From: johnson@sleipnir.mnesouth.corp.mot.com ("Johnson")
Subject: Re: Balls in the urn puzzle (LONG explanat
Reply-To: johnson@sleipnir.mnesouth.corp.mot.com
Organization: Motorola Wireless Enterprises, Schaumburg, IL
Date: Fri, 6 Nov 1992 22:38:28 GMT

In article <1992Nov5.150227.23356@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes:
=)>I'm not sure what your analogy attempts to prove.  I, personally, do not
=)>consider a basket of balls "someone of similar skill."
=)
=)The point was simply that sometimes it does matter when the labels go
=)onto objects.  Instead of humans, imagine the players are computer
=)programs, and then view the game as the execution of some deterministic
=)algorithm (determined by the computer programs) rather than as a contest.
=)Then does it make a difference in the results when you alter slightly
=)the part of the algorithm that decides when to identify the pieces?

This question is, in itself, too general to answer.  I will, however,
contend that how or when you identify which balls are removed, if you
remove and insert the same number, the resultant number of balls is
the same.
=)
=)>Neither the basket nor the balls constitute a proactive participant in
=)>the experiment.
=)
=)No, but try rephrasing the problem ever so slightly so that the person
=)putting the balls in and the person removing the balls are two different
=)people.  Now we have something resembling a "contest," although since
=)the people are constrained to predetermined "moves," we'd expect the
=)same result as if one person "played both sides."
=)
Exactly.  Your chess analogy was faulty in that the preknowledge affected
the strategy of one of the players.  This is clearly not the case in this
situation.

=)>  My point is that the location of a particular ball is
=)>not affected by what is written on it.  If you still feel that the
=)>arrangement of ink on a particular ball affects its location, I am
=)>very interested to know how.
=)
=)The arrangement of ink does not affect the location but it is correlated.
=)Actually "location" seems to me to be a red herring, because when I 
=)am looking for ball number 10 to remove I will search around the urn
=)until I find a ball with the numerals "10" written on it, regardless of
=)where in the urn that ball happens to be located.
=)
Okay, I believe you.  But does your being honest with your choice of which
ball to remove ever affect the *number* of balls remaining?

=)>  To clarify:  in the original experiment,
=)>suppose you don't change which ball goes in or out, but simply change
=)>the writing on the ball which is taken out between steps one and two.
=)
=)If you change the problem by allowing someone to go tampering with
=)the writing on the balls, then of course it is easy to change the
=)solution.  Here's another one:  Every time I remove a ball, someone
=)goes in the urn and changes the number on one ball of their choice.
=)So after my first turn they erase the "0" on ball 10 to make it ball
=)1.  Next they erase the "0" on ball 20 to make it ball 2.  Note that
=)they label the balls after I thought I'd removed them, so these balls
=)never get removed.  Clearly there will be infinitely many balls in
=)the urn at the end.  And just as clearly, this doesn't answer the
=)original problem.
=)
I thought I had implied that you would also change the writing on the
corresponding ball in the urn, so that ball 2 became 20, 1 became 10,
and so on.  I see on rereading that I failed to make this clear, and
apologize.  But given that, you can see that the *numbers* are handled
identically, and that is all that defines a particular ball.  The balls
themselves are interchangeable.  This approach not only shows the
analogy, but the fact that there is *always* a higher-numbered ball
you can play this game with.

And he also writes:
=)The "physical intuition" that some people seem to want to apply to the
=)problem is that the labels on the ball somehow don't matter, or
=)perhaps they are just an arbitrary bookkeeping mechanism.  But the
=)problem clearly states that the algorithm for adding and removing
=)balls depends entirely on the labels, so if you mess around with the
=)labels, you change the algorithm.  You are saying that the thing that
=)really matters in the problem is precisely WHICH LABELS are in the urn
=)(and after we've identified them, we can count them), and this is my
=)opinion also.

Are you saying, then, that writing a number on a ball initiates some
subtle change so that that particular ball can no longer legitimately
bear any other number?  The balls are the medium, here...if the balls
themselves are important, all this set theory is so much rhetoric,
since it deals only with the numbers.  I agree that the labels are
the mechanism, and the only importance of the balls is that it provides
an intuitive means of considering a number of labels.  If this is the
case, then moving labels from one ball to another is entirely legitimate
as long as you keep to the rule of one ball per label.

Unless, of course, your contention is that it is the actual ink
molecules that are important...:-)

tycchow@riesz.mit.edu (Timothy Y. Chow) writes:
=)2. Let S_1 = {2,3,...,9,10}
       =)S_2 = {3,4,...,19,20}
       =)S_3 = {4,5,...,29,30}
   =)etc.  What is a reasonable definition of lim S_n?
=)
=)Phrased in this way, there is unlikely to be any controversy.  The

If you mean that it is obvious that the size of S_n is greater than the
size of S_(n-m) (n and m positive, n>m) then I agree entirely.

=)controversy begins when we introduce physical intuition.  Our physical
=)intuition tells us that
=)
=)a. The balls don't "know" that they're labelled---we could remove the
    =)labels and imagine that they're there while someone performs the
    =)experiment.
Do you disagree with that?

=)b. The two processes of removing the balls therefore are indistinguishable
    =)physically---one person could imagine the balls labelled one way and
    =)someone else could imagine the balls labelled another way.
Or you could put a set of "red" labels on one side of the ball, and a set
of "blue" labels on the other side.  How does this affect the outcome? Do
all the "blue" labels miraculously disappear from the remaining balls at
the end?

=)c. The final state of the urn is well-defined (in particular the number of
    =)balls in the urn)---just "do" the experiment and find out what it is.
Okay, you get the infinite number of numbered balls, and I'll get the
infinitely large basket...maybe someone at WWV can get us the infinitely
long minute.  :-)

=)
=)Clearly these intuitive beliefs generate a paradox.  But who cares?  The
=)error lies in assuming that our physical intuition is applicable to a
=)situation that is clearly physically impossible (as far as we know).

In this case, I contend that the error is in misapplying the notion of
infinity.  See below.

carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick) writes:
=)>Am I the only one on this group that had limit theory in high school?
=)
=)No, but you may be one of the few who took the course but didn't understand it.
Hey, hey!  If you're going to get personal, at least make it funny!

=)>The number of balls in the urn is:
=)>
=)>lim   10i-i  =  lim   9i  =  inf.
=)>i->inf          i->inf
=)
=)No, the LIMIT of the number of balls in the urn is what you've described there. 
=)However, you CANNOT ARBITRARILY ASSUME THAT THE LIMIT OF A FUNCTION AND THE
=)VALUE OF THE FUNCTION ARE THE SAME.  For example, define the function F(x) as
=)follows:
	=)f(x) = 0 if x != 0
	=)f(0) = 1
=)
=)Now, lim  f(x) = 0 != 1 = f(1)
     =)x->0

Yes, I can think of discontinuous functions, too.  I can think of a person
with a mushroom for a head, too, but that doesn't necessarily prove that
you're one!
=)...
=)Perhaps you would be so kind as to show us your proof that the function
=)describing the number of balls in the urn is continuous?

Hmm...let me get this straight...you don't think 9i is continuous?

-----------------------------------------------------------------------------

Here it is.  This whole error is based on the misconception that an infinite
set of integers contains all integers.  This is obviously wrong in the case
of the set of even integers, or the set of integers not divisible by 10 as
in the first problem.  It is also wrong, although not so obviously, in the
case of sequential integers.  There is an infinite set of balls which are
removed, and an infinite set which remain.  The claim that "it can be shown
that any ball has been removed" is hand-waving on the same level as proving
that 1=2 by dividing by zero.  It is easily countered by showing that for
any ball removed, there exist nine higher-numbered balls remaining.

A simple mapping should clarify the point substantially:

Let A be the set of balls inserted in the jar, whether or not they are
removed.  Clearly, A approaches an infinite sequence of integers as i=>inf.

Let B be a mapping of A.  B_i = A_i-{i,i,i,i,i,i.....}

A_1={1,2,3,...,10}            B_1={0,1,2,...,9}
A_2={1,2,3,...,20}            B_2={-1,0,1,2,...18}
A_3={1,2,3,...,30}            B_2={-2,-1,0,1,2,...27}
A_3={1,2,3,...,40}            B_2={-3,-2,-1,0,1,2,...36}

In the set B, the elements <=0 represent the balls removed from the jar, and
the elements >0 represent those remaining.  As i=>inf, both sides of the
set become infinitely large.  Therefore, there are an infinite number of
balls remaining in the jar.

An earlier poster noted the tenacity with which people maintain their
opinions on this problem.  I am also astounded by it--I am entirely
sure I am right, and I gather my detractors feel the same.  If anyone still
disagrees with this conclusion, I am eager to hear additional arguments.
Please try to keep the personal abuse down, though, since I have a nervous
condition.

(No, of course I don't...flame away!  :-)
-- 
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| "Johnson"                              |Just because you're paranoid doesn't|
| johnson@sleipnir.mnesouth.corp.mot.com |   mean they're not out to get you. |
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