From: jokim@jarthur.claremont.edu (John H. Kim)
Newsgroups: comp.os.os2.advocacy
Subject: Re: Good news for 2.0 and other Tidbits.
Date: 1 Oct 92 00:45:46 GMT
Organization: Harvey Mudd College, Claremont, CA 91711

mccall@mksol.dseg.ti.com (fred j mccall 575-3539) writes:
>In <1992Sep30.092259.6407@muddcs.claremont.edu> jokim@jarthur.claremont.edu (John H. Kim) writes:
>>philipla@microsoft.com (Phil Lafornara) writes:
>>>In article <1992Sep28.232210.21615@muddcs.claremont.edu> jokim@jarthur.claremont.edu (John H. Kim) writes:
>>>>
>>>>He doesn't need evidence Phil.  A little logic will prove it.
>>>>
>>>>There are two sets of people.  Set A comprises those who want a
>>>>product badly enough to buy it.  Say that of set A, 75% of the
>>>>purchases are actually used.  Set B comprises those who don't
>>>>want to buy a product.  Obviously, usage in set A is much greater
>>>>than usage in set B, so we'll set B usage at 25%.  (These numbers
>>>>are completely arbitrary, and I use them only to clarify the point)

>>>     The flaw in your logic is the following:  You are taking as
>>>an axiom the idea that usage in set B will always be less that
>>>usage in set A.  I think you are probably correct - it would seem
>>>like common sense to me.  However, there aren't any figures to
>>>back up that assumption, which makes the rest of the (rather nice, 
>>>by the way) proof specious at best.

>>The proof for that axiom follows along the lines of the other proof.
>>Set Y is people who want to use the software.  Set Z is people who
>>don't want to use the software.  Set A is comprised entirely of people
>>from set Y (else why would they buy the software?).  

>Fine to this point.

>>Set B is a mix
>>of people from set Y and set Z.  

>You can't assume this as a certitude, although I'd say it's a pretty
>good bet.  Set B could presumable be an improper subset of Set A and
>not actually include any people who don't want to use the software,
>despite the fact that they did not actively acquire it (i.e., it came
>bundled).

>  The fact that Set B *MAY* contain people from Set Z does
>not mean that it *MUST* contain any.

In a strictly mathematical sense, you're right.  Since the entire set
being considered is a union of A and B, if set B contains no members of
set Z, EVERYBODY must be a member of set Y.  Sounds great up to here.
Now translate that into the real world.  EVERYBODY wants that particular
piece of software.  I think we can safely assume that isn't the case.

>Because one of your premises is falsifiable, your conclusion is
>falsifiable (which I suspect is what Phil was getting at -- although
>it'd be nice if the apparent results of logic training were applied a
>BIT more even-handedly).

One of the first things I learned in engineering was not to get trapped
into thinking if it couldn't be done logically or mathematically, it
couldn't be done at all.  Sometimes you have to make some assumptions
which are entirely reasonable in the real world.  I will give you that
in ONE case (the case you pointed out) my proof doesn't hold.  However,
that one case is so unlikely that I think most people would agree it
doesn't pose a threat to the conclusion.  Remember that the premises
are based on the real world.

Never thought I'd have to argue whether a point was closed or open...
-- 
John H. Kim                 | (This space to be filled when I
jokim@jarthur.claremont.edu | think of something very clever 
uunet!jarthur!jokim         | to use as a disclaimer)
Newsgroups: comp.os.os2.advocacy