🚀 go-pugleaf

: >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
: Ralph, you are dead wrong. Let's look closer at your dice analogy to see why.
: To compare the dice problem to the boy/girl problem, you need to toss the
: dice a bunch of times and write the results down and put them in a hat.
: This is because the two children are already born, and we're not saying
: that if a woman has given birth to a girl, and is pregnant, what is the
: probability that the next child is a girl. We're saying that both children
: have been born and you see one of them on the street, you don't know which
: child). You see a girl, what's the probability the other is a girl. 
: 
: Back to the dice... Roll two dice 100 times. Get 100 pieces of paper and
: write the results of the first die on one side and write the results
: of the second die on the other side of the piece of paper. Put all the
: pieces of paper in a hat. Now pick a piece of paper out of the hat.
: You see a six on the piece of paper, what is the probability that the
: other side has a six? IT IS 1/11!!!!!!! 

If you assume that birth order is important, and add the fact that you don't
know which child was born first, I'd agree with you.  But birth order isn't
mentioned in the puzzle.  You do know the order you see the child, so the
first child seen is the first child, and the unknown child is the second
one.  With this view, the answer is one-half.  If you assume birth order
is important then the answer is one-third.  Those answering one-third have
read more into the puzzle than was written.

As for dice, roll two dice without looking (under a box or something). 
Pull out a die.  It doesn't matter which die came to rest first or 
even was rolled first.  You see a six.  What are the odds that the other is 
a six?  (1/6)  This simulation more closely reflects the question posed this 
time.  If the question was that one of the dice was a six, what's the other
one, then the answer is different, namely the one mentioned above.

(More of a puzzle in poor english and miscommunication than probabilities.)

Jerry Donovan
I wonder what the person doing the FAQ is going to put in there.

In article <C1GuHw.zuF@austin.ibm.com>, bmoore@sunfish.austin.ibm.com (Bryan Moore) writes:
> >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
> Ralph, you are dead wrong. Let's look closer at your dice analogy to see why.
> To compare the dice problem to the boy/girl problem, you need to toss the
> dice a bunch of times and write the results down and put them in a hat.
> This is because the two children are already born, and we're not saying
> that if a woman has given birth to a girl, and is pregnant, what is the
> probability that the next child is a girl. We're saying that both children
> have been born and you see one of them on the street, you don't know which
> child). You see a girl, what's the probability the other is a girl. 
> 
> Back to the dice... Roll two dice 100 times. Get 100 pieces of paper and
> write the results of the first die on one side and write the results
> of the second die on the other side of the piece of paper. Put all the
> pieces of paper in a hat. Now pick a piece of paper out of the hat.
> You see a six on the piece of paper, what is the probability that the
> other side has a six? IT IS 1/11!!!!!!! 

The way you put it, you are still wrong. You are *given* the *fact* that
1 dice is a six. Your p[iece of paper has a six on 1 side. There are
left the combinations of 6-1, 6-2, 6-3, 6-4, 6-5, and 6-6. No other ones
possible, because you allready *know* there's 1 six. So the chance of
the other side being a six is 1/6. You make the mistake of making
a difference on which side of the paper the six is on. When you pick
a paper, and it has a 2 on it, you will also have to check the other
side to see if there's a six on it. The combination of 6-2 and 2-6 are
identical, because it is given that one of the two is six. 

Let's bring this back to the boy/girl problem. 
The (wrong) reasoning is that the possibilities are BG GB GG and BB.
But if you includeboth GB and BG, you should also include an extra
GG and BB, if you want to take birth order into account. Chances stay 
50/50.

This whole thing can be found in any statistical textbook, and is 
easy to understand when you realise that dice do not communicate. They
have no way of telling the other dice "Hey, I'm a six". They are 
independant entities.

--Ralph

In article <C1GuHw.zuF@austin.ibm.com> bmoore@sunfish.austin.ibm.com (Bryan Moore) writes:
 > Ralph, you are dead wrong.
No, strange enough, Ralph's answer is right.
The question was: you meet one girl, what are the chances of the other also
being a girl.  This question has a different solution than when one of the
parents tells you that there is one girl.

 >                                            We're saying that both children
 > have been born and you see one of them on the street, you don't know which
 > child). You see a girl, what's the probability the other is a girl. 
Yup, that is the question.
 > 
 > Back to the dice... Roll two dice 100 times. Get 100 pieces of paper and
 > write the results of the first die on one side and write the results
 > of the second die on the other side of the piece of paper. Put all the
 > pieces of paper in a hat. Now pick a piece of paper out of the hat.
 > You see a six on the piece of paper, what is the probability that the
 > other side has a six? IT IS 1/11!!!!!!! 
But this is the answer to the other question!  (You throw two dice, one of
them shows a 6, what is the probability that the other also shows a 6.)

The correct analogy is: you throw two dice, the first one you see shows a
6, what is the probability that the other one shows a 6.

The other question can also be put as the next analogy: you throw two dice,
the first one you see shows a 6, and/or the second one you see shows a 6,
what is the probability that both show a 6.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

In article <1993Jan26.155709.25082@cs.cornell.edu>, karr@cs.cornell.edu
(David Karr) wrote:
> 
> In article <C1GuHw.zuF@austin.ibm.com> bmoore@sunfish.austin.ibm.com (Bryan Moore) writes:
> >[...]
> >Back to the dice... Roll two dice 100 times. Get 100 pieces of paper and
> >write the results of the first die on one side and write the results
> >of the second die on the other side of the piece of paper. Put all the
> >pieces of paper in a hat. Now pick a piece of paper out of the hat.
> >You see a six on the piece of paper, what is the probability that the
> >other side has a six? IT IS 1/11!!!!!!! 
> 
> No, it's 1/6.  Have you tried it?  Get someone who really understands
> probability to explain it to you.  Or you can simulate it easily on a
> computer, but remember, a paper with two 6's can be selected in two
> different ways but a paper with only one 6 can be selected in only one
> way (if the first side you look at is the other side you have to
> discard the event).
> 
> -- David Karr (karr@cs.cornell.edu)


Well, you're right, but when you ask the problem the CORRECT way, that
is, don't mention meeting someone on the street, then the answer is
1/3, not 1/2.  If there were a club for every family in the US that
had two children at least one of which was a girl, and everyone
qualified joined, then only 1/3 of the member families would have the
other daughter also a girl.

Or, in terms of the paper hat, you have to look at both sides.  If EITHER
side is a six, then the probability that the OTHER side is a six is
1/11.  It's the unintuitiveness of the way you're given the information
that obscures the answer.

Bob Hearn
Spartacus Software

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