NetNews Usenet Archive 1993 #3/NN_1993_3.iso/spool/rec/puzzles/8570

Text File | 1993-01-27 | 3.0 KB | 60 lines

Newsgroups: rec.puzzles
Path: sparky!uunet!pmafire!news.dell.com!swrinde!zaphod.mps.ohio-state.edu!cs.utexas.edu!geraldo.cc.utexas.edu!portal.austin.ibm.com!awdprime.austin.ibm.com!sunfish.austin.ibm.com!bmoore
From: [e-mail] (Bryan Moore)
Subject: Re: RAAARRGGHH!!!! not again!
Sender: [e-mail] (News id)
Message-ID: <[e-mail]>
Date: Tue, 26 Jan 1993 14:49:08 GMT
Organization: IBM, Austin
Lines: 49

From: [e-mail] (Ralph 'Hairy' Moonen)
>>In article <[e-mail]>, [e-mail] (Sean Morgan) writes:
>>
>> (Some of my mail says "not again", and some says "please post a solution".
>>  Since you can't please all of the people all of the time...)
>>
>> Two child families are uniformly distributed between MM,MF,FF,FM, in birth
>> order.  Again, all four possibilities are equally likely.  Stop here.  Make
>> sure you believe that.
>>
>> Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
>> there is only a one third chance that the sibling is a girl.

>Birth order? No one mentioned age here. The question was, that given one
>girl, what are the chances of the other also being a girl. Given a uniform
>distribution, that means for 1-child families, P(girl) = P(boy) = 0.5
>Therefore, given that one child is a girl, the chance that the other child
>is a boy, is 0.5, same as if it were a girl. Try this analogy, and you'll see
>what's wrong with your answer.

>You are throwing 2 consecutive dice. The first one turns up a six. What is
>the chance of the second also being a six. It's easy to see that because dice
>have no memory, the chance is also 1/6. Now, following your line of reasoning,
>there are 36 different ways to throw the dice. Because 1 six already has
>been thrown this leaves 11 combinations, one of which is a double six.
>So, according to you, you only have 1/11 chance of throwing a second six.
>Clearly this is dead wrong.

>--Ralph
Ralph, you are dead wrong. Let's look closer at your dice analogy to see why.
To compare the dice problem to the boy/girl problem, you need to toss the
dice a bunch of times and write the results down and put them in a hat.
This is because the two children are already born, and we're not saying
that if a woman has given birth to a girl, and is pregnant, what is the
probability that the next child is a girl. We're saying that both children
have been born and you see one of them on the street, you don't know which
child). You see a girl, what's the probability the other is a girl.

Back to the dice... Roll two dice 100 times. Get 100 pieces of paper and
write the results of the first die on one side and write the results
of the second die on the other side of the piece of paper. Put all the
pieces of paper in a hat. Now pick a piece of paper out of the hat.
You see a six on the piece of paper, what is the probability that the
other side has a six? IT IS 1/11!!!!!!!
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