confirmed publication date for Marilyn's problem as stated, than the 1980
>date in my original post. That's probably old enough to prove that
>Marilyn had nothing to do with the original problem, which was my main
>objective.
What exactly do you mean by this? That the publication dates
somehow indicate that Marilyn wasn't talking about the same problem?
I do believe this problem has been around a long time before 1980,
if you are trying to claim that Marilyn published it before it was
stated elsewhere. If your are claiming the opposite, that it was
around before her, then how do you conclude that Marilyn had nothing
to do with the original problem?
I think its pretty obvious that Marilyn was re-stating the
"original" problem. The fact that she gave an answer of 1/3 should
pretty much prove that.
>Dennis Rakestraw
Dave Seaman
In article <67olre$4l3$1...@gannett.math.niu.edu>,
Xcott Craver <c...@baker.math.niu.edu> wrote:
>Dave Seaman <a...@seaman.cc.purdue.edu> wrote:
>>In article <67ednd$6a2$1...@gannett.math.niu.edu>,
>
>>
>>I don't agree. If you don't make any assumptions, the problem as
>>stated has no answer. I listed six additional assumptions that make it
>>possible to conclude the probability is 1/3.
>
> But this is a slippery slope. Anyone can introduce an
>unspecified assumption into any story problem which is not entirely
>unrealistic and which would change the answer.
You can't do that if the problem is properly specified. If it isn't,
then some additional assumptions are necessary, and the question
becomes which ones to adopt.
If a problem is underspecified, as this one was, you can fix it by
making reasonable assumptions and then pointing out that you are doing
so. There is nothing wrong with interpreting the problem to mean,
"Given that a man has exactly two children and at least one is a boy,
what is the probability that both are boys?" However, it's best to
make it clear that an assumption is involved.
It was not my idea to bring up the argument about what would happen if
the man were equally likely to say "I have at least one boy," or "I
have at least one girl," when he has one of each. In fact, I believe
you were one of the first in the current thread to say that this
assumption leads to a different answer. It was only after Darrell Ryan
denied this that I even entered the discussion.
drake...@hotmail.com
In article <67omg3$4nc$1...@gannett.math.niu.edu>,
What I meant that she had nothing to do with the CREATION of the problem,
she was simply restating a problem that in all likelihood is much older
than she is. So, why is everyone upset with Marilyn reinventing the
wheel? I was hoping to find a publication date at least 80 years old for
an octogenarian I know who is extremely upset with Marilyn's answer of
1/3 vs. his interpretation of 1/2.
Since I can't find anything older than 1980, is there anyone out there
who could lead me to a website or newsgroup that might deal better with
obscure publications?
David A Karr
Xcott Craver <c...@baker.math.niu.edu> wrote:
>Dave Seaman <a...@seaman.cc.purdue.edu> wrote:
>>I don't agree. If you don't make any assumptions, the problem as
>>stated has no answer. I listed six additional assumptions that make it
>>possible to conclude the probability is 1/3.
> Personally, I would rather read and enjoy mathematical
>puzzles, rather than declaring them all insoluble on the grounds
>that they don't (and likely can't) disclaim every assumption.
The problem with the puzzle was not that it failed to disclaim
assumptions, but that it failed to *give* enough of them.
Dave didn't attempt to change the answer by introducing some
extraneous assumption; he justified what most of us think was
the _intended_ answer by supplying the unstated assumptions.
One could, like Humpty Dumpty, insists that words mean what one
wants them to mean, but apart from that sort of thing there are
puzzles that are well-stated, and those that are less well-stated.
The puzzle as stated was in the latter category. Likewise there
are reasonable assumptions and less reasonable assumptions.
The set of all men with two children of which at least one is a boy is
a quite well-defined set, and if I tell you simply that Mr. X is from
that set, it's reasonable to assume that Mr. X was equally likely to
be selected from any of the members of that set, of which only 1/3
have two sons. You can assume otherwise, but it's a bit of a strain,
and if I use mathematical "given that ..." language, it's a Humpty
Dumpty-ism.
On the other hand, the set of all men who walk up to people and say
simply, "Of my two children at least one is a boy," are a very small
and obscure set. What is the probability space of these?
It seems almost inescapable that the man's statement is an alternative
to some other act (including doing nothing), but what? One assumed
alternative is, "Of my two children neither is a boy," in which case
the answer is 1/3. Another is, "Of my two children at least one is a
girl," in which case the answer is 1/2. I find the latter assumption
a bit more symmetric, but the former is a classic.
Yet another interpretation is to assume the man's statement is an
answer to a question that the puzzle simply neglected to quote,
perhaps, "How many children do you have and is at least one a boy?"
Or perhaps it was, "How many children do you have and could you
please tell me the sex of at least one?"
>Nor would I accept answers on student homework along these lines,
>unless the problem's interpretation is really hard to get.
I don't think the puzzle we saw here is a good student homework
problem. The version is Marilyn's recent column was better, but
perhaps not as good as it should have been.
On a homework paper, I would still like to see some attempt made at
finding a reasonable set of assumptions under which the problem can be
solved. We could even parameterize over a range of assumptions and
express the answer in terms of the parameters. This was done once
already in this thread, but here's my version. Let the priors be:
H = The man has exactly two sons and no daughters.
K = The man has exactly one son and one daughter.
L = (not H) and (not K).
A = The man tells me he has two children, at least one of them a boy.
Let
P(A|H) = p.
P(A|K) = q.
P(A|L) = r.
Then the answer is
P(H|A) = p * P(H) / (p * P(H) + q * P(K) + r * P(L)).
So far no assumptions other than that we have some probability space
in which the probabilities p, q, and r are defined, and that the
denominator above isn't zero, so let's start assuming. A very
reasonable assumption is that P(K) = 2*P(H), since boy-girl families
are about twice as numerous in the real world as two-boy families.
Also, let's assume p > 0, since if p = 0 it's a bit of a strain to
justify any answer other than P(H|A) = 0. Then
P(H|A) = 1 / (1 + 2q/p + (r/p)*(P(L)/P(H))).
I don't see a good way to estimate P(L)/P(H) in this context, so
it seems this is about as far as we get unless r = 0. Fortunately,
r = 0 is a "reasonable" assumption (in the sense that if people can
lie, the puzzle is obviously too wide open), and that gives us
P(H|A) = 1 / (1 + 2q/p).
At this point there are different reasonable additional assumptions
that give different values of q/p and hence of P(H|A). (We've already
seen how to get either 1/3 or 1/2; note that we can get P(H|A) = 1/3
even without q = p = 1.)
Would that be satisfactory for a homework answer?
David A. Karr
Dr D F Holt
In article <882805315...@dejanews.com>,
drake...@hotmail.com writes:
>In article <67h678$t...@crocus.csv.warwick.ac.uk>,
> ma...@csv.warwick.ac.uk (Dr D F Holt) wrote:
>>
>>
>> Let me give another example of a scenario which can lead to the
>> probability 1/2, which actually happened to me once.
>> Somebody told me that Mr X had two children, but he had no idea
>> whether they were boys or girls. A few days later, a girl came up to
>> me and said "Hi, you don't know me, but I'm X's daughter."
>> It occurred to me that I now knew that X had two children and that at
>> least one was a girl, but the probability that they were both girls was
>> clearly 1/2.
>>
>
>This one also works back to the probability of 1/3 as there is only a 2/3
I am not sure what you man by it "works back to 1/3".
The answer is 1/2.
Let me state my assumptions again (I hope) completely unambiguously.
1. All children independently have probability 1/2 of being boy/girl.
2. X has two children.
3. I meet a random child of X, and that child is a girl.
Then the probability that X has two girls is absolutely precisely 1/2.
(Proof: The chance that the child I meet is a girl is 1/2. The chance that
both children are girls and I meet a girl is 1/4 (since if both are girls, I
must meet a girl). Hence the conditional probability of GG is
(1/4)/(1/2) = 1/2.)
The ultimate test of disagreements about probability questions is to
arrange a betting game.
We can simulate boy/girl with heads/tails, and play as follows.
A neutral and honest referee tosses two coins,
and writes the results as B (if heads) or G (if tails) on
two identical pieces of paper and puts them them in a bag. You then draw
out one piece of paper without looking into the bag first. If your selected
piece of paper says B, then the game is void, and we repeat the whole game.
If it says G, then we look at the other one in the bag. If that is G too
(two girls) then I win, and if it is B (boy and girl), then you win.
We could play this game at odds of say 5/12, which you think is favorable
to you and I think is favorable to me. I don't suppose we will ever get to
play this really, but would you be prepared to?
Seasons Greetings to you all!
Derek Holt.
stan...@slu.edu
On Mon, 22 Dec 1997, Koen Samyn wrote:
> I remember having the same discussion in high-school.
> I've picked the possibilities which lead to the 1/3 solution
> from an earlier posting.
>
>
> scenario #1:
> child#1 A
> child#2 A
>
> scenario#2:
> child#1 A
> child#2 B
>
> scenario#3:
> child#1 B
> child#2 A
>
> My remark:
>
> We don't know if the child mentioned is the firstborn or not,
> so in my humble opinion there are four possibilities:
I count five:
scenario#0:
child#1:son mentioned
child#2:son mentioned
>
> scenario#1:
> child#1:son mentioned
> child#2:other son
>
> scenario#2:
> child#1:other son
> child#2:son mentioned
>
> scenario#3:
> child#1:daughter
> child#2:son mentioned
>
> scenario#4:
> child#1:son mentioned
> child#2:daughter
>
So the probability would be 3/5.
> I'm interested to hear valid counterarguments.
> Don't tell me the order doesn't matter because then the
> number of scenario's would be reduced to two.
>
> To me it's clear :the probability of two sons is 50%
>
> Koen Samyn
>
>
drake...@hotmail.com
In article <67rego$b...@crocus.csv.warwick.ac.uk>,
The difference between this and the original problem with a sex change is
the voided game! You must compute the odds on pairs, not individuals.
If the first encountered of a pair is unfavorable you are ignoring the
second member of the pair. The original problem does not, which is why
it yields the 1/3 result. Your problem is a subset of the original
problem which can only yield a probability of 1/2.
"If a man has two children and at least one is a boy, what is the
probability that both are boys?" If you should happen to meet a girl
from this sample space first you can't discard the fact that she has a
brother and thus counts as a member of a failed pair. Your odds of
meeting a girl first are 1/3 and the failure rate of that occurence is
100%.