🚀 go-pugleaf

In article <1993Jan22.131719.36@janus.arc.ab.ca> morgan@arc.ab.ca (Sean Morgan) writes:
>You meet the daughter of a friend of yours on the street.  You know
>that your friend has two children.  You think to yourself, "Ignoring
>the occurrence of twins, and sex-linked differences in birth rates
>and infant mortality, what are the odds of this girl's sibling also
>being a girl?"

Which of my friends did you have in mind?  Then I can tell you whether
the odds are 0% or 100%.

-- David Karr (karr@cs.cornell.edu)

In article <1993Jan22.131719.36@janus.arc.ab.ca> morgan@arc.ab.ca (Sean Morgan)
writes:
>You meet the daughter of a friend of yours on the street.  You know that your
>friend has two children.  You think to yourself, "Ignoring the occurrence of
>twins, and sex-linked differences in birth rates and infant mortality, what
are
>the odds of this girl's sibling also being a girl?"
>-----------------+-------------------------------

(Some of my mail says "not again", and some says "please post a solution".
 Since you can't please all of the people all of the time...)

Two child families are uniformly distributed between MM,MF,FF,FM, in birth
order.  Again, all four possibilities are equally likely.  Stop here.  Make
sure you believe that.

Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
there is only a one third chance that the sibling is a girl.
-----------------+-------------------------------
Sean Morgan      |  ALBERTA  6815 - 8 Street N.E.
                 | RESEARCH  Calgary, AB, Canada
morgan@arc.ab.ca |  COUNCIL  T2E 7H7

In article <1993Jan22.131719.36@janus.arc.ab.ca> morgan@arc.ab.ca (Sean Morgan) writes:
>You meet the daughter of a friend of yours on the street.  You know that your
>friend has two children.  You think to yourself, "Ignoring the occurrence of
>twins, and sex-linked differences in birth rates and infant mortality, what are
>the odds of this girl's sibling also being a girl?"

	Well, I'm new to this group and I'm sure this is "puzzle" is in the
FAQ, but I'll throw in my two cents worth, anyway.
	Of course, the common sense answer is correct: there are equal odds
of the other sibling being a female or a male. The apparent paradox lies
in an inconsistency in the counting procedure described. If one states that
there are four possible gender configurations (FF,MM,FM,MF) then clearly
the order in which the children are born is being taken into account. Otherwise
FM and MF wouldn't be listed separately.
	Therefore, since this order is being taken into account, when you
see the girl on the street there are FOUR possibilities. She is either the
youngest of a FM family, the oldest of a MF family, the youngest of a
FF family, or the oldest of a FF family. Therefore, FF gets counted twice
and the paradox is resolved. There is a 50% chance of the other sibling
being a girl.

				-Ted Quinn
				 quinn@relgyro.stanford.edu

In article <1993Jan25.115908.41@janus.arc.ab.ca> morgan@arc.ab.ca (Sean Morgan) writes:
>In article <1993Jan22.131719.36@janus.arc.ab.ca> morgan@arc.ab.ca (Sean Morgan)
>writes:
>>You meet the daughter of a friend of yours on the street.  You know that your
>>friend has two children.  You think to yourself, "Ignoring the occurrence of
                                                    ++++++++++++++++++++++++++

>>twins, and sex-linked differences in birth rates and infant mortality, what 
  +++++

>>are the odds of this girl's sibling also being a girl?"
>>-----------------+-------------------------------
>

And the answer...


>
>Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
>there is only a one third chance that the sibling is a girl.

Really?  Ignoring the possibility of twins rules out FF, so the chance that
the other sibling is a female is zero!


(Please insert smilies as you see fit).

--Dana

disclaimer: Ohio University ignores my opinions.

(extra lines for my poster....)


In article <1993Jan25.115908.41@janus.arc.ab.ca>, morgan@arc.ab.ca (Sean Morgan) writes:
> 
> (Some of my mail says "not again", and some says "please post a solution".
>  Since you can't please all of the people all of the time...)
> 
> Two child families are uniformly distributed between MM,MF,FF,FM, in birth
> order.  Again, all four possibilities are equally likely.  Stop here.  Make
> sure you believe that.
> 
> Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
> there is only a one third chance that the sibling is a girl.

Birth order? No one mentioned age here. The question was, that given one
girl, what are the chances of the other also being a girl. Given a uniform
distribution, that means for 1-child families, P(girl) = P(boy) = 0.5
Therefore, given that one child is a girl, the chance that the other child
is a boy, is 0.5, same as if it were a girl. Try this analogy, and you'll see
what's wrong with your answer.

You are throwing 2 consecutive dice. The first one turns up a six. What is
the chance of the second also being a six. It's easy to see that because dice
have no memory, the chance is also 1/6. Now, following your line of reasoning,
there are 36 different ways to throw the dice. Because 1 six already has
been thrown this leaves 11 combinations, one of which is a double six.
So, according to you, you only have 1/11 chance of throwing a second six.
Clearly this is dead wrong.

--Ralph

karr@cs.cornell.edu (David Karr) writes:

>>You meet the daughter of a friend of yours on the street.  You know
>>that your friend has two children.  You think to yourself, "Ignoring
>>the occurrence of twins, and sex-linked differences in birth rates
>>and infant mortality, what are the odds of this girl's sibling also
>>being a girl?"

One half - The probability of the other child being a girl is totally
independent of the sex of the first. The first child being
a girl does affect the 50-50 boy-girl distribution of the second (or any 
other) child.

Even if your friend had 10 children, and you met nine of them in the street,
all girls, then the probability of the tenth being female is still one half.

Of course, the probability of having ten girls is 1/1024 (assuming no
genetic oddities), but the probability of any one child being a girl is
one half

Christopher Hicks
--
 ==============================================================================
 Christopher Hicks    |    If it doesn't fit...
 cmh@uk.ac.cam.eng    |                        ...you need a bigger hammer.
 ==============================================================================

cmh@eng.cam.ac.uk (C.M. Hicks) writes:

>karr@cs.cornell.edu (David Karr) writes:

>>>You meet the daughter of a friend of yours on the street.  You know
>>>that your friend has two children.  You think to yourself, "Ignoring
>>>the occurrence of twins, and sex-linked differences in birth rates
>>>and infant mortality, what are the odds of this girl's sibling also
>>>being a girl?"

>One half - The probability of the other child being a girl is totally
>independent of the sex of the first. The first child being
>a girl does affect the 50-50 boy-girl distribution of the second (or any 
>other) child.

Of course, you could say that the possible permutations for a two-child
family are:

        Male    Female
      Female    Male
      Female    Female
        Male    Male

These all have probability 1/4, and since one child is female, one permutation
is discounted (male-male), giving a final probability of a third...

Heck - flame war ahead, and all in my own mind - do I believe Bayes Theorem?

Christopher Hicks

--
 ==============================================================================
 Christopher Hicks    |    If it doesn't fit...
 cmh@uk.ac.cam.eng    |                        ...you need a bigger hammer.
 ==============================================================================

In article <1993Jan26.115545.11556@eng.cam.ac.uk>, cmh@eng.cam.ac.uk (C.M. Hicks) writes:
=karr@cs.cornell.edu (David Karr) writes:
=
=>>You meet the daughter of a friend of yours on the street.  You know
=>>that your friend has two children.  You think to yourself, "Ignoring
=>>the occurrence of twins, and sex-linked differences in birth rates
=>>and infant mortality, what are the odds of this girl's sibling also
=>>being a girl?"
=
=One half - The probability of the other child being a girl is totally
=independent of the sex of the first. The first child being
=a girl does affect the 50-50 boy-girl distribution of the second (or any 
=other) child.
=
=Even if your friend had 10 children, and you met nine of them in the street,
=all girls, then the probability of the tenth being female is still one half.
=
=Of course, the probability of having ten girls is 1/1024 (assuming no
=genetic oddities), but the probability of any one child being a girl is
=one half

OK, maybe if we do this by enumeration, it'll convince people.  Suppose there
are four families.  One has two boys, one two girls, one has an older boy and
youger girl, and the fourth has an older girl and younger boy.  Are you agreed
that each of these four possibilities is equally likely?  Now, suppose as
you're walking around, you run into each of these children separately.  Are you
agreed that the odds of meeting them are identical?  If so, then we have the
following equiprobably possibilities:

Description of child you meet		Gender of sibling
--------------------------------	-----------------
Oldest boy from first family		Male
Youngest boy from first family		Male
Oldest girl from second family		Female
Youngest girl from second family	Female
Boy from third family			Female
Girl from third family			Male
Boy from fourth family			Female
Girl from fourth family			Male

Each of these is equally likely.  Now we note that the problem speficies that
you met a girl.  The possibilities here are:

Description of child you meet		Gender of sibling
--------------------------------	-----------------
Oldest girl from second family		Female
Youngest girl from second family	Female
Girl from third family			Male
Girl from fourth family			Male

Thus there's a 50% chance that the sibling is female.  This is *NOT* the same
as the "I've got two children and at least one of them is a girl" question. 
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.

>>You meet the daughter of a friend of yours on the street.  You know that your
>>friend has two children.  You think to yourself, "Ignoring the occurrence of
>>twins, and sex-linked differences in birth rates and infant mortality, what
>are
>>the odds of this girl's sibling also being a girl?"
>>-----------------+-------------------------------
>Two child families are uniformly distributed between MM,MF,FF,FM, in birth
>order.  Again, all four possibilities are equally likely.  Stop here.  Make
>sure you believe that.
>
>Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
>there is only a one third chance that the sibling is a girl.

But you meet the F on the street. So MF and FM are the same in this case
(but not the way the problem is usually stated). I say 50%.

In article <1993Jan25.115908.41@janus.arc.ab.ca>, morgan@arc.ab.ca (Sean Morgan) writes:
>(Some of my mail says "not again", and some says "please post a solution".
> Since you can't please all of the people all of the time...)
>
>Two child families are uniformly distributed between MM,MF,FF,FM, in birth
>order.  Again, all four possibilities are equally likely.  Stop here.  Make
>sure you believe that.

Okay, got it.

>Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
>there is only a one third chance that the sibling is a girl.

But what are the chance of you meeting one of these three pairs?  Their
birth chance does not imply your meeting chance.  Let me show it:

We are talking about two-kids families.  4 equal combinations but 8
possible kids, mm, MF, FM, ff.  Assume you have equal chance meeting
one of them.  But you are limiting your calculation ONLY if you meet a
girl, i.e. only if you meet the "F" of {FM,MF} or either "f" of "ff".
So, total four possible girls.  Half of them have a femal sibling, the
other half male sibling.  Thus the answer of your question is 0.5.

Nothing personal but here is a joke i remember reading the "Born Loser"
comic.  He figured the chance of boarding a plain having a bomb is
1 out of a million.  Having two bombs on the same plane should be
1 out of a trillion.  So, he decided to bring along a bomb when
he was reading to board it.

In article <6670016@hplsla.hp.com> barryg@hplsla.hp.com (Barry Gunn) writes:
>>>You meet the daughter of a friend of yours on the street.  You know that your
>>>friend has two children.  You think to yourself, "Ignoring the occurrence of
>>>twins, and sex-linked differences in birth rates and infant mortality, what
>>are
>>>the odds of this girl's sibling also being a girl?"
>>>-----------------+-------------------------------
>>Two child families are uniformly distributed between MM,MF,FF,FM, in birth
>>order.  Again, all four possibilities are equally likely.  Stop here.  Make
>>sure you believe that.
>>
>>Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
>>there is only a one third chance that the sibling is a girl.
>
>But you meet the F on the street. So MF and FM are the same in this case
>(but not the way the problem is usually stated). I say 50%.

I disagree. I made this mistake once myself and promptly realized the flaw.
Why don't you consider this metaphor : you have these teeming
billions of sibling-pairs in the Earth (yeah, I mean all continents,
islands, places hitherto undiscovered or unknown, etc. etc. taken
together ad nauseam :-) ) ?  Now, it's quite easily seen that FF 
pairs comprise only 25 % of _this_ population (assuming that the
birth events of a pair of siblings are independent, thereby precluding 
identical twins, etc.). 50 % of this population will comprise MF
sibling pairs, and the remaining 25 %, MM pairs.

Now, you the witness of one of a pair of siblings, are a Martian
and have just floated down on the earth from Mars on a spaceship,
on a special invitation from Presidents Clinton and Yeltsin,
and see one female member of a pair of siblings, who has been sent
by her Dad to receive you.  You have had numerous inter-planetary 
conversations with her Dad (mostly on new and exciting ways of
proving that P equals NP :-) ), and know that he has two children, 
but you've never asked him their sexes. Now without any other info, 
it's a fair assumption that you'll encounter sibling pairs on the Earth
in proportion to their relative frequencies of occurrence.  Hence, 
the female sibling you encounter is twice as likely to be a part of a
mixed sex pair than of an FF pair.  Hence, the other sibling being a
female has probability 1/3.

Hope it helps.

Kanad

The difficulty in this problem is that you can't apply Bayes' theorem without
knowing the prior probability that, if your friend has a boy and a girl, the
girl is the one that you meet.  Half of the people have been arguing that the
probability is 50% because it's equally likely that you'll meet either child.
The other half have been arguing (wrongly, I think) that the probability is
100% because you did meet the girl.  But probability is not about single
events.  Probability answers the question, "If I saw my friend with one of
her children a large number of times, about how many of those times would
the child with her be female?"  I don't think it's necessarily correct to
assume that your friend would be equally likely to meet you with her son
as with her daughter, so the 50% camp is also on shaky ground.

-- 
John Dalbec (dalbec@cs.cornell.edu) Speak for yourself--I do!

In article <1993Jan22.131719.36@janus.arc.ab.ca>, morgan@arc.ab.ca (Sean
Morgan) wrote:
> 
> You meet the daughter of a friend of yours on the street.  You know that your
> friend has two children.  You think to yourself, "Ignoring the occurrence of
> twins, and sex-linked differences in birth rates and infant mortality, what are
> the odds of this girl's sibling also being a girl?"
> -----------------+-------------------------------
> Sean Morgan      |  ALBERTA  6815 - 8 Street N.E.
>                  | RESEARCH  Calgary, AB, Canada
> morgan@arc.ab.ca |  COUNCIL  T2E 7H7


I don't think you asked that quite right.  It's supposed to be, "A man
has two children.  One of them is a boy.  What are the odds that the
other one is also a boy?"
The answer, which surprises most people, is 1/3.  There are four
possibilities: boy/boy, boy/girl, girl/boy, and girl/girl, all equally
likely.  Since one child is a boy, the last possibility is eliminated.
That leaves two out of three cases where the other child is a girl.

However, meeting one of the children on the street changes the odds,
because if they are both girls you'll be twice as likely to meet one
of them.  That is, after meeting the girl you do know that there are
only three possibilties, and that they are not both boys, but the
three possibilities are not equally likely.  Assuming, that is, that
your odds off meeting someone of type X on the street depend on how
populous people of type X are.


Bob Hearn
Spartacus Software

Opinions expressed here are those of my empolyer, since that's me.

In article <1993Jan26.204422.15093@zip.eecs.umich.edu> kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:

[Original problem deleted]

>Why don't you consider this metaphor : you have these teeming
>billions of sibling-pairs in the Earth (yeah, I mean all continents,
>islands, places hitherto undiscovered or unknown, etc. etc. taken
>together ad nauseam :-) ) ?  Now, it's quite easily seen that FF 
>pairs comprise only 25 % of _this_ population (assuming that the
>birth events of a pair of siblings are independent, thereby precluding 
>identical twins, etc.). 50 % of this population will comprise MF
>sibling pairs, and the remaining 25 %, MM pairs.

[Metaphor deleted]

>Now without any other info, 
>it's a fair assumption that you'll encounter sibling pairs on the Earth
>in proportion to their relative frequencies of occurrence.  Hence, 
>the female sibling you encounter is twice as likely to be a part of a
>mixed sex pair than of an FF pair.  Hence, the other sibling being a
>female has probability 1/3.

	Nobody seems to have noticed my original argument for 1/2, so I'll
respond to this one. Contrary to the author's intention, this is actually
a good way to see that the answer is 1/2. The key, once again, is that
there are two females in an FF pair and only one in a MF or FM pair.
Taking a cue from his argument, assume that there are 1000 sibling pairs
in the world. Of these pairs, 250 are MM, 250 are MF, 250 are FM, and 250
are FF. Now you see a female, so you know that she can't be part of an
MM. This limits you to the other 750 pairs. While it is true that there
are 500 mixed pairs remaining and 250 FF pairs remaining, there are 2 females
in each FF pair. Therefore, of the 1000 females in the mixed pairs and FF
pairs, 500 are in mixed pairs and 500 are in FF pairs. There is a 50%
chance the female you're looking at is from an FF pair, and hence her
sibling is female.
				-Ted Quinn
				 quinn@relgyro.stanford.edu

Sean Morgan (morgan@arc.ab.ca) wrote:
: You meet the daughter of a friend of yours on the street.  You know that your
: friend has two children.  You think to yourself, "Ignoring the occurrence of
: twins, and sex-linked differences in birth rates and infant mortality,what are
: the odds of this girl's sibling also being a girl?"

Suppose there are just four families.  Each family has two children.
One family has two boys, a second has two girls, and the other two
have one of each.  Denote these families as BB, GG, BG, and GB.

There are four girls, and two of them have sisters.  If you pick
one of the girls at random, then the probability is 1/2 that she
has a sister.

Now suppose you pick one of the mothers at random.  You ask her if
she has at least one daughter, and she says that she does.  The probability
that both of her children are daughters is 1/3.

Do you see why the answers are different?  The woman must belong to
GG, GB, or BG; each of these cases are equally likely.  The girl must
also be a member of GG, GB, or BG; but GG is twice as likely as the
other two cases. 

Let us consider another situation.  Suppose you wish to find the
average family size in a community.  If you pick children at random
and record the number of members of their families, then large families
will be overrepresented.  You would get a more accurate result if
you pick mothers at random.  (I have ignored certain sociological
realities here, of course.)

--
David Radcliffe                                
radcliff@csd4.csd.uwm.edu            

the probability of A, given B, is the probability of A and B, divided
by the probability of B.  does anyone dare dispute me?

no? good.  to continue, let's dissect the question: "i have two children.
at least one is a daughter. what's the chance that the other is a daughter?"
is the same as "what is the chance that both of my children are female, given
that at least one is female?" so we have that A is "both of my children are
daughters", and B is "at least one of my two children is a daughter."

so the chance that both are daughters given that one is a daughter is
the chance that both are daughters AND at least one is a daughter, divided
by the chance that at least one is a daughter.

the chance that both are daughters and at least one is a daughter is
the same as the chance that both are daughters (i.e., "at least one is
a daughter" is a foregone conclusion, if both are daughters), which is
one out of four.

the chance that at least one is a daughter is one minus the chance that
both are sons, which is one minus a quarter, which is three quarters.

one quarter divided by three quarters is one third.  last i checked,
a third and a half were not the same.

to all you proponents of the so-called "common sense" argument,
remember the archaic definition of "common".

bob vesterman.

In article <1993Jan26.095218.17387@cbfsb.cb.att.com>,
cooper@cbnewsg.cb.att.com (Ralph 'Hairy' Moonen) wrote:
>Try this analogy, and you'll see
> what's wrong with your answer.
> 
> You are throwing 2 consecutive dice. The first one turns up a six. What is
> the chance of the second also being a six. It's easy to see that because dice
> have no memory, the chance is also 1/6. Now, following your line of reasoning,
> there are 36 different ways to throw the dice. Because 1 six already has
> been thrown this leaves 11 combinations, one of which is a double six.
> So, according to you, you only have 1/11 chance of throwing a second six.
> Clearly this is dead wrong.

Yes, your analogy is dead wrong.  The analogy for the original problem is:

You are throwing two dice, and at least one of them turns up a six.  What
are the chances that you rolled a total of twelve?  If you believe that the
answer is 1/6 and not 1/11, please come over and bet me anytime. :-)
NEVIN ":-)" LIBER, RISC Meanie, Mac System Software
 email:     nevin@apple.com        paper:  Apple Computer, Inc.
 voice:     (408) 974-MIX1                 20525 Mariani Avenue, MS: 81-GC
 AppleLink: BADENOV                        Cupertino, CA 95014