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Article #2500Re: Are you sure?
From: Theodore C. Quinn
Date: Mon, 25 Jan 1993 23:16
Date: Mon, 25 Jan 1993 23:16
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In article <1993Jan22.131719.36@janus.arc.ab.ca> morgan@arc.ab.ca (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 quinn@relgyro.stanford.edu
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Newsgroups: rec.puzzles Path: arc1-ca.pugleaf.net!sparky!uunet!stanford.edu!nntp.Stanford.EDU!camus From: camus@leland.Stanford.EDU (Theodore C. Quinn) Subject: Re: Are you sure? Message-ID: <1993Jan25.231618.15311@leland.Stanford.EDU> Sender: news@leland.Stanford.EDU (Mr News) Organization: DSG, Stanford University, CA 94305, USA References: <1993Jan22.131719.36@janus.arc.ab.ca> Date: Mon, 25 Jan 93 23:16:18 GMT Lines: 25