NetNews Usenet Archive 1992 #30/NN_1992_30.iso/spool/sci/math/16956

Internet Message Format | 1992-12-15 | 5.1 KB

Xref: sparky sci.math:16956 rec.puzzles:7932
Newsgroups: sci.math,rec.puzzles
Path: sparky!uunet!spool.mu.edu!uwm.edu!rpi!batcomputer!cornell!karr
From: [e-mail] (David Karr)
Subject: Re: Marilyn Vos Savant's error?
Message-ID: <[e-mail]>
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
References: <[e-mail]> <[e-mail]> <[e-mail]>
Date: Tue, 15 Dec 1992 16:00:00 GMT
Lines: 123

Below, I try to say something interesting on the following question,
which otherwise seems to get beaten to death every few months on this
newsgroup:

In article <[e-mail]> [e-mail] (Steven Newman) writes:
>
>   1) You meet a man on the street and ask him how many children he has.
>      He replies "two, and one is a boy."  What is the probability that
>      his other child is also a boy?

It depends entirely on how your probability space is constructed.
Specifically, what is the distribution of numbers and kinds of
children had by people in the population that you are sampling, and
what is the probability, given that a person has a certain set of
children, that he or she will give the above response?

Assume for simplicity that there are no twins and that the sexes of
different children in the same family are independently distributed.
Then of people who have two children, the following four events are
equally likely:

(a) The children are both boys
(b) The older child is a boy and the younger is a girl
(c) The older child is a girl and the younger is a boy
(d) Both children are girls

(We can discard all encounters with people who have more or fewer than
two children on the reasonable assumption that they would not respond
"two" in that case.  That is, all probabilities are implicitly
conditioned on the event "exactly two children.")

Here are some possible assumptions that might determine the
distribution of responses by a parent of two children.  Use the
letter X to represent the response, "Two, and one is a boy."
In all cases, we'll assume people don't lie (though why shouldn't
they?), so in case (d), X never occurs.

(1) The important thing is to have at least one boy in your family,
    it doesn't matter how many.  So in cases (a), (b), and (c) the
    response is always X.  In formal terms:

    P(X|a) = P(X|b) = P(X|c) = 1

(2) It is customary to pick one of one's children at random and report
    the child's sex.  Then in case (a) the response is always X, while
    in cases (b) and (c) the response is X exactly 1/2 the time.
    Formally:

    P(X|a) = 1, and P(X|b) = P(X|c) = 1/2.

(3) It is customary to report the older child's age (but not to say
    that it is the older child).  Then (going directly to the formal
    statement):

    P(X|a) = P(X|b) = 1, P(X|c) = P(X|d) = 0.

(4) When people tell you the sex of some of their children, they
    always tell you the exact number of that sex.  So for two boys,
    the response is either, "Two, and both are boys" (or an equivalent
    statement) or no information about the sex.  For a boy and a girl,
    possible responses are "one is a boy," "one is a girl," or no sex
    information.  Formally,

    P(X|a) = P(X|d) = 0,  P(X|b) > 0,  P(X|c) > 0.

Personally, I regard (4) as the set of assumptions closest to reality,
but this can be argued as a matter of opinion since all are
unrealistic to some degree, starting with the assumptions of no lying
and no twins.

In any of the above cases, the events (a), (b), (c), and (d) are
mutually exclusive, P(a) = P(b) = P(c) = P(d) = 1/4, and the
probability of X can be computed by:

   P(X) = P(X|a)*P(a) + P(X|b)*P(b) + P(X|c)*P(c) + P(X|d)*P(d)
    = (P(X|a) + P(X|b) + P(X|c) + P(X|d))/4

But we want P(a|X), since (a) is the only event in which the "other"
child is also a boy.  Use the rule

   P(R and S) = P(R|S)*P(S)

Then

   P(a|X) = P(a and X)/P(X)
      = P(X and a)/P(X)
      = P(X|a)*P(a)/P(X)
      = P(X|a)/(4*P(X))

You can then work out the results for the four sets of assumptions:

(1)    P(X) = 3/4
    P(a|X) = 1/3

(2,3)    P(X) = 1/2
    P(a|X) = 1/2

(4)    P(X) > 0
    P(a|X) = 0
    (Yes, a big fat whopping ZERO, since P(X|a) = 0.)

>   2) You meet a man on the street and ask him how many children he has.
>      He replies "two, and the older one is a boy."  What is the
>      probability that his other child is also a boy?
>
>The answer to problem 1 is 1/3, while the answer to problem 2 is 1/2.
>(One must be careful in the interpretation of the statement in problem
>1; it means "I have two children, and it is not the case that both of
>them are girls.")

You have to be more careful than that; for problem 1 I get 1/3 under
only one of my four proposed sets of assumptions, and one which in my
opinion is relatively unreasonable.  Problem 2 runs into similar
difficulties; the only set of assumptions I can think of offhand to
get the answer 1/2, is that nobody cares about younger children so
they always tell you about their oldest.  Under more reasonable
assumptions (i.e. if both are boys, the person will say so), I still
get the answer 0.

-- David Karr ([e-mail])