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From: mmar@sphinx.UChicago.UUCP (Mitchell Marks)
Newsgroups: net.puzzle
Subject: Re: Mixing Liquids (SPOILER)
Message-ID: <1815@sphinx.UChicago.UUCP>
Date: Wed, 19-Mar-86 16:55:52 EST
Article-I.D.: sphinx.1815
Posted: Wed Mar 19 16:55:52 1986
Date-Received: Sat, 22-Mar-86 04:14:44 EST
References: <1907@trwrba.UUCP> <184@winston.UUCP>
Reply-To: mmar@sphinx.UUCP (Mitchell Marks)
Organization: U Chicago
Lines: 55
Summary: 

In article <184@winston.UUCP> foreman@winston.UUCP (Alastair Foreman) writes:
>
>Here is an interesting puzzle, which I must confess I already know the
>answer to, but would like to know the ELEGANT solution.  I will post a
>summary if there is sufficient response.
>
>Puzzle:
>
>    Take a half a cup of tea, and a half a cup of coffee.
>    Take one tablespoon of the tea and mix it in with the coffee.
>    Take one tablespoon of the mixture and mix it back in with the tea.
>
>    The question is, which of the two cups (if either) contains more of 
>    its original contents, and WHY.
>
>Remember, I'm looking for simple, elegant solutions here....
>
>enjoy
>
>-- 

If the amounts transferred are exactly equal, so that each container
ends up with its original volume, then the amount of coffee in the tea
and the amount of tea in the coffee have to be the same (or, as A.F.
poses the question, the amount of coffee left in the cup that was originally
pure coffee, and the amount of tea left in the cup that was originally
pure tea, are the same.)

Since coffee and tea aren't pure substances, and this should be a puzzle
about logic, not physical chemistry and the geometry of molecules, let's
make this two urns that start out with all red marbles in one and all
blue marbles in the other.

What you *don't* want to do is start asking how well it's stirred, and get
into probabilities.  The answer is independent of how well you stir.

So we start out with R red marbles and B blue marbles in separate urns,
(they can be different, it doesn't affect the answer); we transfer X
red marbles into the blue urn, stir perhaps well or perhaps not, and transfer
X marbles of the mixture back to the red urn.

Now the red urn has R-X+X = R total marbles again.  A certain number of them,
Y, are blue, 0 <= Y <= X.  So the red urn still has R-Y red marbles.  It
is missing Y of the original R red marbles.  These red marbles have to be
over in the blue urn; and they are the only red marbles in the blue urn.
So the blue urn has B-Y blue marbles left.  Each urn has Y marbles of the
non-original color, and [original number] - Y of the original color 
marbles.  And this holds regardless of the exact value of Y.

We would bring in probability and discuss stirring only if we wanted to
estimate Y -- but we don't need Y.
-- 

            -- Mitch Marks @ UChicago 
               ...ihnp4!gargoyle!sphinx!mmar