🚀 go-pugleaf

In article <1993Jan25.115908.41@janus.arc.ab.ca>, morgan@arc.ab.ca (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.

Okay, got it.

>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.

But what are the chance of you meeting one of these three pairs?  Their
birth chance does not imply your meeting chance.  Let me show it:

We are talking about two-kids families.  4 equal combinations but 8
possible kids, mm, MF, FM, ff.  Assume you have equal chance meeting
one of them.  But you are limiting your calculation ONLY if you meet a
girl, i.e. only if you meet the "F" of {FM,MF} or either "f" of "ff".
So, total four possible girls.  Half of them have a femal sibling, the
other half male sibling.  Thus the answer of your question is 0.5.

Nothing personal but here is a joke i remember reading the "Born Loser"
comic.  He figured the chance of boarding a plain having a bomb is
1 out of a million.  Having two bombs on the same plane should be
1 out of a trillion.  So, he decided to bring along a bomb when
he was reading to board it.