NetNews Usenet Archive 1993 #3/NN_1993_3.iso/spool/rec/puzzles/8537

Text File | 1993-01-25 | 1.7 KB | 37 lines

Newsgroups: rec.puzzles
Path: sparky!uunet!stanford.edu!nntp.Stanford.EDU!camus
From: [e-mail] (Theodore C. Quinn)
Subject: Re: Are you sure?
Message-ID: <[e-mail]>
Sender: [e-mail] (Mr News)
Organization: DSG, Stanford University, CA 94305, USA
References: <[e-mail]>
Date: Mon, 25 Jan 93 23:16:18 GMT
Lines: 25

In article <[e-mail]> [e-mail] (Sean Morgan) writes:
>You meet the daughter of a friend of yours on the street.  You know that your
>friend has two children.  You think to yourself, "Ignoring the occurrence of
>twins, and sex-linked differences in birth rates and infant mortality, what are
>the odds of this girl's sibling also being a girl?"

    Well, I'm new to this group and I'm sure this is "puzzle" is in the
FAQ, but I'll throw in my two cents worth, anyway.
    Of course, the common sense answer is correct: there are equal odds
of the other sibling being a female or a male. The apparent paradox lies
in an inconsistency in the counting procedure described. If one states that
there are four possible gender configurations (FF,MM,FM,MF) then clearly
the order in which the children are born is being taken into account. Otherwise
FM and MF wouldn't be listed separately.
    Therefore, since this order is being taken into account, when you
see the girl on the street there are FOUR possibilities. She is either the
youngest of a FM family, the oldest of a MF family, the youngest of a
FF family, or the oldest of a FF family. Therefore, FF gets counted twice
and the paradox is resolved. There is a 50% chance of the other sibling
being a girl.

                -Ted Quinn
                 [e-mail]