Re: infinity

Tony Orlow (aeo6) wrote:
> Virgil said:
> > In article <MPG.1d62983bf135ca84989ffd@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> >
> > > I would propose a slightly different problem:
> > >
> > > We have two vases, A and B, both initially empty at 1 minute before
> > > noon. At 1/2 a minute before noon we add 10 balls to vase A, then
> > > remove a ball from vase A and place it vase B, and repeat this
> > > process every 1/2^n minutes until noon. In this case, every ball is
> > > either in vase A or vase B. At noon, how many balls are in vase A and
> > > how many are in vase B?
> >
> > This is essentially the same problem as before, so any balls not in vase
> > A before will now be in vase B at noon, instead of merely being ignored.
> >
> > Under all circumstances, there will be infinitely many in vase B at
> > noon. Which and how many will be in vase A will depend on the rule by
> > which balls are removed from A in the same way as in the former case
> > where the removed balls were merely ignored..
> >
> > Why TO should think that adding a garbage bin for the removed balls to
> > the problem would change it is a puzzlement.
> >
> So, it is your position that, even though 9 balls are added to A for every one
> added to B, that B could end up more balls than A, and could even end up with
> all the balls? Whew! It boggles the mind.
>
> Okay, what if you have white ping pong balls, and you put 10 in, and then take
> out a white one, and paint it black, and put it back? How many will be white
> and how many will be black, at noon? Can't they all just be labelled "white"
> and "black"? What ratio will you have given this scenario?

As usual an attempt at counting will get us nowhere.

We make some simple definitons:

B_n is the set of balls added at each step
(all of these are painted black)
W_n is the set of balls painted at each step

B, the set of balls added to the vase,
is the union of the B_n

W, the set of white balls in the vase at noon,
is the union of the W_n

The set of black balls in the vase at noon
is B\W.

(if you want to use labels rather
than paint, change the words white and black to white labelled
and black labelled)

It is immediately clear that to answer the question, which
elements are in B\W, we need to know not "how many balls are
painted white?" but "which balls are painted white". The
problem presented does not give this information, so the
problem presented does not have a solution.

If we add to the problem the information that, at step n
we take out ball n, paint it white and put it back, we find
that there are no elements of B\W, so no black balls
in the vase at noon. This despite the fact that at
every step, we add 10 black balls and only paint one white.
Yes, I think we can call this answer counterintuitive.
It is also correct.

-William Hughes

.

Relevant Pages

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