Re: physics/pedagogy of coffee-mixing

[brian whatcott <inet@INTELLISYS.NET>]

At 14:00 6/25/00 -0400, you wrote:

> ....[Ludwik]
 my answer is that
> > there is more coffee in the tea than tea in the coffee.


[John D]
> ...Oops, sorry, this is going down the wrong track.  But I suppose it
> illustrates my assertion that the wrong answers are very attractive.
>
> Here's my analysis of the problem
>
> 1) The incisive qualitative analysis:  There is a conservation law:  In the
> final state there are 2000 balls total (red+black) and 1000 balls in each
> urn.  Each red ball in the black urn displaces one black ball, which must
> (by conservation) be in the other urn.  Therefore there is _exactly_ the
> same amount of coffee in the tea as tea in the coffee.
>
> You don't even need to assume perfect mixing.
>
> 2) You can do numerical examples until you're convinced that the
> concentrations always wind up the same.  This alas loses some insights,
> such as the (non)role of mixing....

> Anybody else have thoughts on why this riddle is more challenging than it
> looks?


Hehehe....John is providing an amusing illustration of the dangers of using
 simplifying assumptions, another favorite device of physicists....

And that is how I show that his incisive qualititative analysis is wrong.

 John models the puzzle as a zero sum game with an equal partition
of two shares. This is a simplification of the puzzle as given (which
does not in fact demand equipartition in each container, as is evident from
the first sharing.)

Take a look: this is how the puzzle was presented:
"Suppose you have a cup of coffee and a cup of tea.  In step 1, you transfer
one spoonful of liquid from the coffee-cup to the tea-cup.  In step 2, you
transfer one spoonful of liquid from the tea-cup back to the
coffee-cup.  Question:  Is there more tea in the coffee, or more coffee in
the tea?"

His misdirection here, is in taking "a spoonful" to mean a fixed number
 of molecules of tea or coffee.

For any partition of a two part mixture which provides an unequal
 quantity in each of two containers, it is trivial to show that the
mix ratios can differ. Here's a numerical example to demonstrate that
the unequal mixes at the first sharing can
continue to any number of sharings which result in unequal quantities.

Cup 1 500 red balls    Cup 2 1000 black balls, 500 red balls.

Only the probabilities are in question, and hence the legitimate
concern for mixing (even if the need for unequal shares is not made
explicit)

Accordingly, where we take the quantities to represent real world measures,
we can use John's qualitative analysis to be the one and only case where
the mix ratios would in fact be identical: for the galaxy of other unequal
 partitions, the mixes can be unequal to some degree.
 In statistical terms therefore, his answer has zero probability of
occuring in the real world.    :-)

Maurice Barnhill, I notice, picked up on the misdirective equipartitioning
 assumption, and made it explicit.

 Bravo!


brian whatcott <inet@intellisys.net>
Altus OK