Re: The "two child problem"

[twayburn@juno.com (Thomas L Wayburn)]

On Mon, 28 Jul 1997 08:37:21 -0800 (PST) John Mallinckrodt
<ajmallinckro@CSUPomona.Edu> writes:

> A man and a woman each have two children.  The man's older child is a
son and at 
> least one of the woman's children is a son.  Is either more likely 
> than the other to have two sons?
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Suppose that neither boys nor girls enjoy even the slightest preference.
The probabilities are the same, I think.    Both parents have had their
children and nothing can be done now to change the outcome.  From this
point of view the first born and second born occupy the same portions of
the space.  We could, if we wished, define the woman's first born to be
the second born and vice versa.  In particular, we can define the birth
of the boy about which we are certain to have been the first born.  Now,
the woman's and the man's situations are the same.
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Suppose contrariwise that boys are preferred by 90% of all parents and
the modus operandi is identical throughout that majority of the
population, namely, no second birth occurs unless the first born be a
girl.   (Basically, we are in China.  I guess we are in China with
probability 1.2/8.4 = 0.143, say.)   It could have occurred that the
woman continued to have children until a boy were born.   In which case,
with the probability that this is the case, which we are not ruling out,
the woman has a first-born girl.   The man had the boy first, therefore
the probability that the second be a boy is just 0.49 as usual.   The
probability that the woman have two boys is 0.1 X 0.49 X 0.49 / 0.7399 =
0.03245; whereas the probability that the man have two boys is 0.49.  The
sum of the probabilities of all possible outcomes sums to 1.0.  The
reader is aware that the probability of having a boy is less (0.49) than
that of having a girl (0.51).

(0.1/0.7399) [0.49 X 0.49 + 0.49 X 0.51 + 0.51 X 0.49]  +  ( 0.9/0.49) [
0.49] = 1.0 

I'll have to admit that I would have to look this up in the probability
book and it ain't worth it.   I don't believe one can normalize out
prohibited cases as I have done.
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If we take the problem as a combination situation with probability 0.143
that we are in China and 0.857 that we are not, the probabilities are
different again and the man's probability of having two boys is greater.
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When it comes to probability I am not ever very confident.  But, I'm a
g--d-- engineer and engineers live in mortal terror of their own
fallibility.  I can't imagine how many different ways I would need to
solve it before I was at all confident.   Actually, I believe I am wrong!
- A

P.S.  What did you expect from such a dummy at thermo?

P.P.S.  I peeked at some of the other postings, so I am almost certain to
be wrong.  However, I post this for the wrinkle.
************************************************************

> What's your answer, how would you justify it, and how confident are 
> you that your answer is correct?
>
> John
> -----------------------------------------------------------------
> A. John Mallinckrodt      http://www.intranet.csupomona.edu/~ajm
> Professor of Physics    mailto:ajmallinckro@csupomona.edu
> Physics Department       voice:909-869-4054
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