🚀 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.
> 
> 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