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Re: The "two child problem"



There are four possible cases for two children (oldest listed first):
1. SS
2. SD
3. DS
4. DD

In the case of the man we know we have either case 1 or 2, and we have
used up all our information when we say that we are dealing with one of
those two cases. In the absence of further information either case is
equally likely, and since one case involves a son (S) and one case
doesn't, there is a 1/2 probability that the man's other child is a
son. [This result agrees with the obvious 1/2 probability that any
given child will be male.]

In the case of the woman we have less information, and the information
is exhausted by saying that we have case 1, 2, or 3. In the absence of
further information any of the three cases is equally likely, and since
two of the cases involve an additional son and one doesn't, there is a
2/3 probability that the woman's other child is a son. Hence the woman
is more likely to have two sons than the man is.

I am completely satisfied with this conclusion. However, that does not
mean that I have a completely satisfactory *intuitive* explanation of
the result.

John Mallinckrodt wrote:
[snip]

Marilyn had previously dealt with the following problem: 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?

[snip]


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
Cal Poly Pomona fax:909-869-5090
Pomona, CA 91768-4031 office:Building 8, Room 223

--
Maurice Barnhill, mvb@udel.edu
http://www.physics.udel.edu/~barnhill/
Physics Dept., University of Delaware, Newark, DE 19716