NetNews Usenet Archive 1992 #30/NN_1992_30.iso/spool/sci/math/16936

Internet Message Format | 1992-12-15 | 3.6 KB

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From: [e-mail] (Brian Cash)
Subject: Re: Marilyn Vos Savant's error?
Message-ID: <[e-mail]>
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Reply-To: [e-mail] (Brian Cash)
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Date: Tue, 15 Dec 1992 06:32:13 GMT
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In article <[e-mail]>,
[e-mail] (Steven Newman) writes:
|>> ... but while I'm on Bayes' theorem let me recall a problem similar
|>> to the above one.
|>>
|>> 1)  You draw 4 cards froma well-shuffled standard deck.  You turn one
|>> over and it's an ace.  What's the probability that they are all aces?
|>>
|>> 2)   You draw 4 cards from a well-shuffled standard deck.  You turn one
|>> over and it's the ace of hearts.  What's the probability that they are
|>> all aces?
|>>
|>> The answers to problems 1) and 2) are NOT THE SAME.
|>>
|>> Probability theory, like quantum mechanics, takes a while to get used to.
|>
|>Actually, both of these problems have answer 6/(49*50*51) = 1/20825.  There
|>are 52! possible orderings for a well-shuffled deck.  In 4! * 48! orderings
|>the first four cards are an ace; in 4 * 51! orderings the first card is an
|>ace; and the first set of orderings is a subset of the other.  Hence the
|>answer to problem 1 is (4! * 48!) / (4 * 51!) = 6 / (49*50*51).
|>
|>Similarly, for problem 2, in 51! orderings the first card is the Ace of
|>Hearts, and in 6 * 48! of those orderings the next three cards are also
|>aces.  Hence the answer to this problem is (6 * 48!) / 51! = 6 / (49*50*51)
|>as well.
|>
|>There are problems that sound very similar to these in which the answers
|>are not the same.  For example:
|>
|>   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?
|>
|>   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.")
|>
|> - Steve Newman ([e-mail])

Sorry!  You made the same mistake as the first poster!
The odds of the other child being a boy is 1/2 in both cases.
Observe:

Case 1)  Child = boy   Other = ?
Yes  ->  a) Child = boy , Other = boy
No   ->  b) Child = boy , Other = girl
         c) Child = girl, Other = boy
         d) Child = girl, Other = girl
         Obviously, c and d are invalid.

Case 2)  Replace "Child" with "Oldest Child" and
                 "Other" with "Youngest Child".

I think what you wanted was:

  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
  the older one is a boy.

Here the odds are 2/3.


    Yes     -> a) Older = Boy , Younger = Girl
    Yes     -> b) Older = Boy , Younger = Boy
    No      -> c) Older = Girl, Younger = Boy
           d) Older = Girl, Younger = Girl


Brian /-|-\