Re: The "two child solution"

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

Richard Langer wrote:

> The man has one of the following
>     older son      older son
>     younger son	   younger daughter
>
> The woman has one of the following
>     older son      older son           older daughter
>     younger son	   younger daughter    younger son

Others (e.g., Stanley McCaslin, Maurice Barnhill, David Bowman) 
presented similar evidence and all went on to use these equal 
probability outcomes to answer the question.

Nevertheless, despite the fact that this line of reasoning is 
absolutely *correct* (neglecting the very minor "real world" 
considerations that some mentioned and the philosophical concerns 
that Leigh and I have expressed), I do not find it at all 
*convincing* at a gut level as Maurice--one of very few who responded 
to my request and expressed a level of confidence in his answer--also 
confessed.  Moreover, I think the response from Marilyn's readers 
proves my point.  To them (and, obviously, to some if not most of 
us), whether or not the known son is older *seems* irrelevant; the 
*only* question in either case seems to be, "What is the gender of 
the other child?"  (I find it interesting to note that this *is* the 
only question *only* because the order of birth info is *not* 
irrelevant and only *when* the order of birth info is given.)

Richard also wrote

> In my opinion, Ms. Vos Savant's appeal to her readers to send her data
> about their children is a statistically flawed experiment ...

I agree.  I worry that her results will not be any more convincing 
than the theoretical argument and that they may even be misleading.   
As a result I sent her the following note Sunday afternoon:

  I always find that the most convincing argument for the correct
  answer in situations like this involves simply imagining a very
  large number of individual events and keeping track of outcomes.
  So...
  
  Consider 1,000,000 men who will have two children.  Clearly we
  expect 500,000 of these men to have a son as their first child and
  250,000 of those 500,000 to have another as their second child.
  So 250,000/500,000 or 1 out of every 2 men with older sons have 2
  sons.  No surprise.
  
  Now consider 1,000,000 women who will have two children.  Clearly
  we expect 500,000 of these women to have a son as their first
  child; the other 500,000 will have a daughter.  Of the 500,000 who
  had a son first, 250,000 will have another son and of the 500,000
  who had a daughter first, 250,000 will have another daughter.  Now
  simply look at the results--250,000 had two sons, 250,000 had two
  daughters, and the other 500,000 had one of each.  It doesn't take
  a rocket scientist to see that 750,000 had at least one son and
  250,000 had two.  Clearly then, 250,000/750,000 or 1 out of every
  3 women with at least one son have two sons.

Of course, this is just a not very elegant way of talking about 
conditional probabilities, but I also think it is far more 
convincing.  Some might ask, "Why a *million*?  Why not just use 4?"  
Interestingly enough I think the answer is that people have a pretty 
good innate understanding that the "law of averages" applies better 
to large rather than to small populations; the fractions are sharply 
determined with a million instances and not at all so with only four.

Donald Simanek's "coin toss" analogy is equally convincing when 
extended to the same large number limit, but it requires one to 
evaluate the appropriateness of the analogy--something most are not 
adept at.

John
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