Mark Galecki wrote regarding "Fun With Bayes' Theorem":
>Lee Corbin says:
>> 2. A little girl's father discovers that his wife is
>> pregnant again (but they don't know the sex of the
>> unborn child). The man decides to visit a mathematician.
>> "I have two children, sir", he says, "and one of them
>> is a girl. What is the probability that the other is
>> a boy?" What did the mathematician tell him?
>>
>> ***Answers*** For problems 1, 2, and 3, the answer is 2/3.
>One has to be careful about formulation of the problem. Lee wants to ask
>"given that there is at least one girl, what is the probability that there
>is one boy and one girl". The answer to this is of course 2/3, just as Lee
>claims.
>
>However, this is not what Lee is really asking. Note the pair of phrases:
>"one", and "the other". What the father is really asking is "I have picked
>one of my children (not necessarily at random). That child is a girl. What
>is the probability that the other child is a boy". To this question the
>answer is 1/2.
The flaw in your argument is that the mathematician is obligated to
entertain no hypotheses concerning the procedure by which the parent
determined the truth or falsity of the statement "one of my children
is a girl". The only thing that matters is whether the statement is
true. It's not the case that the father is saying "I have picked...".
The mathematician's sample space in this case consists firstly of
all parents in the universe, then that set restricted to all parents
having two children, and then finally that set in turn restricted to
all parents who have two children one of whom is a girl. If the
mathematician then assumes that each such parent in the universe
has an equal chance of being the particular parent in question,
it follows that in two cases out of three the parent's other
child will be a boy.
It may help to consider that from the mathematician's perspective
there are actually three types of parents, not four. There are
those with two girls, those with a girl and boy each, and those
with two boys. It can be misleading to think that there are four
cases, namely GG, GB, BG, and BB, because one is then tempted to
infer some sort of order (although many people do go on from that
analysis to obtain the correct solution).
Of the three actual kinds, their weights are 1/4, 1/2, and 1/4
respectively. Since the last case has been excluded, it follows
that the probability is 2 in 3 that the mathematician is dealing
with a parent from the second category. I've inflicted this
problem on numerous acquaintances since the time years ago
when I made it up. But you have formatted the most substantial-
sounding objection I've heard. Thanks for the critical thinking.
Lee Corbin
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