infinity
Theo Jacobs
I'm having an argument with a friend about the following problem:
Suppose you have a giant vase and a bunch of ping pong balls with an
integer written on each one, e.g. just like the lottery, so the balls
are numbered 1, 2, 3, ... and so on. At one minute to noon you put
balls 1 to 10 in the vase and take out number 1. At half a minute to
noon you put balls 11 - 20 in the vase and take out number 2. At one
quarter minute to noon you put balls 21 - 30 in the vase and take out
number 3. Continue in this fashion. Obviously this is physically
impossible, but you get the idea. Now the question is this: At noon,
how many ping pong balls are in the vase?
An 2nd experiment goes as follows:
put no. 1 - 10 in, take no. 10 out, put no. 11-20 in, take no. 20 out, put
no. 21-30 in, take no. 30 out, etc.
(of course at the same moments as above)
My friend claims both experiments end up with an empty vase.
I think however the 2nd experiment ends up with a vase with an infinite
number of balls:
1-9, 11-19, 21-29 etc. are definitely in the vase.
He says it all has to do with Cantor's set theory, cardinality etc..., but
browsing the internet didn't really help me much.
Any information or relevant links are very welcome,
Thanks, Theo
Stuart M Newberger
pong,vases and time -I hope their natures are not essentials when this
problem is modeled in set theory ruled mathematics.
Theo Jacobs wrote:
> Hello everyone,
Hi-this should be safe
> I'm having an argument with a friend about the following problem:
>
> Suppose you have a giant vase and a bunch of ping pong balls with an
> integer written on each one, e.g. just like the lottery, so the balls
> are numbered 1, 2, 3, ... and so on. At one minute to noon you put
> balls 1 to 10 in the vase and take out number 1. At half a minute to
> noon you put balls 11 - 20 in the vase and take out number 2. At one
> quarter minute to noon you put balls 21 - 30 in the vase and take out
> number 3. Continue in this fashion. Obviously this is physically
> impossible, but you get the idea. Now the question is this: At noon,
> how many ping pong balls are in the vase?
How big are the ping pong balls-ok I withdraw the question.But no
action is taken at or after noon .Ok,I see that you both agree that in
experiment 1 the vase is empty at and after noon.I agree so lets forget
this part.
> An 2nd experiment goes as follows:
> put no. 1 - 10 in, take no. 10 out, put no. 11-20 in, take no. 20 out, put
> no. 21-30 in, take no. 30 out, etc.
> (of course at the same moments as above)
>
> My friend claims both experiments end up with an empty vase.
> I think however the 2nd experiment ends up with a vase with an infinite
> number of balls:
> 1-9, 11-19, 21-29 etc. are definitely in the vase.
Here is really all I have to say.I do not see the difference between
the 2nd experiment and a third one where you just put in 1-9,11-19 ,etc
(into the vase,into the vase) at the indicated times and never mention
the other balls.
Why wouldnt they all be there -is there a hole in the bottom
(sorry).This means I agree with you what is your friends problem ?
Wait,how many,Answer-the countable disjoint union of those finite sets
1-9,11-19.etc of sets in the vase at noon is a denumerable infinite
set.It is said to have cardinality aleph naught.
> He says it all has to do with Cantor's set theory, cardinality etc..., but
> browsing the internet didn't really help me much.
Learning something about cardinality and set theory is a noble and
worthwhile undertaking but very little is needed for this problem
except as mentioned above,well I guess I don't see the
problem.Regards,Stuart M Newberger
Jeroen Boschma
>
> Hello everyone,
>
> I'm having an argument with a friend about the following problem:
>
> Suppose you have a giant vase and a bunch of ping pong balls with an
> integer written on each one, e.g. just like the lottery, so the balls
> are numbered 1, 2, 3, ... and so on. At one minute to noon you put
> balls 1 to 10 in the vase and take out number 1. At half a minute to
> noon you put balls 11 - 20 in the vase and take out number 2. At one
> quarter minute to noon you put balls 21 - 30 in the vase and take out
> number 3. Continue in this fashion. Obviously this is physically
> impossible, but you get the idea. Now the question is this: At noon,
> how many ping pong balls are in the vase?
Each time that you do something with the balls, you add 10 to the vase and take one out, so the
number of balls in the vase increases by 9.
Let T(n) be the time left until it's noon at the n-th time that you add balls to the vase, then T(n)
= 2^(1-n). This function never reaches noon (T=0), but only approaches it for n->infinity. So the
number of balls in the vase keep growing to infinity as you approach noon.
>
> An 2nd experiment goes as follows:
> put no. 1 - 10 in, take no. 10 out, put no. 11-20 in, take no. 20 out, put
> no. 21-30 in, take no. 30 out, etc.
> (of course at the same moments as above)
Same experiment, who cares what the number is on the ball that's taken out of the vase: each time
that you do something with the balls, you add 10 to the vase and take one out, so the number of
balls in the vase increases by 9.
>
> My friend claims both experiments end up with an empty vase.
So each time you add more balls to the vase (10) then you take out (1), and that gives him an empty
vase? Not in my universe...
> I think however the 2nd experiment ends up with a vase with an infinite
> number of balls:
> 1-9, 11-19, 21-29 etc. are definitely in the vase.
>
> He says it all has to do with Cantor's set theory, cardinality etc..., but
> browsing the internet didn't really help me much.
For somebody who wants to deal with these subjects, he's giving quite a 'naive' (I'm just being
nice...) answer to the simple experiments above.
David C. Ullrich
wrote:
>Hello everyone,
>
>I'm having an argument with a friend about the following problem:
>
>Suppose you have a giant vase and a bunch of ping pong balls with an
>integer written on each one, e.g. just like the lottery, so the balls
>are numbered 1, 2, 3, ... and so on. At one minute to noon you put
>balls 1 to 10 in the vase and take out number 1. At half a minute to
>noon you put balls 11 - 20 in the vase and take out number 2. At one
>quarter minute to noon you put balls 21 - 30 in the vase and take out
>number 3. Continue in this fashion. Obviously this is physically
>impossible, but you get the idea. Now the question is this: At noon,
>how many ping pong balls are in the vase?
None.
>An 2nd experiment goes as follows:
>put no. 1 - 10 in, take no. 10 out, put no. 11-20 in, take no. 20 out, put
>no. 21-30 in, take no. 30 out, etc.
>(of course at the same moments as above)
>
>My friend claims both experiments end up with an empty vase.
>I think however the 2nd experiment ends up with a vase with an infinite
>number of balls:
>1-9, 11-19, 21-29 etc. are definitely in the vase.
Of course that's correct.
>He says it all has to do with Cantor's set theory, cardinality etc..., but
>browsing the internet didn't really help me much.
>Any information or relevant links are very welcome,
Cardinality has very little to do with this problem.
The _reason_ cardinality has very little to do with the problem,
if you want to get technical, is that the cardinality of a limit
of a sequence of sets need not be the limit of the cardinalities.
>Thanks, Theo
>
************************
David C. Ullrich
David C. Ullrich
<jeroen....@tno.nl> wrote:
>Theo Jacobs wrote:
>>
>> Hello everyone,
>>
>> I'm having an argument with a friend about the following problem:
>>
>> Suppose you have a giant vase and a bunch of ping pong balls with an
>> integer written on each one, e.g. just like the lottery, so the balls
>> are numbered 1, 2, 3, ... and so on. At one minute to noon you put
>> balls 1 to 10 in the vase and take out number 1. At half a minute to
>> noon you put balls 11 - 20 in the vase and take out number 2. At one
>> quarter minute to noon you put balls 21 - 30 in the vase and take out
>> number 3. Continue in this fashion. Obviously this is physically
>> impossible, but you get the idea. Now the question is this: At noon,
>> how many ping pong balls are in the vase?
>
>Each time that you do something with the balls, you add 10 to the vase and take one out, so the
>number of balls in the vase increases by 9.
>
>Let T(n) be the time left until it's noon at the n-th time that you add balls to the vase, then T(n)
>= 2^(1-n). This function never reaches noon (T=0), but only approaches it for n->infinity. So the
>number of balls in the vase keep growing to infinity as you approach noon.
Your formula for T(n) is wrong. But that doesn't matter. It's true
that the limit of T(n) as n -> infinity is infinite. And that
doesn't change the fact that the vase is empty at noon.
(You're argument would be correct if cardinality were "continuous",
in the sense that the cardinality of the limit of a sequence of sets
is the limit of the cardinailities. But cardinality is simply not
continuous in this sense.)
>> An 2nd experiment goes as follows:
>> put no. 1 - 10 in, take no. 10 out, put no. 11-20 in, take no. 20 out, put
>> no. 21-30 in, take no. 30 out, etc.
>> (of course at the same moments as above)
>
>Same experiment, who cares what the number is on the ball that's taken out of the vase: each time
>that you do something with the balls, you add 10 to the vase and take one out, so the number of
>balls in the vase increases by 9.
>
>>
>> My friend claims both experiments end up with an empty vase.
>
>So each time you add more balls to the vase (10) then you take out (1), and that gives him an empty
>vase? Not in my universe...
In the first experiment, in your universe, can you name a ball that is
still in the vase at noon?
>> I think however the 2nd experiment ends up with a vase with an infinite
>> number of balls:
>> 1-9, 11-19, 21-29 etc. are definitely in the vase.
>>
>> He says it all has to do with Cantor's set theory, cardinality etc..., but
>> browsing the internet didn't really help me much.
>
>For somebody who wants to deal with these subjects, he's giving quite a 'naive' (I'm just being
>nice...)
Uh, you're also being wrong about the answer.
>answer to the simple experiments above.
>
>> Any information or relevant links are very welcome,
>>
>> Thanks, Theo
************************
David C. Ullrich
David C. Ullrich
<ull...@math.okstate.edu> wrote:
>On Wed, 03 Aug 2005 12:20:06 +0200, Jeroen Boschma
><jeroen....@tno.nl> wrote:
>
>>Theo Jacobs wrote:
>>>
>>> Hello everyone,
>>>
>>> I'm having an argument with a friend about the following problem:
>>>
>>> Suppose you have a giant vase and a bunch of ping pong balls with an
>>> integer written on each one, e.g. just like the lottery, so the balls
>>> are numbered 1, 2, 3, ... and so on. At one minute to noon you put
>>> balls 1 to 10 in the vase and take out number 1. At half a minute to
>>> noon you put balls 11 - 20 in the vase and take out number 2. At one
>>> quarter minute to noon you put balls 21 - 30 in the vase and take out
>>> number 3. Continue in this fashion. Obviously this is physically
>>> impossible, but you get the idea. Now the question is this: At noon,
>>> how many ping pong balls are in the vase?
>>
>>Each time that you do something with the balls, you add 10 to the vase and take one out, so the
>>number of balls in the vase increases by 9.
>>
>>Let T(n) be the time left until it's noon at the n-th time that you add balls to the vase, then T(n)
>>= 2^(1-n). This function never reaches noon (T=0), but only approaches it for n->infinity. So the
>>number of balls in the vase keep growing to infinity as you approach noon.
>
>Your formula for T(n) is wrong. But that doesn't matter. It's true
>that the limit of T(n) as n -> infinity is infinite. And that
>doesn't change the fact that the vase is empty at noon.
Strike that last paragraph - I read your definition of T(n) wrong,
sorry.
Say S(n) is the number of balls in the vase at time T(n). Then it's
true that S(n) tends to infinity as n -> infinity. Which doesn't
change the fact that there are _no_ balls in the vase at noon
(see the next paragraph.)
************************
David C. Ullrich
Rusty
"Theo Jacobs" <t.ja...@chello.nl> wrote in message
news:Vj%He.2121$jA4.1934@amstwist00...
I think a physicist would say that the time scale is getting exponentially
smaller and smaller and eventually Heisenburg's uncertanty principle will
kick in. You will not be certain whether the balls are inside or outside as
the situation will be unobservable at the quantum level.
rusty
Jeroen Boschma
Hi David,
Always nice to be corrected by a true mathematician :)
When I think about the problem again, it gets more interesting. To my defence, I used the phrase
"the number of balls in the vase keeps growing to infinity as you _approach_ noon". To be honest, I
forgot to give an answer on the number of balls in the vase at noon exactly.
So here's what I think about it. It's obvious that the number of balls in the vase keeps growing as
you approach noon. However, the algorithm which describes the filling of the vase with balls never
reaches noon. Therefore, I think it is not even correct to ask the original question 'how many balls
at noon' (do mathemathicians call that ill-defined, -posed or something like that?). The algorithm
just cannot answer that question because noon is not covered by the algorithm. And for that reason I
also think that it is not fair just to state that the vase is empty at noon, it is just 'not
defined'. Or do mathematicians have another view on this?
Jeroen
The Qurqirish Dragon
Rusty wrote:
> I think a physicist would say that the time scale is getting exponentially
> smaller and smaller and eventually Heisenburg's uncertanty principle will
> kick in. You will not be certain whether the balls are inside or outside as
> the situation will be unobservable at the quantum level.
Although this doesn't at all help in answering the question, I must say
that I think this is one of the best responses I have heard to this
question (other than the proper answer, of course)
David C. Ullrich
<jeroen....@tno.nl> wrote:
Nobody said anything about _algorithms_ - an algorithm terminates in
finitely many steps, so whatever we're talking about it's not an
algorithm. And of course the problem has no actual physical meaning.
But if we do assume what the problem says we're supposed to assume
then there's nothing poorly defined about the answer: At noon every
ball that was inserted has been removed, so there are no balls left.
> Jeroen
************************
David C. Ullrich
Dave Seaman
> So here's what I think about it. It's obvious that the number of balls in the vase keeps growing as
> you approach noon. However, the algorithm which describes the filling of the vase with balls never
> reaches noon. Therefore, I think it is not even correct to ask the original question 'how many balls
> at noon' (do mathemathicians call that ill-defined, -posed or something like that?). The algorithm
> just cannot answer that question because noon is not covered by the algorithm. And for that reason I
> also think that it is not fair just to state that the vase is empty at noon, it is just 'not
> defined'. Or do mathematicians have another view on this?
> Jeroen
Which ball do you consider to have an undefined state at noon? If there
is more than one, then which is the lowest-numbered ball whose state is
undefined?
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
cbr...@cbrownsystems.com
<snip>
>
> So here's what I think about it. It's obvious that the number of balls in
> the vase keeps growing as you approach noon.
> However, the algorithm which describes the filling of the vase with balls
> never reaches noon. Therefore, I think it is not even correct to ask the
> original question 'how many balls at noon' (do mathemathicians call that
> ill-defined, -posed or something like that?).
There is some sense to what you say.
There is still some room IMO for claiming that the question itself is
"ill-defined". It's certainly not physically possible for such an
operation to take place in the real world to start with; so the
question of "what would happen at noon?" is really best interpreted as
"what would be a logically consistent answer to the question, assuming
that it /could/ be done in practice?"
Suppose instead of your experiment 1, we put a ball into the vase on
odd steps, and then take it out the even steps.
Then it seems obvious that there is no single well-defined answer to
"how many balls are there at noon?". Neither "empty" nor "containing a
ball" seems to be correct - there is no answer that is somehow
/logically preferable/ to all others; and that is the basic sense of
"not well-defined".
On the other hand, if you did /nothing/ at each step, that there would
somehow be a ball in the vase "at noon" seems obviously wrong. The vase
would "obviously" be empty, because we intuitively require that there
can't be a ball in the vase unless it was put there at some finite step
n.
Similarly, it seems intuitive that if at step 1, you put ball "1" in
the vase, and then at no later step do you remove it, then that ball
"must be" in the vase "at noon".
These two intuitions are the assumptions we are making when we ask
"assuming it is possible in the real world, what balls are in the vase
at noon?"; and a well-defined answer would be one which is consistent
with these intuitions, and also be the /only/ one consistent with these
intutions.
Thus, in experiment 1, we can agree that, if at some definite step t_n,
we remove the ball labelled "n", and then there is no later step that
we put the ball back in, then it is perfectly reasonable to insist
that, /whatever/ we mean by "the balls in the vase at noon", it can't
mean that ball "n" is in the vase at noon.
Since, in your construction, the above is true for every n, the only
/possible/ answer consistent with our assumptions is that there are no
balls in the vase at noon.
And since that /is/ "the" single, consistent answer to the question, we
say "the vase is empty at noon" is a well-defined answer to the
question.
Cheers - Chas
guenther vonKnakspot
on your answer, and have further questions for you. What would happen
if the labels were inside the balls, that is if the numbers on the
balls were not readable to the person performing the experiment? would
the vase be empty at noon? Suppose some person A goes ever morning and
starts putting ping pong balls with the numbers painted on the inside
into the vase and taking one out of every batch of ten that go in as
described in experiment 1. Person B has a device which enables him to
read the numbers inside the balls, so person B can actually determine
when the sequence of experiment 1 actually took place. Would the vase
appear full to A and empty to B on the evening of that day? What
happens if you set up the following experiment you have vases v1 and v2
and you have numbered balls and unnumbered balls. Then you perform
experiment 1 with the numbered balls and v1. At the same time, for each
numbered ball that you put into v1, you put an unnumbered one into v2
and for each numbered ball you take out of v1, you take an unnumbered
one out of v2. If v1 is empty at noon, is v2 empty at noon too?
Thanks a lot in advance,
Regards.
Theo Jacobs
news:1123107478.9...@g49g2000cwa.googlegroups.com...
I'd like to thank everybody for their input so far.
Guenther: your description is not the same as experiment 1:
in this experiment no. 1-10 are put in and no.1 is taken out
no. 11-20 are put in, no. 2 is taken out etc.....
So, no. 2 is _not_ the same 'batch' as balls 11-20, in this experiment the
balls taken out come from a batch that was put in earlier (except the 1st
batch)
What makes this so remarkable is that -eventually- every ball, and so of
course every batch too, is taken out again, before noon.
For unnumbered balls the outcome -as far as I can tell - is also an empty
vase! This is not my invention however, I read this here:
http://www.cut-the-knot.org/Probability/infinity.shtml
(scroll down a bit for the relevant part, although you might like to read it
all)
In your 'unnumbered experiments' the chance of a ball eventually being taken
out again is calculated as being 1 ! - caused of course by the fact that you
do this an infinite number of times. This applies for every ball, so the
vase will be empty!!
Regards, Theo
ste...@nomail.com
>>number of balls:
>>1-9, 11-19, 21-29 etc. are definitely in the vase.
> Of course that's correct.
>>He says it all has to do with Cantor's set theory, cardinality etc..., but
>>browsing the internet didn't really help me much.
>>Any information or relevant links are very welcome,
> Cardinality has very little to do with this problem.
> The _reason_ cardinality has very little to do with the problem,
> if you want to get technical, is that the cardinality of a limit
> of a sequence of sets need not be the limit of the cardinalities.
This is a good example of why subtraction is not defined
for the transfinite cardinals. In both cases aleph_0 balls
are added and removed, but the answer is not aleph_0 - aleph_0.
Someone somewhat familiar with cardinality may make the
mistake of thinking that aleph_0 - aleph_0 equals 0, and
that the number of balls left in the vase must be 0 in
both cases.
Stephen
cbr...@cbrownsystems.com
<snip>
> For unnumbered balls the outcome -as far as I can tell - is also an empty
> vase! This is not my invention however, I read this here:
> http://www.cut-the-knot.org/Probability/infinity.shtml
> (scroll down a bit for the relevant part, although you might like to read it
> all)
I found his argument not /quite/ so convincing as the argument for
numbered balls.
His argument seems to be essentially that, assuming that the ball to be
removed at each step is selected with a uniform probability from the
set of balls in the vase at that step, that therefore the /probability/
that any /particular/ ball will still be in the vase tends to 0 as the
number of steps tends toward infinity. Therefore there is probability 1
that the vase is empty "at noon".
I have two (slight) problems with this.
First, the assumption that the balls are distinguishable but selected
at random is quite reasonable but not neccessary; we could equally well
say that the balls are merely chosen by some /unknown/ method, not
neccessarily with random distribution (for example, maybe the balls at
the "bottom" of the vase are naturally chosen with somewhat less
frequently than the balls at the "top" of the pile). If we only assume
that "some" method is used and that the balls are essentially
indistinguishable, I think a better answer is "a countably infinite
number of balls are in the vase at noon".
But even accepting the premise, the fact that an event has probability
0 of occuring is not (in general) the same as saying that it is
logically /impossible/ for that event to occur.
Thus, the statement that "the vase is empty" is not the same as "the
vase is empty, with probability 1" (the latter being all that is
asserted at the link you gave).
Still, it is pretty slick... :)
Cheers - Chas
Dave Seaman
> Theo Jacobs wrote:
><snip>
>> For unnumbered balls the outcome -as far as I can tell - is also an empty
>> vase! This is not my invention however, I read this here:
>> http://www.cut-the-knot.org/Probability/infinity.shtml
>> (scroll down a bit for the relevant part, although you might like to read it
>> all)
> I found his argument not /quite/ so convincing as the argument for
> numbered balls.
> His argument seems to be essentially that, assuming that the ball to be
> removed at each step is selected with a uniform probability from the
> set of balls in the vase at that step, that therefore the /probability/
> that any /particular/ ball will still be in the vase tends to 0 as the
> number of steps tends toward infinity. Therefore there is probability 1
> that the vase is empty "at noon".
The probability of any particular ball being left in the vase at noon
does not "tend to" 0. The probability of any particular ball being left
in the vase at noon is exactly 0.
And since there are countably many balls, it follows by countable
additivity of probability measure that the probability of the vase being
nonempty at noon is also 0.
> I have two (slight) problems with this.
> First, the assumption that the balls are distinguishable but selected
> at random is quite reasonable but not neccessary; we could equally well
> say that the balls are merely chosen by some /unknown/ method, not
> neccessarily with random distribution (for example, maybe the balls at
> the "bottom" of the vase are naturally chosen with somewhat less
> frequently than the balls at the "top" of the pile). If we only assume
> that "some" method is used and that the balls are essentially
> indistinguishable, I think a better answer is "a countably infinite
> number of balls are in the vase at noon".
If the balls are merely chosen by "some unknown method", then for all we
know, the balls may be removed in last-in, first-out order, which
guarantees that infinitely many balls will be left at noon. That's why
more information is needed. The assumption of a uniform distribution at
each time step is sufficient, but not necessary, to conclude that the
vase is emptied with probability 1.
> But even accepting the premise, the fact that an event has probability
> 0 of occuring is not (in general) the same as saying that it is
> logically /impossible/ for that event to occur.
Nobody has claimed otherwise.
> Thus, the statement that "the vase is empty" is not the same as "the
> vase is empty, with probability 1" (the latter being all that is
> asserted at the link you gave).
> Still, it is pretty slick... :)
--
cbr...@cbrownsystems.com
Yes, well I suppose I shouldn't just mumble my thoughts. :)
The tie-the-knot site uses explicitly numbered balls, just as the OP's
first example did; but most recently, the topic seems to have changed
to essemtially "unlabelled" balls.
If the balls are chosen by some unknown method, but the balls are
indistinguishable (which is what I "essentially" said), then it seems
to me that we are within our rights to "relabel" the balls at each
step, since at each step all we /really/ know about the balls in the
vase is the cardinality of the set of those balls.
The sequence is not then some function from the naturals to the
naturals indicating what which ball is removed at that step, as
distingushed by its labelling.
Instead we have a function from the naturals to some set of balls,
about which all we know is its resultant cardinality (i.e., that they
/can be/ labelled distinctly in a certain way).
This sequence would seem to have as its limit a countable set of balls,
not an empty or finite one.
> The assumption of a uniform distribution at
> each time step is sufficient, but not necessary, to conclude that the
> vase is emptied with probability 1.
>
> > But even accepting the premise, the fact that an event has probability
> > 0 of occuring is not (in general) the same as saying that it is
> > logically /impossible/ for that event to occur.
>
> Nobody has claimed otherwise.
>
The poster to whom I was replying said:
> >> For unnumbered balls the outcome -as far as I can tell - is also an empty
> >> vase!
To me, saying the outcome is an empty vase is not the same as saying
that thye probability that the vase would be empty is 1.
Cheers - chas
cbr...@cbrownsystems.com
> Dave Seaman wrote:
> > cbr...@cbrownsystems.com wrote:
<snip>
No, no, no.
The problem statement requires that there be a total of a countable
number of balls which can ever be in the urn at some step n in the
sequence (if we had a step where we added an uncountable number of
balls, then the original "vase is empty at noon" argument wouldn't
hold).
Therefore there is a bijection betwen the set of naturals and the balls
which allows us, in principle, to say that they are "distinguishable"
(or at least labelled in some definite way). Whether I know what this
bijection is or not, it certainly exists; call it T.
Then over all possible functions which at step n select a ball from the
subset of N represented by the balls in the vase at step n, there are
an equal number (cardinality) of such functions which select one of
these balls as any another; that translates into "select a ball from
the set of balls in the vase, with uniform distribution", and then the
original probability argument holds.
I stand corrected (even if by myself).
Cheers - Chas
Jeroen Boschma
>
> Jeroen Boschma wrote:
>
> <snip>
> >
> > So here's what I think about it. It's obvious that the number of balls in
> > the vase keeps growing as you approach noon.
>
> > However, the algorithm which describes the filling of the vase with balls
> > never reaches noon. Therefore, I think it is not even correct to ask the
> > original question 'how many balls at noon' (do mathemathicians call that
> > ill-defined, -posed or something like that?).
>
> There is some sense to what you say.
>
> There is still some room IMO for claiming that the question itself is
> "ill-defined". It's certainly not physically possible for such an
> operation to take place in the real world to start with; so the
> question of "what would happen at noon?" is really best interpreted as
> "what would be a logically consistent answer to the question, assuming
> that it /could/ be done in practice?"
The basic idea I had was that we can make a function B(n) which describes the number of balls in the
vase from one minute to noon to every time less than noon. But the description of the experiment
does not allow us to reach noon. It does not have a sense of a continuously progressing time which
is required to reach noon. So I would say that the experiment directly implies that 'noon' is not
within the domain of B(n), and therefore the function, or description of the experiment, cannot give
an answer to the question.
*** For nitpickers :) , as B(n) is a function of variable 'n', statements about the domain should
be based on 'n' only, I guess. But every 'n' implies a certain time t_n, so I think you get the
idea... I hope... else beat me again :(
***
From the reasoning in this thread, experiment 1 should result in an empty vase, and experiment 2 in
infinite balls in the vase at noon. Different answers while the function B(n) that describes the
number of balls in the vase at any 'n' is the same for both... The question is a physical one
(number of balls), and for that reason the answer cannot depend on how the balls are numbered or
what color they have.
>
> Suppose instead of your experiment 1, we put a ball into the vase on
> odd steps, and then take it out the even steps.
>
> Then it seems obvious that there is no single well-defined answer to
> "how many balls are there at noon?". Neither "empty" nor "containing a
> ball" seems to be correct - there is no answer that is somehow
> /logically preferable/ to all others; and that is the basic sense of
> "not well-defined".
For the same reason as I described above: for each n we can find the number of balls in the vase,
but there is no n which equals 'noon'. Therefore, because the number of balls change at every step
n, we should not want to force an answer. We can just say that no answer is possible because 'noon'
is not in the domain of the function B(n) which describes the number of balls in the vase.
>
> On the other hand, if you did /nothing/ at each step, that there would
> somehow be a ball in the vase "at noon" seems obviously wrong. The vase
> would "obviously" be empty, because we intuitively require that there
> can't be a ball in the vase unless it was put there at some finite step
> n.
Because there are no steps defined where the number of balls in the vase changes, we don't need to
stick to those discrete moments t_n, but we can directly describe the number of balls in 'continuous
time': nothing happens. So we are not bound by a function B(n) in which 'noon' is not element of the
domain.
>
> Similarly, it seems intuitive that if at step 1, you put ball "1" in
> the vase, and then at no later step do you remove it, then that ball
> "must be" in the vase "at noon".
>
> These two intuitions are the assumptions we are making when we ask
> "assuming it is possible in the real world, what balls are in the vase
> at noon?"; and a well-defined answer would be one which is consistent
> with these intuitions, and also be the /only/ one consistent with these
> intutions.
Again, IMHO, forgetting that the answer must be given by a function which describes the number of
balls in the vase, and that 'noon' is not in the domain of that function. We don't _need_ to give a
number as an answer.
In this experiment, we try to connect some logical reasoning (about putting balls with number in the
vase and getting some out) with a physical entity (number of balls in the vase). That connection
cannot be made because we have no description or function which can make the connection!
>
> Thus, in experiment 1, we can agree that, if at some definite step t_n,
> we remove the ball labelled "n", and then there is no later step that
> we put the ball back in, then it is perfectly reasonable to insist
> that, /whatever/ we mean by "the balls in the vase at noon", it can't
> mean that ball "n" is in the vase at noon.
>
> Since, in your construction, the above is true for every n, the only
> /possible/ answer consistent with our assumptions is that there are no
> balls in the vase at noon.
So your explanation is that for each ball n in the vase, there is a step t_n where that ball is
taken out. So each ball is taken out eventually which results in an empty vase at noon. This is
based on _physical_ observations: every ball put in the vase is taken out at some time. Another
_physical_ observation I made in my original reply is: at every 'n' the number of balls in the vase
increases by 9. So there is no interval between one minute to noon and noon where the number of
balls in the vase decreases. So how can the vase be empty at noon?
IMO, both reasonings are OK, but they _seem_ to end up with different answers. Reason for this
contradiction is that we're not allowed to formulate a numeric answer to the original question,
because noon is not in the domain of the experiment.
snapdr...@gmail.com
on infinite series first and then explain to you why the vase is not
empty at the noon time.
Let S = 1 + 2 + 3 + 4 + 5 + 6 + ... + oo oo = infinity
Then 2*S = 2 + 4 + 6 + 8 + .... + oo
Subtract S from 2*S, we have
2*S - S = (2 + 4 + 6 + ... + oo) - (1 + 2 + 3 + 4 + 5 + 6 + ... +
oo)
==> S = (2 + 4 + 6 + ... + oo) - (1 + 3 + 5 + ... + oo + 2 + 4 + 6 +
... + oo)
==> S = (2 + 4 + 6 + ... + oo) - (1 + 3 + 5 + ... + oo) - (2 + 4 + 6 +
... + oo)
==> S = -(1 + 3 + 5 + 7 .... + oo)
What? The summation of all integers is a negative number? Obviously
there must be something wrong. The error is that we cannot rearrange
numbers like that for an infinite series.
Now go back to your question
Let S1 be the balls being put in the vase
Let S2 be the balls being taken out from the vase
At time = noon
S1 = {1, 2, 3, ... , oo} n(S1) = oo
S2 = {1, 2, 3, ... , oo} n(S2) = oo
# of balls in the vase would be n(S1) - n(S2) = oo - oo which is
undefined.
As seen from the previous example, we cannot just cancel out same terms
from each set and come up with result zero. S1 and S2 are infinite
series with different properties. We cannot simply say every number in
the set S1 also exists in S2 then S1 - S2 = 0. If you do that then you
are rearrangeing the numbers in the infinite series.
In mathematics, usually if at 'noon' the question is undefined then we
would take the limit of the time --> noon to see the tendency of the
result
As time --> noon
S1 = {1, 2, 3, ... , 10n) Number of S1 = 10n
S2 = {1, 2, 3, ... , n) Number of S2 = n
The remaining balls are {n+1, n+2, ... , 10n}
# of balls remained in the vase would be n(S1) - n(S2) = 10n - n = 9n >
0
Therefore, the vase is not empty and actually it is growing rapidly.
>From a physical point of view, the number of balls in the vase are
growing rapidly as the time approaches noon. Even though, right at
the noon time, the problem is undefined but we would expect the vase is
not empty.
guenther vonKnakspot
guenther vonKnakspot wrote:
<snip>
> starts putting ping pong balls with the numbers painted on the inside
> into the vase and taking one out of every batch of ten that go in as
That should of course, as pointed out by Jacobs, read taking one out
FOR every batch of ten that go in.
Theo Jacobs
<snapdr...@gmail.com> schreef in bericht
news:1123144560.3...@g43g2000cwa.googlegroups.com...
Hello,
Thanks for your contribution. The thing is that there's something to be said
for both lines of reasoning (full & empty).
However, even your set of 'remaining balls' are taken out before noon.
In general ball no. n is taken out at 2^(n-1) minutes before noon, so ball
no. 10n is simply taken out at 2^(10n-1) minutes before noon.
An argument that I read about is 'where the ball finally comes to rest', for
_every_ ball this is _outside_ the vase.
Anyway, I don't claim to be an expert, far from it, all I contributed is
what I read on the internet, this problem is also known as the
'Ross-Littlewood paradox', you can read more about it on the following
sites:
http://www.csc.twu.ca/byl/little.fin.doc
http://www.cut-the-knot.com/Probability/infinity.shtml
http://www.cs.uu.nl/wais/html/na-dir/puzzles/archive/logic/part5.html
(and scroll a bit down for the problem)
Another very similar (but better known) problem you might like to read about
is the Tristram-Shandy paradox, simple google for that and you'll find a lot
of links, much more than I could find about the vase problem.
Theo
guenther vonKnakspot
Theo Jacobs wrote:
<snip>
> I'd like to thank everybody for their input so far.
>
> Guenther: your description is not the same as experiment 1:
> in this experiment no. 1-10 are put in and no.1 is taken out
> no. 11-20 are put in, no. 2 is taken out etc.....
> So, no. 2 is _not_ the same 'batch' as balls 11-20, in this experiment the
> balls taken out come from a batch that was put in earlier (except the 1st
> batch)
Excuse my blunder Jacobs, your objection is quite correct. What I had
meant to say however, is that for each batch of ten that goes into the
vase one ball is taken out. In that case, we have the same setup as in
experiment 1 and we get contradicting results.
> What makes this so remarkable is that -eventually- every ball, and so of
> course every batch too, is taken out again, before noon.
> For unnumbered balls the outcome -as far as I can tell - is also an empty
> vase! This is not my invention however, I read this here:
> http://www.cut-the-knot.org/Probability/infinity.shtml
> (scroll down a bit for the relevant part, although you might like to read it
> all)
> In your 'unnumbered experiments' the chance of a ball eventually being taken
> out again is calculated as being 1 ! - caused of course by the fact that you
> do this an infinite number of times. This applies for every ball, so the
> vase will be empty!!
> Regards, Theo
I don´t think it´s such a good idea to introduce probabilities into
the discussion. Fact is that you can manipulate the outcome of the
experiment by fiddling around with the lables. Let´s take experiment 1
again and modify it in such a manner that instead of taking out the
ball marked n you take out the ball marked 2*n then no matter what, it
is obvious that the uneven numbered balls never get taken out, and the
vase is definitely not empty. I gave examples of situations in which
the outcome of the same experiment should be different for each of two
persons involved. I think this gives a strong indication that the
problem is not being modelled correctly, and would like to hear
Ullrich´s opinion on this.
Regards
Martin Shobe
>Suppose instead of your experiment 1, we put a ball into the vase on
>odd steps, and then take it out the even steps.
>
>Then it seems obvious that there is no single well-defined answer to
>"how many balls are there at noon?". Neither "empty" nor "containing a
>ball" seems to be correct - there is no answer that is somehow
>/logically preferable/ to all others; and that is the basic sense of
>"not well-defined".
>
>On the other hand, if you did /nothing/ at each step, that there would
>somehow be a ball in the vase "at noon" seems obviously wrong. The vase
>would "obviously" be empty, because we intuitively require that there
>can't be a ball in the vase unless it was put there at some finite step
>n.
>
>Similarly, it seems intuitive that if at step 1, you put ball "1" in
>the vase, and then at no later step do you remove it, then that ball
>"must be" in the vase "at noon".
>
>These two intuitions are the assumptions we are making when we ask
>"assuming it is possible in the real world, what balls are in the vase
>at noon?"; and a well-defined answer would be one which is consistent
>with these intuitions, and also be the /only/ one consistent with these
>intutions.
In a couple of other threads, I've said that I didn't know what the
natural topology of sets would be. I think I know now.
Martin
Martin Shobe
wrote:
>"guenther vonKnakspot" <apa...@gmail.com> schreef in bericht
I'm going to disagree slightly here. When dealing with an infinite
process (such as this one), a probability of one does not necessarily
mean that it is certain to happen. Likewise, a probability of zero
doesn't necessarily mean that it can't happen. From here, the
conculsion would not be that the vase is empty.
Martin
Dave Seaman
>> know, the balls may be removed in last-in, first-out order, which
>> guarantees that infinitely many balls will be left at noon. That's why
>> more information is needed.
> Yes, well I suppose I shouldn't just mumble my thoughts. :)
> The tie-the-knot site uses explicitly numbered balls, just as the OP's
> first example did; but most recently, the topic seems to have changed
> to essemtially "unlabelled" balls.
> If the balls are chosen by some unknown method, but the balls are
> indistinguishable (which is what I "essentially" said), then it seems
> to me that we are within our rights to "relabel" the balls at each
> step, since at each step all we /really/ know about the balls in the
> vase is the cardinality of the set of those balls.
The whole point of numbering the balls is to illustrate the fact that the
order of removal makes a difference. Taking away the numbers does not
change this fact. Whether *we* can distinguish the balls or not, the
fact remains that each ball has an individual identity.
> The sequence is not then some function from the naturals to the
> naturals indicating what which ball is removed at that step, as
> distingushed by its labelling.
> Instead we have a function from the naturals to some set of balls,
> about which all we know is its resultant cardinality (i.e., that they
> /can be/ labelled distinctly in a certain way).
> This sequence would seem to have as its limit a countable set of balls,
> not an empty or finite one.
Limits are irrelevant, because the limit of the cardinalities is not
necessarily equal to the cardinality of the limit.
> The poster to whom I was replying said:
>> >> For unnumbered balls the outcome -as far as I can tell - is also an empty
>> >> vase!
> To me, saying the outcome is an empty vase is not the same as saying
> that thye probability that the vase would be empty is 1.
Agreed. But when the balls are chosen randomly, a probabilistic outcome
is the best we can do.
Dave Seaman
> cbr...@cbrownsystems.com wrote:
>> Jeroen Boschma wrote:
>> <snip>
>> > So here's what I think about it. It's obvious that the number of balls in
>> > the vase keeps growing as you approach noon.
>> > However, the algorithm which describes the filling of the vase with balls
>> > never reaches noon. Therefore, I think it is not even correct to ask the
>> > original question 'how many balls at noon' (do mathemathicians call that
>> > ill-defined, -posed or something like that?).
>> There is some sense to what you say.
>> There is still some room IMO for claiming that the question itself is
>> "ill-defined". It's certainly not physically possible for such an
>> operation to take place in the real world to start with; so the
>> question of "what would happen at noon?" is really best interpreted as
>> "what would be a logically consistent answer to the question, assuming
>> that it /could/ be done in practice?"
> The basic idea I had was that we can make a function B(n) which describes the number of balls in the
> vase from one minute to noon to every time less than noon. But the description of the experiment
> does not allow us to reach noon. It does not have a sense of a continuously progressing time which
> is required to reach noon. So I would say that the experiment directly implies that 'noon' is not
> within the domain of B(n), and therefore the function, or description of the experiment, cannot give
> an answer to the question.
On the contrary, the description does allow us to reach noon if correctly
formulated.
For each ball n, there is a time a_n when the ball is added to the vase,
and a time b_n when the ball is removed, never to be returned. The exact
values of a_n and b_n do not matter, but for the original version of the
problem we may assume that a_n <= b_n < 0 for each n, and that the a_n
form a nondecreasing sequence that approaches 0 (where t=0 represents
noon).
For each n there is a characteristic function B_n: R -> {0,1} such that
B_n(t) is 1 if ball n is in the vase at time t, and 0 otherwise. In
fact, we have
Case I. Ball n is added at time n and removed at time b_n.
-----------------------------------------------------------
B_n(t) = 1, if a_n <= t < b_n,
= 0, otherwise.
We now define B(t) = sum_{n_1^oo} B_n(t). Thus B(t) expresses the number
of balls in the vase at time t, and it is perfectly well defined for all
t.
In the modified problem, there may be some n such that ball n is not
removed, and therefore b_n is undefined for those n. In that case we
have
Case II. Ball n is added at time n and never removed.
------------------------------------------------------
B_n(t) = 1, if a_n <= t,
= 0, otherwise.
With this modification we define B(t) as before, and again it is defined
for all t.
> *** For nitpickers :) , as B(n) is a function of variable 'n', statements about the domain should
> be based on 'n' only, I guess. But every 'n' implies a certain time t_n, so I think you get the
> idea... I hope... else beat me again :(
> ***
On the contrary, B_n(t) depends on both n and t.
> From the reasoning in this thread, experiment 1 should result in an empty vase, and experiment 2 in
> infinite balls in the vase at noon. Different answers while the function B(n) that describes the
> number of balls in the vase at any 'n' is the same for both... The question is a physical one
> (number of balls), and for that reason the answer cannot depend on how the balls are numbered or
> what color they have.
>> Suppose instead of your experiment 1, we put a ball into the vase on
>> odd steps, and then take it out the even steps.
>> Then it seems obvious that there is no single well-defined answer to
>> "how many balls are there at noon?". Neither "empty" nor "containing a
>> ball" seems to be correct - there is no answer that is somehow
>> /logically preferable/ to all others; and that is the basic sense of
>> "not well-defined".
Here we have only one ball, whose characteristic function we can call
B_1. The problem here is that B_1(t) is defined only for t < 0, and
therefore this version of the problem is ill-posed.
> For the same reason as I described above: for each n we can find the number of balls in the vase,
> but there is no n which equals 'noon'. Therefore, because the number of balls change at every step
> n, we should not want to force an answer. We can just say that no answer is possible because 'noon'
> is not in the domain of the function B(n) which describes the number of balls in the vase.
Here we have B(t) = B_1(t) because there is only one ball. Yes, it's
true that noon is not in the domain of this function, but that objection
does not apply to earlier formulations of the problem, as I have shown.
>> On the other hand, if you did /nothing/ at each step, that there would
>> somehow be a ball in the vase "at noon" seems obviously wrong. The vase
>> would "obviously" be empty, because we intuitively require that there
>> can't be a ball in the vase unless it was put there at some finite step
>> n.
Each of the B_n is identically zero, and therefore so is B.
> Because there are no steps defined where the number of balls in the vase changes, we don't need to
> stick to those discrete moments t_n, but we can directly describe the number of balls in 'continuous
> time': nothing happens. So we are not bound by a function B(n) in which 'noon' is not element of the
> domain.
Certainly not, since noon *is* in the domain of each B_n and therefore of
B as well.
>> Similarly, it seems intuitive that if at step 1, you put ball "1" in
>> the vase, and then at no later step do you remove it, then that ball
>> "must be" in the vase "at noon".
See Case II above, in which B_n(0) = 1.
>> These two intuitions are the assumptions we are making when we ask
>> "assuming it is possible in the real world, what balls are in the vase
>> at noon?"; and a well-defined answer would be one which is consistent
>> with these intuitions, and also be the /only/ one consistent with these
>> intutions.
> Again, IMHO, forgetting that the answer must be given by a function which describes the number of
> balls in the vase, and that 'noon' is not in the domain of that function. We don't _need_ to give a
> number as an answer.
Noon is in the domain as long as either Case I or Case II applies for
each n. More generally, noon is in the domain as long as each ball is
moved only finitely many times.
> In this experiment, we try to connect some logical reasoning (about putting balls with number in the
> vase and getting some out) with a physical entity (number of balls in the vase). That connection
> cannot be made because we have no description or function which can make the connection!
You have not proved that assertion, and my model clearly disproves it.
>> Thus, in experiment 1, we can agree that, if at some definite step t_n,
>> we remove the ball labelled "n", and then there is no later step that
>> we put the ball back in, then it is perfectly reasonable to insist
>> that, /whatever/ we mean by "the balls in the vase at noon", it can't
>> mean that ball "n" is in the vase at noon.
>> Since, in your construction, the above is true for every n, the only
>> /possible/ answer consistent with our assumptions is that there are no
>> balls in the vase at noon.
> So your explanation is that for each ball n in the vase, there is a step t_n where that ball is
> taken out. So each ball is taken out eventually which results in an empty vase at noon. This is
> based on _physical_ observations: every ball put in the vase is taken out at some time. Another
> _physical_ observation I made in my original reply is: at every 'n' the number of balls in the vase
> increases by 9. So there is no interval between one minute to noon and noon where the number of
> balls in the vase decreases. So how can the vase be empty at noon?
Because the function B is discontinuous at t=0. That is a direct
consequence of the definition B(t) = sum_n B_n(t).
> IMO, both reasonings are OK, but they _seem_ to end up with different answers. Reason for this
> contradiction is that we're not allowed to formulate a numeric answer to the original question,
> because noon is not in the domain of the experiment.
It's not a contradiction. Assuming functions to be continuous in the
face of contrary evidence is not a valid argument.
David C. Ullrich
<apa...@gmail.com> wrote:
If the person inserted the same balls and removed the same
balls as before then the fact that he can't read the numbers
makes no difference.
However, it seems clear that the question you meant to ask
was not the same as what you actually asked. Say he puts in
ten balls and removes one ball every cycle, and because he
can't see the numbers, the balls he removes are chosen
at random. What would be the status of the vase at noon?
Answer: impossible to say, because you haven't specified
exactly what happens.
>Suppose some person A goes ever morning and
>starts putting ping pong balls with the numbers painted on the inside
>into the vase and taking one out of every batch of ten that go in as
>described in experiment 1. Person B has a device which enables him to
>read the numbers inside the balls, so person B can actually determine
>when the sequence of experiment 1 actually took place. Would the vase
>appear full to A and empty to B on the evening of that day? What
>happens if you set up the following experiment you have vases v1 and v2
>and you have numbered balls and unnumbered balls. Then you perform
>experiment 1 with the numbered balls and v1. At the same time, for each
>numbered ball that you put into v1, you put an unnumbered one into v2
>and for each numbered ball you take out of v1, you take an unnumbered
>one out of v2. If v1 is empty at noon, is v2 empty at noon too?
There's nothing we can tell you about the status of v2, because
you have not specified what happens at each stage.
>Thanks a lot in advance,
>Regards.
************************
David C. Ullrich
David C. Ullrich
<dse...@no.such.host> wrote:
>On 3 Aug 2005 19:01:50 -0700, cbr...@cbrownsystems.com wrote:
>> Theo Jacobs wrote:
>
>><snip>
>
>>> For unnumbered balls the outcome -as far as I can tell - is also an empty
>>> vase! This is not my invention however, I read this here:
>>> http://www.cut-the-knot.org/Probability/infinity.shtml
>>> (scroll down a bit for the relevant part, although you might like to read it
>>> all)
>
>> I found his argument not /quite/ so convincing as the argument for
>> numbered balls.
>
>> His argument seems to be essentially that, assuming that the ball to be
>> removed at each step is selected with a uniform probability from the
>> set of balls in the vase at that step, that therefore the /probability/
>> that any /particular/ ball will still be in the vase tends to 0 as the
>> number of steps tends toward infinity. Therefore there is probability 1
>> that the vase is empty "at noon".
>
>The probability of any particular ball being left in the vase at noon
>does not "tend to" 0.
Look again - he didn't say it did.
> The probability of any particular ball being left
>in the vase at noon is exactly 0.
That seems right to me. But it's not quite as obvious as
people seem to be taking it to be (for example, say that
at the n-th stage we insert n balls instead of 10 balls,
and then remove one at random. The probability that a
given ball is still inside at noon is now greater than 0.)
************************
David C. Ullrich
David C. Ullrich
<apa...@gmail.com> wrote:
>
>Theo Jacobs wrote:
>
><snip>
>> I'd like to thank everybody for their input so far.
>>
>> Guenther: your description is not the same as experiment 1:
>> in this experiment no. 1-10 are put in and no.1 is taken out
>> no. 11-20 are put in, no. 2 is taken out etc.....
>> So, no. 2 is _not_ the same 'batch' as balls 11-20, in this experiment the
>> balls taken out come from a batch that was put in earlier (except the 1st
>> batch)
>
>Excuse my blunder Jacobs, your objection is quite correct. What I had
>meant to say however, is that for each batch of ten that goes into the
>vase one ball is taken out. In that case, we have the same setup as in
>experiment 1
We don't have _any_ setup until you tell us _which_ ball is removed.
************************
David C. Ullrich
David C. Ullrich
Exactly right.
Another way to put the same thing: One of the experiments
would seem to indicate that aleph_0 - aleph_0 = 0,
while the other would seem to indicate that it's aleph_0.
>Stephen
************************
David C. Ullrich
David C. Ullrich
Reading the start of this post I was all set to reply that it
was all nonsense. But the fact that the calculations above
are wrong was exactly your point, great.
>In mathematics, usually if at 'noon' the question is undefined then we
>would take the limit of the time --> noon to see the tendency of the
>result
>As time --> noon
>S1 = {1, 2, 3, ... , 10n) Number of S1 = 10n
>S2 = {1, 2, 3, ... , n) Number of S2 = n
>The remaining balls are {n+1, n+2, ... , 10n}
># of balls remained in the vase would be n(S1) - n(S2) = 10n - n = 9n >
>0
>Therefore, the vase is not empty and actually it is growing rapidly.
Alas, this is wrong. The number of balls in the vase at noon is
simply not equal to the limit as n -> infinity of the number of
balls at the n-th stage. It's exactly right that the number of
balls tends to infinity as we approach noon - it does not follow
that the vase is not empty at noon. It _is_ empty at noon.
(Which one of the balls does remain at noon, in your opinion?)
>>From a physical point of view, the number of balls in the vase are
>growing rapidly as the time approaches noon. Even though, right at
>the noon time, the problem is undefined but we would expect the vase is
>not empty.
************************
David C. Ullrich
guenther vonKnakspot
This is not quite true. At every stage the same amout of balls is being
removed from v2 as as the amount of balls being removed from v1. The
only difference being that you cannot identify the balls bein removed.
I am not contending with you Ullrich, I am merely asking you to
explain more elaborately how a mathematician should go about modelling
the problem in order to arrive at a mathematically correct model. If
the lables on the balls are a problem then there is something wrong
going on at the modelling end. My guess is that you either made a
blunder answering the way you did in the first place, or that there is
something to be learnt from you, which is being occluded by your
succint answers. I would certainly prefer case two and would be really
thankfull for clarification.
Regards.
Dave Seaman
> On Thu, 4 Aug 2005 03:47:53 +0000 (UTC), Dave Seaman
><dse...@no.such.host> wrote:
>>On 3 Aug 2005 19:01:50 -0700, cbr...@cbrownsystems.com wrote:
>>> I found his argument not /quite/ so convincing as the argument for
>>> numbered balls.
>>> His argument seems to be essentially that, assuming that the ball to be
>>> removed at each step is selected with a uniform probability from the
>>> set of balls in the vase at that step, that therefore the /probability/
>>> that any /particular/ ball will still be in the vase tends to 0 as the
>>> number of steps tends toward infinity. Therefore there is probability 1
>>> that the vase is empty "at noon".
>>The probability of any particular ball being left in the vase at noon
>>does not "tend to" 0.
> Look again - he didn't say it did.
You're right. He didn't.
>> The probability of any particular ball being left
>>in the vase at noon is exactly 0.
> That seems right to me. But it's not quite as obvious as
> people seem to be taking it to be (for example, say that
> at the n-th stage we insert n balls instead of 10 balls,
> and then remove one at random. The probability that a
> given ball is still inside at noon is now greater than 0.)
No, it's not entirely obvious, but the argument depends on the fact that
a particular infinite product diverges to 0. Taking logarithms gives
a sum that is asymptotic to the negative harmonic series.
Anon
<ull...@math.okstate.edu> wrote:
>On 4 Aug 2005 01:36:00 -0700, snapdr...@gmail.com wrote:
>
>>Hi Theo. Before answering your question, I would like to do some math
>>on infinite series first and then explain to you why the vase is not
>>empty at the noon time.
[snip]
>>As seen from the previous example, we cannot just cancel out same terms
>>from each set and come up with result zero. S1 and S2 are infinite
>>series with different properties. We cannot simply say every number in
>>the set S1 also exists in S2 then S1 - S2 = 0. If you do that then you
>>are rearrangeing the numbers in the infinite series.
>
>Reading the start of this post I was all set to reply that it
>was all nonsense. But the fact that the calculations above
>are wrong was exactly your point, great.
>
>>In mathematics, usually if at 'noon' the question is undefined then we
>>would take the limit of the time --> noon to see the tendency of the
>>result
>>As time --> noon
>>S1 = {1, 2, 3, ... , 10n) Number of S1 = 10n
>>S2 = {1, 2, 3, ... , n) Number of S2 = n
>>The remaining balls are {n+1, n+2, ... , 10n}
>># of balls remained in the vase would be n(S1) - n(S2) = 10n - n = 9n >
>>0
>>Therefore, the vase is not empty and actually it is growing rapidly.
>
>Alas, this is wrong. The number of balls in the vase at noon is
>simply not equal to the limit as n -> infinity of the number of
>balls at the n-th stage. It's exactly right that the number of
>balls tends to infinity as we approach noon - it does not follow
>that the vase is not empty at noon. It _is_ empty at noon.
>
The OP has belatedly shared with us that he is talking about the
Littlewood-Ross paradox, and has provided (among other references)
a pointer to a paper by John Byl, ON RESOLVING THE LITTLEWOOD-ROSS
PARADOX at
www.csc.twu.ca/byl/little.fin.doc
[So that no one will have to deal with a .doc file, just Google
'littlewood ross paradox byl' and read the HTML]
Where do you say, if you are, that Byl has it wrong?
William Hughes
More completely, at any time before noon, v1 and v2 have exactly the
same
number of balls and as you approach noon this number gets arbitrarily
large.
At noon v1 is empty, but we dont know the status of v2, because
this depends on exactly which balls have been removed. (If you argue
that
the term unnumbered means that the balls cannont be labelled, even
in theory, then there is no meaning to the question "How many balls
are in v2 at noon?".)
> This is not quite true. At every stage the same amout of balls is being
> removed from v2 as as the amount of balls being removed from v1. The
> only difference being that you cannot identify the balls bein removed.
The problem here is that while the number of balls is useful in the
finite
case, is it not useful in the infinite case (where despite the fact
that an infinite number of balls has been added and an infinite number
removed, there can be any number of balls (from 0 to infinity) left)
> I am not contending with you Ullrich, I am merely asking you to
> explain more elaborately how a mathematician should go about modelling
> the problem in order to arrive at a mathematically correct model. If
> the lables on the balls are a problem then there is something wrong
> going on at the modelling end.
No. In the infinite case it is necessary to know more than how many
balls
have been removed, you also have to know exactly which balls have been
removed. This means labelling. The labelling scheme is arbitrary, but
if we change the subset of the integers we use for labelling we must
change are arguments (e.g. if all balls are labelled 2n for some n, it
does
not make sense to talk about the balls with odd labels).
-William Hughes
Dave Seaman
> On Thu, 04 Aug 2005 09:28:31 -0500, David C. Ullrich
> The OP has belatedly shared with us that he is talking about the
> Littlewood-Ross paradox, and has provided (among other references)
> a pointer to a paper by John Byl, ON RESOLVING THE LITTLEWOOD-ROSS
> PARADOX at
> www.csc.twu.ca/byl/little.fin.doc
> [So that no one will have to deal with a .doc file, just Google
> 'littlewood ross paradox byl' and read the HTML]
I looked at the html version of Byl's paper at
<http://64.233.167.104/search?q=cache:y6vy_XmaHvsJ:www.csc.twu.ca/byl/little.fin.doc+littlewood+ross+paradox&hl=en&client=safari>,
but that paper is so full of nonsense that it is difficult to know where
to begin.
> Where do you say, if you are, that Byl has it wrong?
Byl talks about "continuity of the number N(t) of balls in the urn", and
also about "continuity of the position functions". Neither of those is
continuous. I think he needs to start over and rethink his entire
presentation.
The position functions, which I called B_n(t) elsewhere in the thread,
are discontinuous at each time when balls are added to or removed from
the urn. As the problem is usually formulated, no balls are actually
moved at noon, which means that each B_n is continuous at t=0, but not
necessarily at other times.
From the continuity of each B_n at time t=0, it does *not* follow that
the sum B(t) = sum_n B_n(t) is continuous at t=0. In fact, the sum of
those functions is demonstrably discontinuous at t=0 under the conditions
described in the original experiment. Notice that the argument here does
not refer in any way to balls or urns or anything physical, but only to
the mathematical description B(t) = sum_n B_n(t). The discontinuity may
be a "paradox" in the sense that it defies intuition, but mathematics is
full of facts that defy intuition. There is nothing that needs to be
reconciled here. The mathematics speaks for itself.
>>(Which one of the balls does remain at noon, in your opinion?)
>>>>From a physical point of view, the number of balls in the vase are
>>>growing rapidly as the time approaches noon. Even though, right at
>>>the noon time, the problem is undefined but we would expect the vase is
>>>not empty.
Why is the problem undefined? For which n is the value of B_n(0)
undefined?
ste...@nomail.com
> Theo Jacobs wrote:
> <snip>
>> I'd like to thank everybody for their input so far.
>>
>> Guenther: your description is not the same as experiment 1:
>> in this experiment no. 1-10 are put in and no.1 is taken out
>> no. 11-20 are put in, no. 2 is taken out etc.....
>> So, no. 2 is _not_ the same 'batch' as balls 11-20, in this experiment the
>> balls taken out come from a batch that was put in earlier (except the 1st
>> batch)
> Excuse my blunder Jacobs, your objection is quite correct. What I had
> meant to say however, is that for each batch of ten that goes into the
> vase one ball is taken out. In that case, we have the same setup as in
> experiment 1 and we get contradicting results.
There is no contradiction. In the infinite case exactly how the
balls are added and removed matters. It is not just simply
a matter of "add an infinite number of balls" and "remove
an infinite number of balls". Once you describe exactly
how the balls are added and removed there is only one answer.
The fact that different methods of adding and removing balls
have different answers is not a contradiction. It may be
unintuitive, but most anything to do with infinity is unintuitive.
Stephen
Virgil
"guenther vonKnakspot" <apa...@gmail.com> wrote:
Wouldn't it be easier to paint numbers on the *outsides* of those ping
pong balls?
guenther vonKnakspot
persons performing the experiment at the same time, get different
results because one person knows which balls are being removed and the
other one does not. Remember it is the same experiment, that is there
is a single vase a single set of balls and two persons performing.
Thanks in advance.
Regards.
guenther vonKnakspot
delivers one single vase which is empty for person A and simultaneously
is not empty for person B, because A does not know which balls were
removed while B does?
Regards.
William Hughes
The paper by Byl contains so much nonsense it is not possible
to pick a single point where he "has it wrong". However his major
errors seem to me to be:
-he tries to argue continuity of the number of balls in the
urn using physical intuition, when it is very clear that
task described is physically impossible.
-he talks about the state of the urn the instant
before noon. There is no such instant. Either the time is
noon or there remains an infinite number of steps.
- William Hughes
ste...@nomail.com
No. How did you come to that conclusion?
A ball is either removed or it is not. If you describe
exactly how the balls are removed than there is one and
only one answer
Stephen
ste...@nomail.com
> Thanks in advance.
A ball is either removed or not. It does not matter if one
person does not know which ball is being removed. All it
means is that person will be unable to predict the outcome
of the experiment, whereas the person with full knowledge
will be able to predict the outcome. That is true in
any experiment.
Stephen
guenther vonKnakspot
step...@nomail.com wrote:
<snip>
> > Are you saying then that one single performance of the experiment
> > delivers one single vase which is empty for person A and simultaneously
> > is not empty for person B, because A does not know which balls were
> > removed while B does?
>
> No. How did you come to that conclusion?
>
> A ball is either removed or it is not. If you describe
> exactly how the balls are removed than there is one and
> only one answer
Then it would appear that you did not follow my articles. Let me repeat
for you. Suppose the balls are not labeled on the outside but inside,
so that person A who is performing the experiment does not see which
balls he is putting out, and which ones he is taking out. Person B has
a device which enables him to see which balls are being put into the
vase and which ones are being taken out. Person A performs the
experiment every morning, untill person B determines that the
experiment performed is exactly the one described as experiment 1 by
the OP. Now we have a situation in which A would end up with a non
empty vase (he has been dropping 10 indistinguishable balls into the
vase and taking one out all morning long) while B knows that for every
natural number n the ball labelled n has been taken out of the vase. So
which one is it, is the vase empty or not?
Regards.
> Stephen
William Hughes
<SNIP>
> Thank you. Now, what do you say about the examples I built, where two
> persons performing the experiment at the same time, get different
> results because one person knows which balls are being removed and the
> other one does not. Remember it is the same experiment, that is there
> is a single vase a single set of balls and two persons performing.
>
> Thanks in advance.
>
> Regards.
You have provide no such example. Remember the number of balls
remaining at noon depends on exactly which balls are removed.
Case 1: One urn. A can read the labels, B cannot
A and B agree as to how many balls remain at noon. A
knows exactly what happened; B does not.
Case 2: Two urns. A can read the labels, B cannot
Depending on the strategy A uses, there can be any number
of balls from 0 to infinity in A's urn at noon.
B cannot use a strategy that depends on reading labels.
(If B can pick a ball at random at each step, we can
say that B's urn is empty with probability 1, whatever
that means)
A and B do not use the same strategy
for removing balls, thus the experiments
are not the same, and we do not expect the
outcomes to be the same.
-William Hughes
guenther vonKnakspot
experiment but about determining it. This is quite a different thing.
We have a situation wher e the outcome of the experiment is dependent
upon the information available while it is being performed, and we can
end up with contradictory results being determined, allow me to refer
you to my response to you in another branch of this thread.
Regards.
William Hughes
Why do you think that A will end up with a non empty vase?
Depending on which balls A removes he may end up with any
number of balls in his vase from 0 to infinity. A does not
know what is going to happen. B waits for a day when A uses
(unintentionally since A cannot read labels) a strategy that
leads to an empty vase. Both A and B agree that the vase
is empty. (B knows the order in which A removed the balls,
A doesn't, but the fact that the balls can be labelled means
that this order exists.)
-William Hughes
ste...@nomail.com
Why would A end up with a non empty vase? A does not know
how the balls are being removed and therefore cannot make
any prediction about the outcome.
The observers are irrelevant. Either a ball is removed
or not. In the experiment you described each ball is
removed at some time before noon. Just because A does
not know this does not mean it does not happen.
Stephen
guenther vonKnakspot
Because there is no reason for A to end up with an empty vase. A has
been dropping 10 balls into the vase and taking one out. The whole
contention that B finds the vase to be empty is that B knows that for
any n in the set of natural numbers, the ball labelled n is not inside
the vase.
Regards.
ste...@nomail.com
What is the difference in this case? It is entirely
a thought experiment.
> This is quite a different thing.
> We have a situation wher e the outcome of the experiment is dependent
> upon the information available while it is being performed, and we can
> end up with contradictory results being determined, allow me to refer
> you to my response to you in another branch of this thread.
The final outcome depends on how the balls are removed.
It does not matter if the person removing the balls does
not know which ball they are removing. That will not
affect which balls are removed and will not affect the outcome.
Suppose there is a set of weights, some of which weigh 1 kg
and some of which weight 5kg, and a structure that can
support 20 kg. Person A puts 10 weights on the structure,
but does not know which type of weights they are.
The structure will collapse or not. The fact that A does
not know will not change the outcome. Similarly when
A removes a ball, the ball is removed, even if
A does not know which ball it was.
Stephen
William Hughes
But the whole point is that it is possible to perform the operation
"drop 10 balls into the vase, take one out" an infinite number of
times and end up with an empty vase. A knows this.
So A is not surprised to see an empty vase, even though he
does not know exactly how this happened. It is not necessary
for A to know how he got an empty vase for A to know that the
vase is empty.
-William Hughes
ste...@nomail.com
What does that have to do with anything? 10*infinity-infinity
is not defined.
> The whole
> contention that B finds the vase to be empty is that B knows that for
> any n in the set of natural numbers, the ball labelled n is not inside
> the vase.
Because A removed each ball labelled n. Are you saying that
because A could not see the label when she removed each
ball labelled n that she in fact did not remove each
ball labelled n?
Stephen
cbr...@cbrownsystems.com
Dave Seaman wrote:
> On 3 Aug 2005 21:35:51 -0700, cbr...@cbrownsystems.com wrote:
> > If the balls are chosen by some unknown method, but the balls are
> > indistinguishable (which is what I "essentially" said), then it seems
> > to me that we are within our rights to "relabel" the balls at each
> > step, since at each step all we /really/ know about the balls in the
> > vase is the cardinality of the set of those balls.
>
> The whole point of numbering the balls is to illustrate the fact that the
> order of removal makes a difference. Taking away the numbers does not
> change this fact. Whether *we* can distinguish the balls or not, the
> fact remains that each ball has an individual identity.
>
Yes, I mea culpa'd that on my drive home. You are correct.
Cheers - Chas
David R Tribble
>> Why do you think that A will end up with a non empty vase?
>> Depending on which balls A removes he may end up with any
>> number of balls in his vase from 0 to infinity. A does not
>> know what is going to happen. B waits for a day when A uses
>> (unintentionally since A cannot read labels) a strategy that
>> leads to an empty vase. Both A and B agree that the vase
>> is empty. (B knows the order in which A removed the balls,
>> A doesn't, but the fact that the balls can be labelled means
>> that this order exists.)
Actually, it would appear that A ends up with either zero or
infinity (aleph-0) balls in his vase. I can't see how he could
end up with, say, a million balls.
guenther vonKnakspot wrote:
> Because there is no reason for A to end up with an empty vase. A has
> been dropping 10 balls into the vase and taking one out. The whole
> contention that B finds the vase to be empty is that B knows that for
> any n in the set of natural numbers, the ball labelled n is not inside
> the vase.
That's the best answer I've seen so far in this thread that seems
to come the closest to the answer (zero).
Unless I missed something, no one has pointed out that the problem
appears to be
aleph-0 x 10 - aleph_0 = n
where n is the number of balls left in the vase.
Conventional cardinal arithmetic says that
aleph_0 x k = aleph_0, for any finite k
so one would be led to believe that n should be zero.
The other approach, which appears to have been mentioned in
passing, is that there exists a mapping (bijection) from N
to Out, the set of all balls taken out, and a mapping from
N to In, the set of all balls left in the vase.
Out_i = 10i+1, for all i in N
In_i = i, for all i in N and i mod 10 not= 1
Thus
card(Out) = aleph_0
card(In) = aleph_0
So it appears, at least to me, that both sets Out and In have the
same number of elements. So by noon exactly the same number of
balls have been added to the vase, card(In), as have been taken
from the vase, card(Out). So the vase is empty at noon.
Or am I missing something?
-drt
guenther vonKnakspot
into the vase, take one out" and end up with a non empty vase. This is
the case, for example, if you take out the ball labelled 2*n, in which
you end up with a vase full of balls labelled with uneven numbers. A
would not be surprised to end up with an empty vase if performing the
experiment with unlabelled (to him/her) balls if (she/he) knew of a
proof that this is possible in advance. The proof that you can put ten
unlabelled balls into a vase take one out, repeatedly for an infinite
number of times and end up with an empty vase is eluding me. I would be
very thankful for this proof. I am not saying that any answer given up
to now is wrong, I am only trying to understand where the paradox stems
from, and wether it can be resolved.
Regards.
David Kastrup
> Yes, and it is also possible to perform the operation "drop 10 balls
> into the vase, take one out" and end up with a non empty vase. This
> is the case, for example, if you take out the ball labelled 2*n, in
> which you end up with a vase full of balls labelled with uneven
> numbers. A would not be surprised to end up with an empty vase if
> performing the experiment with unlabelled (to him/her) balls if
> (she/he) knew of a proof that this is possible in advance. The proof
> that you can put ten unlabelled balls into a vase take one out,
> repeatedly for an infinite number of times and end up with an empty
> vase is eluding me. I would be very thankful for this proof. I am
> not saying that any answer given up to now is wrong, I am only
> trying to understand where the paradox stems from, and wether it can
> be resolved. Regards.
The first time you do something an infinite number of times is always
the hardest. You'll get the knak after a few times.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
guenther vonKnakspot
reason at all for the vase to be empty. There are several scenarios
under which A can drop ten balls into the vase, take one out an
infinite amount of times, and end up with a non empty vase. One such
case would be if the labels on the balls he/she took out were all even
numbers. I cannot find any reason for the vase to be empty if the balls
are not labelled at all. Can you?
guenther vonKnakspot
William Hughes wrote:
> guenther vonKnakspot wrote:
>
> <SNIP>
>
> > Thank you. Now, what do you say about the examples I built, where two
> > persons performing the experiment at the same time, get different
> > results because one person knows which balls are being removed and the
> > other one does not. Remember it is the same experiment, that is there
> > is a single vase a single set of balls and two persons performing.
> >
> > Thanks in advance.
> >
> > Regards.
>
> You have provide no such example. Remember the number of balls
> remaining at noon depends on exactly which balls are removed.
>
> Case 1: One urn. A can read the labels, B cannot
>
> A and B agree as to how many balls remain at noon. A
> knows exactly what happened; B does not.
two persons agree on anything at all. In this case, A an B do not even
need to know of each other´s existence. There is no reason for them to
agree on the amount of balls in the vase because they do not have to be
in contact with each other at all.
Regards.
Regards.
guenther vonKnakspot
done. There seems to be no end to it though ;P
Regards.
georgie
Dave Seaman wrote:
<snip>
>
> For each n there is a characteristic function B_n: R -> {0,1} such that
> B_n(t) is 1 if ball n is in the vase at time t, and 0 otherwise. In
> fact, we have
>
> Case I. Ball n is added at time n and removed at time b_n.
> -----------------------------------------------------------
> B_n(t) = 1, if a_n <= t < b_n,
> = 0, otherwise.
>
> We now define B(t) = sum_{n_1^oo} B_n(t). Thus B(t) expresses the number
> of balls in the vase at time t, and it is perfectly well defined for all
> t.
Except some people would claim that an infinite sum that diverges is
not
well defined. B(t) diverges as t approaches noon from the left.
Dave Seaman
> Dave Seaman wrote:
><snip>
What sum is not well defined? The one-sided limit lim_{t->0-} is a
divergent limit, but it is not a sum at all, and therefore certainly not
a divergent sum.
For which t does B(t) diverge?
The only sum involved here is B(t) = sum_n B_n(t) that defines the
function B, and the fact of the matter is that this is a convergent sum
over n and is finite for each and every t.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
William Hughes
> William Hughes wrote:
> >> Why do you think that A will end up with a non empty vase?
> >> Depending on which balls A removes he may end up with any
> >> number of balls in his vase from 0 to infinity. A does not
> >> know what is going to happen. B waits for a day when A uses
> >> (unintentionally since A cannot read labels) a strategy that
> >> leads to an empty vase. Both A and B agree that the vase
> >> is empty. (B knows the order in which A removed the balls,
> >> A doesn't, but the fact that the balls can be labelled means
> >> that this order exists.)
>
> Actually, it would appear that A ends up with either zero or
> infinity (aleph-0) balls in his vase. I can't see how he could
> end up with, say, a million balls.
>
If A removes balls 1,2,3... he ends up with no balls
If A removes balls 2,3,4... he ends up with 1 ball
If A removes balls 2,3,4,...,10,12,13 ...
he ends up with 2 balls
If A removes balls 2,3,4,...,10,12,13,...,20,22,23...
he ends up with 3 balls
Clearly we can generalize to any number of balls. A keeps the
number of balls he wants from the set {1,11,21,31,...}.
If he wants to end up with a million balls, he keeps the
first million balls of this set. Note that A never tries
to remove a ball before it is added.
Your are missing the fact that the cardnality argument only works for
finite sets. It is quite possible for a infinite set to have a proper
subset with the same cardnality. (Indeed this is the definition
of an infinite set.) So the fact that card(Out) = Card(In) does not
mean that Out is equal to In.
-William Hughes
georgie
>divergent limit
limit of what? B(t)? B(t) is an infinite sum (by definition).
Timothy Little
> Actually, it would appear that A ends up with either zero or
> infinity (aleph-0) balls in his vase. I can't see how he could end
> up with, say, a million balls.
If A picks uniformly randomly, then like every other nonzero result it
has probability zero but is possible. For example, A could happen to
miss exactly every odd-numbered ball less than 2000000.
If we remove the assumption that A picks uniformly randomly, then we
can end up with other distributions. They include distributions
where every possibility for the cardinality at noon has nonzero
probability.
> Unless I missed something, no one has pointed out that the problem
> appears to be
> aleph-0 x 10 - aleph_0 = n
> where n is the number of balls left in the vase.
>
> Conventional cardinal arithmetic says that
> aleph_0 x k = aleph_0, for any finite k
> so one would be led to believe that n should be zero.
Conventional cardinal arithmetic says that n + aleph_0 = 10 aleph_0
for all n <= aleph_0, so one would be led to believe that n could be
any finite cardinal or aleph_0.
That's why subtraction of infinite cardinals isn't defined.
> So it appears, at least to me, that both sets Out and In have the
> same number of elements. So by noon exactly the same number of
> balls have been added to the vase, card(In), as have been taken
> from the vase, card(Out). So the vase is empty at noon.
>
> Or am I missing something?
The fact that there are (uncountably many!) subsets X of N, such that
|X| = |N|, and yet |N \ X| = |N|.
- Tim
Timothy Little
> The proof that you can put ten unlabelled balls into a vase take one
> out, repeatedly for an infinite number of times and end up with an
> empty vase is eluding me.
Without labelling the balls at all: after adding the balls in step n,
A can remove one of the balls added in step (n/10 rounded up). If A
does so for all n, then every ball added will be removed before noon.
Note that A doesn't have to do so deliberately. Accidently will
suffice, if you merely want to show that it is possible.
We can actually show more than that though: we can demonstrate that
even if A loses track of which balls were added when and just picks
one at random from all the balls in the vase, then the vase will
"almost always" be empty at noon.
However, the proof in that case does rely on how the balls are added
and removed. If we add n balls at step n, then it is still possible
to have an empty vase at noon, but randomly picking out balls will
have probability 0 of doing so.
Here's a nasty twist to the problem: suppose the balls are labelled,
put in the vase in sequence, and A always removes the lowest numbered
ball from the vase. However, B is a miser who doesn't want to buy any
more balls than necessary. He has been removing the label from each
ball taken out and relabelling it as the first ball yet to be put in.
Does it still make sense to ask how many balls are in the vase at
noon?
- Tim
georgie
Obviously, to get to noon, we must transcend the
integers. Most mathematicians would know this.
snapdr...@gmail.com
infinity as we approach noon does not imply that the vase is not empty
at noon. That is why I say 'from a physical point of view' (or in
reality). If we have balls in the vase before noon and there are more
balls going in than coming out. In reality, at noon time the vase
would not be empty.
According to this problem, noon is not reachable. The operation is
undefined exactly at noon. Therefore, even no ball is being taken out
from the vase, you still can claim that there are no ball in the vase
at noon.
Take a look at function Y = (X^2 - 4)/(X - 2)
We know that Y = X + 2 when X not = 2 and Y is undefined at X = 2.
You can put whatever value to Y at X = 2 as you wish. It is perfectly
okay say Y = 0 at X = 2.
If Y = (X^2 - 4)/(X - 2) is an equation to represent a real event then
we would like to find out what value of Y would be when X approaches 2.
As X approaches to 2, Y approaches to 4. Then we may claim that the
likelyhood of Y would be 4 when X = 2. If you claim anything else then
you have to prove it.
In the vase example, if we only look at the withdrawal portion of the
argument then the claim is not strong. If both putting in and taken
out of balls are taken into account. In order to prove that the vase
is empty at noon, you must have to show at certain time the rate of
taking out is greater than the rate of putting in.
Timothy Little
> In order to prove that the vase is empty at noon, you must have to
> show at certain time the rate of taking out is greater than the rate
> of putting in.
You're implicitly assuming a differentiable function from ... what,
exactly? Naturals? Reals? To ... what exactly? Cardinals? With
what precise meaning of "rate"?
The short answer is: No we don't. A slightly longer answer appends:
because that wouldn't make any sense at all.
- Tim
Dave Seaman
B(t) is a convergent sum for every t. There is no divergent sum involved.
snapdragon31
In order to claim the vase is empty, not only all the balls in the vase
are removed but also all the newly added balls have to be removed - not
at a later time. All the newly added balls can only be removed at a
later time. Therefore, we cannot claim that the vase is empty anytime
before noon.
William Hughes
> Hi David. It is correct to say that the number of balls tends to
> infinity as we approach noon does not imply that the vase is not empty
> at noon. That is why I say 'from a physical point of view' (or in
> reality). If we have balls in the vase before noon and there are more
> balls going in than coming out. In reality, at noon time the vase
> would not be empty.
>
There is no reality here. It is not possible to have a real experiment
with an infinite number of balls. Nor is it possible to approximate
this using a large finite number of balls, no finite number, however
large, acts like an infinite number.
> According to this problem, noon is not reachable.
Wrong. The experiment described is quite well defined. We know
which balls are added to the vase and when and we can consider
the different outcomes by choosing which balls are removed from
the vase and when. Noon is not reachable in any physical sense,
but we know that the experiment described has no physical realization
or approximation.
> The operation is
> undefined exactly at noon. Therefore, even no ball is being taken out
> from the vase, you still can claim that there are no ball in the vase
> at noon.
>
> Take a look at function Y = (X^2 - 4)/(X - 2)
> We know that Y = X + 2 when X not = 2 and Y is undefined at X = 2.
> You can put whatever value to Y at X = 2 as you wish. It is perfectly
> okay say Y = 0 at X = 2.
>
> If Y = (X^2 - 4)/(X - 2) is an equation to represent a real event then
> we would like to find out what value of Y would be when X approaches 2.
> As X approaches to 2, Y approaches to 4. Then we may claim that the
> likelyhood of Y would be 4 when X = 2. If you claim anything else then
> you have to prove it.
>
Yes, everything in physics (at least in the classical approximation) is
continuous. Thus if a function defines a physical event then the
function
must be continous. So what? The experiment we are considering is
not a physical experiment.
> In the vase example, if we only look at the withdrawal portion of the
> argument then the claim is not strong. If both putting in and taken
> out of balls are taken into account. In order to prove that the vase
> is empty at noon, you must have to show at certain time the rate of
> taking out is greater than the rate of putting in.
No. The rate at which we add balls is infinite. So is the rate at
which
we remove them. We cannot conclude anything from this fact. True, at
any
time before noon the rate at which we are adding balls is greater than
the rate at which we are removing them. This tells us nothing about
what
happens at noon.
-William Hughes
David R Tribble
>> Depending on which balls A removes he may end up with any
>> number of balls in his vase from 0 to infinity.
David R Tribble <da...@tribble.com> wrote:
>> Actually, it would appear that A ends up with either zero or
>> infinity (aleph-0) balls in his vase. I can't see how he could
>> end up with, say, a million balls.
ste...@nomail.com wrote:
> A million is possible, but it is easier to explain how you
> could end up with 5 balls. If you remove ball 6 on the
> first step, ball 7 on the second step, etc, every ball will
> be removed except for the first five.
>
> To get a million I think the following works: remove ball n on the
> first 200000 steps. Then remove ball 1200000+(n-200000) on the
> remaining steps.
William Hughes wrote:
> If A removes balls 1,2,3... he ends up with no balls
> If A removes balls 2,3,4... he ends up with 1 ball
> If A removes balls 2,3,4,...,10,12,13 ...
> he ends up with 2 balls
> If A removes balls 2,3,4,...,10,12,13,...,20,22,23...
> he ends up with 3 balls
>
> Clearly we can generalize to any number of balls. A keeps the
> number of balls he wants from the set {1,11,21,31,...}.
> If he wants to end up with a million balls, he keeps the
> first million balls of this set. Note that A never tries
> to remove a ball before it is added.
Timothy Little wrote:
> If A picks uniformly randomly, then like every other nonzero result it
> has probability zero but is possible. For example, A could happen to
> miss exactly every odd-numbered ball less than 2000000.
>
> If we remove the assumption that A picks uniformly randomly, then we
> can end up with other distributions. They include distributions
> where every possibility for the cardinality at noon has nonzero
> probability.
Yeah, I realize all that. But that's not what A does in the problem
originally posed. He puts 10 balls in (10k+1, 10k+2, 10k+3, ...,
10k+10), and takes out the first one he just put in (10k+1).
The original statement of the problem does not seem to allow arbitrary
or random putting-in and taking-out.
-drt
David R Tribble
>> appears to be
>> aleph-0 x 10 - aleph_0 = n
>> where n is the number of balls left in the vase.
>>
>> Conventional cardinal arithmetic says that
>> aleph_0 x k = aleph_0, for any finite k
>> so one would be led to believe that n should be zero.
ste...@nomail.com wrote:
> aleph_0 - aleph_0 is undefined. That is why the
> answer to the question is not aleph_0* 10 - aleph_0.
William Hughes wrote:
> Your are missing the fact that the cardnality argument only works for
> finite sets. It is quite possible for a infinite set to have a proper
> subset with the same cardnality. (Indeed this is the definition
> of an infinite set.) So the fact that card(Out) = Card(In) does not
> mean that Out is equal to In.
Mea culpa - I realized that after I posted.
Also, my mapping has an error:
>> Out_i = 10i+1, for all i in N
>> In_i = i, for all i in N and i mod 10 not= 1
>> Thus
>> card(Out) = aleph_0
>> card(In) = aleph_0
My mapping for In_i excludes all balls labeled i, where i mod 10 = 1.
In fact, this excludes an infinite number of balls from In, which
implies that In contains less members than Out!
But even if we devise a complete mapping of N to In and N to Out,
all this proves is that In and Out are both countably infinite
sets (card(In) = card(Out) = card(N) = aleph_0). This follows
the rules that if two sets can be placed in one-to-one
correspondence to another set, then the first two sets must be
the same size.
However, this does not help us to determine the difference of the
two sets, which is the point of the whole problem.
-drt
ste...@nomail.com
The number of balls that remain in the vase depends on
how you remove the balls. In the original problem
you always end up with 0 balls remaining. If you remove
the balls in a different order you can end up with
a different number. That is why the answer is not simply
#balls added - #balls removed.
Stephen
Dave Seaman
> David R Tribble <da...@tribble.com> wrote:
>> Yeah, I realize all that. But that's not what A does in the problem
>> originally posed. He puts 10 balls in (10k+1, 10k+2, 10k+3, ...,
>> 10k+10), and takes out the first one he just put in (10k+1).
>> The original statement of the problem does not seem to allow arbitrary
>> or random putting-in and taking-out.
Actually, that's not how the original problem was specified. The original
version had ball k removed at step k, which leads to the conclusion that the
vase is empty at noon.
The way David Tribble describes it here, nine balls out of each ten are
never removed from the vase at all, and therefore infinitely many must be
left at noon.
> The number of balls that remain in the vase depends on
> how you remove the balls. In the original problem
> you always end up with 0 balls remaining. If you remove
> the balls in a different order you can end up with
> a different number. That is why the answer is not simply
> #balls added - #balls removed.
Of which the above misstatement of the problem provides a perfect
example.
snapdragon31
Problem 1
Suppose you have a giant vase and a bunch of ping pong balls with an
integer written on each one, e.g. just like the lottery, so the balls
are numbered 1, 2, 3, ... and so on. At one minute to noon you put
balls 1 to 10 in the vase and take out number 1. At half a minute to
noon you put balls 11 - 20 in the vase and take out number 2. At one
quarter minute to noon you put balls 21 - 30 in the vase and take out
number 3. Continue in this fashion. Obviously this is physically
impossible, but you get the idea. Now the question is this: At noon,
how many ping pong balls are in the vase?
We can change the above question into an equivalent problem.
Problem 2.
Suppose you have a giant vase and a bunch of ping pong balls with an
integer written on each one, e.g. just like the lottery, so the balls
are numbered 1, 2, 3, ... and so on.
At minute 1 you put balls 1 to 10 in the vase and take out number 1.
At minute 2 you put balls 11 - 20 in the vase and take out number 2.
At minute 3 you put balls 21 - 30 in the vase and take out number 3.
Continue in this fashion. Obviously this is physically impossible, but
you get the idea. Now the question is this: At an infinite time from
now, how many ping pong balls are in the vase?
William Hughes wrote:
> > According to this problem, noon is not reachable.
>
> Wrong. The experiment described is quite well defined. We know
> which balls are added to the vase and when and we can consider
> the different outcomes by choosing which balls are removed from
> the vase and when. Noon is not reachable in any physical sense,
> but we know that the experiment described has no physical realization
> or approximation.
>
I hope you can see that 'noon' in problem 1 is equivalent to 'an
infinite time from now' in problem 2. They are un-reachable. Again I
have to emphasis that in order to prove the vase is empty, we have to
prove that the last ball is taken out from the vase.
Yes, you can claim from problem 2 that ball n is removed from the vase
at the nth minute. The vase can only be claimed to be empty if ball n
is the last ball in the vase. Obviously by the time the nth ball is
removed there are many more balls in the vase. So the vase cannot be
empty even though every ball currently in the vase will be removed
later in the future.
By mathematical induction. n=1 is true. n=n+1 is also true. It implies
that it is true for all n. The vase is not empty for all n > 0.
Mathematical induction still cannot tell you what happens exactly when
n is infinite which does not exist physically or mathematically. But
it can claim that the vase is not empty as n approaches to infinity.
> > In the vase example, if we only look at the withdrawal portion of the
> > argument then the claim is not strong. If both putting in and taken
> > out of balls are taken into account. In order to prove that the vase
> > is empty at noon, you must have to show at certain time the rate of
> > taking out is greater than the rate of putting in.
>
> No. The rate at which we add balls is infinite. So is the rate at
> which we remove them. We cannot conclude anything from this fact. True, > at any time before noon the rate at which we are adding balls is greater > than the rate at which we are removing them. This tells us nothing about
> what happens at noon.
> -William Hughes
In the equivalent problem (problem 2), the rate of adding the balls to
the vase is 10 balls/minute. The withdrawal rate is 1 ball/minute.
Assume f(x) is an increasing function and f(0) > 0. Even though f(oo)
is undefined but I am quite sure that f(oo) cannot be 0.
Dave Seaman
Noon may be un-reachable in a physical sense, but this is not a physical
problem. It's a mathematical problem. The question makes perfect sense
from a mathematical standpoint.
Your final sentence above makes no sense at all. There is no such thing
as a "last ball". All that is necessary to show that the vase is empty
is to show that each ball is removed from the vase before noon and is not
subsequently replaced.
> Assume f(x) is an increasing function and f(0) > 0. Even though f(oo)
> is undefined but I am quite sure that f(oo) cannot be 0.
Why?
snapdragon31
> Noon may be un-reachable in a physical sense, but this is not a physical
> problem. It's a mathematical problem. The question makes perfect sense
> from a mathematical standpoint.
>
> Your final sentence above makes no sense at all. There is no such thing
> as a "last ball". All that is necessary to show that the vase is empty
> is to show that each ball is removed from the vase before noon and is not
> subsequently replaced.
>
> > Assume f(x) is an increasing function and f(0) > 0. Even though f(oo)
> > is undefined but I am quite sure that f(oo) cannot be 0.
>
> Why?
>
Thanks Dave,
This is a paradox because it is not a physical nor a mathematical
problem but rather a logical problem!!!
Physically or mathematical it is not difficult to prove that the vase
is not empty at noon. Mathematically, # of balls in the vase can be
expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the
time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO.
f(1) = 9
f(1/2) = 18
f(1/4) = 27
f(1/8) = 36
Number of balls in the vase at noon is f(0) = OO. Try it :)
The statement: "All that is necessary to show that the vase is empty is
to show that each ball is removed from the vase before noon and is not
subsequently replaced." is a logical argument. Unfortunately, if
infinity gets involved, this statement alone is not sufficient to claim
the vase is empty before noon. The vase can only be claimed to be
empty at a particular time t is when all the balls are removed at that
time. At t = 1, ball 1 is removed but ball 10 (the last ball) is added
to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last
ball) is in the vase. There is no such time that all balls are
removed. Therefore, the vase cannot be empty.
If the problem is changed to remove the last ball each time then the
argument would be much simplier - ball 1 is always in the base.
Dave Seaman
>> Noon may be un-reachable in a physical sense, but this is not a physical
>> problem. It's a mathematical problem. The question makes perfect sense
>> from a mathematical standpoint.
>> Your final sentence above makes no sense at all. There is no such thing
>> as a "last ball". All that is necessary to show that the vase is empty
>> is to show that each ball is removed from the vase before noon and is not
>> subsequently replaced.
>> > Assume f(x) is an increasing function and f(0) > 0. Even though f(oo)
>> > is undefined but I am quite sure that f(oo) cannot be 0.
>> Why?
>> Dave Seaman
> Thanks Dave,
> This is a paradox because it is not a physical nor a mathematical
> problem but rather a logical problem!!!
Saying that it's a "paradox" merely means that the result is
counterintuitive. It doesn't mean there is actually a contradiction.
> Physically or mathematical it is not difficult to prove that the vase
> is not empty at noon. Mathematically, # of balls in the vase can be
> expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the
> time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO.
> f(1) = 9
> f(1/2) = 18
> f(1/4) = 27
> f(1/8) = 36
> Number of balls in the vase at noon is f(0) = OO. Try it :)
Incorrect. You are confusing f(0) with lim_{t->0-} f(t). The former is
0, but the latter is +oo. The former is what the problem asks for. Once
you get your notation straight, the paradox disappears.
> The statement: "All that is necessary to show that the vase is empty is
> to show that each ball is removed from the vase before noon and is not
> subsequently replaced." is a logical argument. Unfortunately, if
> infinity gets involved, this statement alone is not sufficient to claim
> the vase is empty before noon. The vase can only be claimed to be
> empty at a particular time t is when all the balls are removed at that
> time.
Incorrect. Change "at that time" to "at or before that time". Then the
time t=noon fills the bill.
> At t = 1, ball 1 is removed but ball 10 (the last ball) is added
> to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last
> ball) is in the vase. There is no such time that all balls are
> removed. Therefore, the vase cannot be empty.
All balls are removed before noon.
> If the problem is changed to remove the last ball each time then the
> argument would be much simplier - ball 1 is always in the base.
If the problem is changed so that all the balls are in the vase at 11:00
and from then on balls are only removed, never added, then do you agree
that it is possible for the vase to be empty at noon?
William Hughes
Yes.
>They are un-reachable.
Why? An infinite time from now is un-reachable in any physical
sense, but there is no problem defining it in a mathematical
sense.
> Again I
> have to emphasis that in order to prove the vase is empty, we have to
> prove that the last ball is taken out from the vase.
No, there is no such thing as "the last ball". To prove that the vase
is empty we have to prove that every ball that is put into the vase
is taken out of the vase.
>
> Yes, you can claim from problem 2 that ball n is removed from the vase
> at the nth minute. The vase can only be claimed to be empty if ball n
> is the last ball in the vase. Obviously by the time the nth ball is
> removed there are many more balls in the vase. So the vase cannot be
> empty even though every ball currently in the vase will be removed
> later in the future.
>
And thus at any finite t the vase will not be empty. This does not
tell
us what will hapen after an infinite amount of time.
> By mathematical induction. n=1 is true. n=n+1 is also true. It implies
> that it is true for all n. The vase is not empty for all n > 0.
>
Yes, after any finite amount of time the vase is not empty.
However, this does not tell us what will happen after an infinite
amount of time.
> Mathematical induction still cannot tell you what happens exactly when
> n is infinite which does not exist physically or mathematically.
Small error here. Inifinity does not exist physically, but it
does exist mathematically.
> But
> it can claim that the vase is not empty as n approaches to infinity.
>
And as this does not tell us what happens at infinity, it does
not help.
> > > In the vase example, if we only look at the withdrawal portion of the
> > > argument then the claim is not strong. If both putting in and taken
> > > out of balls are taken into account. In order to prove that the vase
> > > is empty at noon, you must have to show at certain time the rate of
> > > taking out is greater than the rate of putting in.
> >
> > No. The rate at which we add balls is infinite. So is the rate at
> > which we remove them. We cannot conclude anything from this fact. True, > at any time before noon the rate at which we are adding balls is greater > than the rate at which we are removing them. This tells us nothing about
> > what happens at noon.
> > -William Hughes
> In the equivalent problem (problem 2), the rate of adding the balls to
> the vase is 10 balls/minute. The withdrawal rate is 1 ball/minute.
>
So we can have an infinite rate and a finite time, or a finite rate
and an infinite time. We still have an infinite number of balls.
> Assume f(x) is an increasing function and f(0) > 0. Even though f(oo)
> is undefined but I am quite sure that f(oo) cannot be 0.
Why?
-William Hughes
William Hughes
> >
> > Noon may be un-reachable in a physical sense, but this is not a physical
> > problem. It's a mathematical problem. The question makes perfect sense
> > from a mathematical standpoint.
> >
> > Your final sentence above makes no sense at all. There is no such thing
> > as a "last ball". All that is necessary to show that the vase is empty
> > is to show that each ball is removed from the vase before noon and is not
> > subsequently replaced.
> >
> > > Assume f(x) is an increasing function and f(0) > 0. Even though f(oo)
> > > is undefined but I am quite sure that f(oo) cannot be 0.
> >
> > Why?
> >
> > Dave Seaman
>
> Thanks Dave,
>
> This is a paradox because it is not a physical nor a mathematical
> problem but rather a logical problem!!!
>
> Physically or mathematical it is not difficult to prove that the vase
> is not empty at noon. Mathematically, # of balls in the vase can be
> expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the
> time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO.
> f(1) = 9
> f(1/2) = 18
> f(1/4) = 27
> f(1/8) = 36
> Number of balls in the vase at noon is f(0) = OO. Try it :)
O.K, f(0) = 9 + 9*log(1/0)/log(2) which is undefined.
I guess we will have to use a different way to
find out how many balls are in the vase
at noon. If this were a physical problem we could use the limit
of f(t) as t approaches 0. However, this is not a physical problem.
>
> The statement: "All that is necessary to show that the vase is empty is
> to show that each ball is removed from the vase before noon and is not
> subsequently replaced." is a logical argument.
And a sound one.
> Unfortunately, if
> infinity gets involved, this statement alone is not sufficient to claim
> the vase is empty before noon.
True, but only because the vase is not empty at any time before
noon. The statement is sufficient to claim the vase is empty at noon.
> The vase can only be claimed to be
> empty at a particular time t is when all the balls are removed at that
> time.
No, we need
The vase can only be claimed to be
empty at a particular time t is when all the balls
are removed at or before that time.
---------
In there are only a finite number of steps, adding "or before"
is not needed, we can always find the last time at which balls
are removed. However, if there are a infinite number of steps there
may not be a last time at which balls are removed.
> At t = 1, ball 1 is removed but ball 10 (the last ball) is added
> to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last
> ball) is in the vase. There is no such time that all balls are
> removed. Therefore, the vase cannot be empty.
>
Yes, there is no time before noon at which all balls are removed.
However, there are an infinite number of steps and this statement
does not tell us anything about what happens after an infinite number
of steps.
> If the problem is changed to remove the last ball each time then the
> argument would be much simplier - ball 1 is always in the base.
As has been noted many times, if you change the order in which
you remove the balls you can get any result from 0 to infinity.
- William Hughes
William Hughes
Ouch very nasty. Note there are two seemingly correct arguments.
i) A cannot tell where B get the balls from, so the answer
is not changed. The vase is empty at noon.
ii) The first ball (indeed every ball) gets added to and
removed from the vase an infinite number of times
in any interval (noon-t,noon). Thus we cannot say
whether or not the first ball is in the vase at noon.
Thus the number of balls in the vase at noon is
indeterminate.
I would attempt to resolve the seeming paradox as follows. If
we consider not balls but labels, it is clear that there is
no label in the vase at noon (labels are not reused). Can we
conclude from this that there are no balls in the vase at noon?
i) if every ball has one and only one label, Yes.
i') if every ball has only a finite number of labels, Yes.
ii) if any ball has an inifinite number of labels, No.
In the original problem we make the assumption that every
label corresponds to exactly one ball, i.e. we assume that
the balls are not reused. With this assumption the arguments
used are correct. If we make the assumption that every ball
is reused an infinite number of times, then every ball will
correspond to an infinite number of labels and the arguments
used are not correct.
Still, if the original result is counterintuitive, this is
counterintuitive squared.
-William Hughes
snapdragon31
> > Physically or mathematical it is not difficult to prove that the vase
> > is not empty at noon. Mathematically, # of balls in the vase can be
> > expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the
> > time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO.
> > f(1) = 9
> > f(1/2) = 18
> > f(1/4) = 27
> > f(1/8) = 36
> > Number of balls in the vase at noon is f(0) = OO. Try it :)
>
> Incorrect. You are confusing f(0) with lim_{t->0-} f(t). The former is
> 0, but the latter is +oo. The former is what the problem asks for. Once
> you get your notation straight, the paradox disappears.
>
Hi Dave, thanks for your response,
According to the fomula f(t) = 9+9*log(1/t)/log(2),
f(0) = 9 + 9 log(1/0)/log(2)
= 9 + 9 log(OO)/log(2)
= 9 + 9 * OO / log(2)
= OO not 0.
In mathematical term, usually we refer infinity as undefined
e.g 1/0 = OO (infinity) but very often we say 1/0 is undefined.
I think you do not mean t->0- above. It should be t->0+, right? f(t)
is undefined as t->0-.
f(0+) = 9 + 9 log(1/0+)/log(2)
= 9 + 9 log(OO)/log(2)
= 9 + 9 * OO / log(2)
= OO
f(0-) = 9 + 9 log(1/0-)/log(2)
= 9 + 9 log(-OO)/log(2)
Cannot proceed because log(-n) is not valid
> > The statement: "All that is necessary to show that the vase is empty is
> > to show that each ball is removed from the vase before noon and is not
> > subsequently replaced." is a logical argument. Unfortunately, if
> > infinity gets involved, this statement alone is not sufficient to claim
> > the vase is empty before noon. The vase can only be claimed to be
> > empty at a particular time t is when all the balls are removed at that
> > time.
>
> Incorrect. Change "at that time" to "at or before that time". Then the
> time t=noon fills the bill.
>
> > At t = 1, ball 1 is removed but ball 10 (the last ball) is added
> > to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last
> > ball) is in the vase. There is no such time that all balls are
> > removed. Therefore, the vase cannot be empty.
>
> All balls are removed before noon.
Again this argument alone is not sufficient to claim that the vase is
empty at noon. Do you have any other argument?
> > If the problem is changed to remove the last ball each time then the
> > argument would be much simplier - ball 1 is always in the base.
>
> If the problem is changed so that all the balls are in the vase at 11:00
> and from then on balls are only removed, never added, then do you agree
> that it is possible for the vase to be empty at noon?
If 'all the balls' refers to only a finite number of balls, say a
million, a billion, a trillion or a google balls then yes the vase will
be empty before noon. Let n be the total number of ball at 11:59
t minute be the time before noon
f(t) = n - log(1/t)/log(2)
f(1) = n
f(1/2) = n - 1
f(1/4) = n - 2
...
f(1/2^n) = n - n = 0
You can see that one of the requirements for the vase to be empty at
noon is the ball withdrawal rate is greater than the adding rate. In
this case the adding rate is 0 but the removal rate is greater than 0.
The last ball (ball n) is to be removed at a time 1/2^n minute before
noon. By the time the last ball is being removed, no other ball is
left in the vase. This is exactly the criteria for the vase to be
empty.
Unfortunately, 'all the balls' in this problem refers to an infinite
number of balls.
f(t) = OO - log(1/t)/log(2)
f(1) = OO
f(1/2) = OO - 1 = OO
f(1/4) = OO - 2 = OO
...
There is no such a time that the vase would be empty.
---------------------------------------------------------
Case 1. If the removal rate is 10 balls/each time and the adding rate
is 1 ball/each time then the vase will be empty at noon provide that
there are not infinitely number of balls in the vase initially.
Case 2. If the removal rate is 1 ball/each time and the adding rate is
10 balls/each time then the vase will not be empty at noon.
Case 3. If the removal rate is 1 ball/each time and the adding rate is
1 ball/each time then the vase will be empty at noon only if there is
no ball in the vase initially.
I hope you can see that the argument "All balls are removed before
noon." alone is not sufficient to claim that the vase is empty at noon.
Torkel Franzen
> He says it all has to do with Cantor's set theory, cardinality etc..., but
> browsing the internet didn't really help me much.
> Any information or relevant links are very welcome,
This is an old philosophical favorite from the fifties and sixties.
Googling for supertask+philosophy will lead you to a number of papers
and discussions.
Torkel Franzen
> Googling for supertask+philosophy will lead you to a number of papers
> and discussions.
The Stanford Encyclopedia gives a good overview:
snapdragon31
> O.K, f(0) = 9 + 9*log(1/0)/log(2) which is undefined.
Mathematically: f(0) = OO (infinity). In usual case we would say it is
undefined but because we are dealing with infinities here so f(0) = OO
is well defined.
Physically: Infinity is undefined physically.
> I guess we will have to use a different way to
> find out how many balls are in the vase
> at noon. If this were a physical problem we could use the limit
> of f(t) as t approaches 0. However, this is not a physical problem.
> >
> > The statement: "All that is necessary to show that the vase is empty is
> > to show that each ball is removed from the vase before noon and is not
> > subsequently replaced." is a logical argument.
>
> And a sound one.
Agree, a sound one but still not sufficient to make the claim.
>
> > Unfortunately, if
> > infinity gets involved, this statement alone is not sufficient to claim
> > the vase is empty before noon.
>
> True, but only because the vase is not empty at any time before
> noon. The statement is sufficient to claim the vase is empty at noon.
The operation of adding or removal of balls is undefined at noon. In
order for the vase to be empty at noon, the vase has to be empty before
noon. It is proved that the vase is not empty at any time before noon.
How can the vase be empty at noon!
>
> > The vase can only be claimed to be
> > empty at a particular time t is when all the balls are removed at that
> > time.
>
> No, we need
> The vase can only be claimed to be
> empty at a particular time t is when all the balls
> are removed at or before that time.
> ---------
>
> In there are only a finite number of steps, adding "or before"
> is not needed, we can always find the last time at which balls
> are removed. However, if there are a infinite number of steps there
> may not be a last time at which balls are removed.
>
> > At t = 1, ball 1 is removed but ball 10 (the last ball) is added
> > to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last
> > ball) is in the vase. There is no such time that all balls are
> > removed. Therefore, the vase cannot be empty.
> >
>
> Yes, there is no time before noon at which all balls are removed.
> However, there are an infinite number of steps and this statement
> does not tell us anything about what happens after an infinite number
> of steps.
Again, the operation of putting in or taking out balls is undefined at
noon. You are not allow to add or remove any ball right at the noon
time. In order for the vase to be empty at noon, the vase has to be
empty some time before noon. You already agree that no time before
noon at which all balls are removed. How can the vase be empty at
noon?
> > If the problem is changed to remove the last ball each time then the
> > argument would be much simplier - ball 1 is always in the base.
>
> As has been noted many times, if you change the order in which
> you remove the balls you can get any result from 0 to infinity.
No, the result can never be 0.
Lee Rudolph
>> infinity gets involved,
badly working minds stop working entirely.
We see it all the time.
Lee Rudolph
David C. Ullrich
wrote:
>William Hughes wrote:
>> O.K, f(0) = 9 + 9*log(1/0)/log(2) which is undefined.
>
>Mathematically: f(0) = OO (infinity). I
No, mathematically f(0) is undefined.
But never mind that. There's a much more important point
that you're missing: The given formula for f(t) is valid
_only_ for t > 0. Whether that formula is infinite,
undefined or whatever when t = 0 is simply _irrelevant_,
because the formula simply does not apply in the case
t = 0.
>[...]
>
>Again, the operation of putting in or taking out balls is undefined at
>noon. You are not allow to add or remove any ball right at the noon
>time. In order for the vase to be empty at noon, the vase has to be
>empty some time before noon. You already agree that no time before
>noon at which all balls are removed. How can the vase be empty at
>noon?
This doesn't make much sense. You seem to be saying that it cannot
happen that something is true at noon unless it was already true
at some time before noon. What would make you believe such a silly
thing?
(Like: at noon it is _noon_, right? How can it be noon at
noon when it was never noon at any time before noon?
I imagine you agree that that makes no sense.)
>> > If the problem is changed to remove the last ball each time then the
>> > argument would be much simplier - ball 1 is always in the base.
>>
>> As has been noted many times, if you change the order in which
>> you remove the balls you can get any result from 0 to infinity.
>
>No, the result can never be 0.
************************
David C. Ullrich
Dave Seaman
> Dave Seaman wrote:
>> > Physically or mathematical it is not difficult to prove that the vase
>> > is not empty at noon. Mathematically, # of balls in the vase can be
>> > expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the
>> > time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO.
>> > f(1) = 9
>> > f(1/2) = 18
>> > f(1/4) = 27
>> > f(1/8) = 36
>> > Number of balls in the vase at noon is f(0) = OO. Try it :)
>> Incorrect. You are confusing f(0) with lim_{t->0-} f(t). The former is
>> 0, but the latter is +oo. The former is what the problem asks for. Once
>> you get your notation straight, the paradox disappears.
> Hi Dave, thanks for your response,
> According to the fomula f(t) = 9+9*log(1/t)/log(2),
That formula does not apply to the case t=0. You also have a sign error
if the formula is intended to apply to times before noon (t<0).
> f(0) = 9 + 9 log(1/0)/log(2)
> = 9 + 9 log(OO)/log(2)
> = 9 + 9 * OO / log(2)
> = OO not 0.
Your math is faulty. Division by 0 is not allowed, and oo is not a
number. Your formula can be made to say something about the limit as
t->noon, but the limit is not what the problem asks for.
> In mathematical term, usually we refer infinity as undefined
> e.g 1/0 = OO (infinity) but very often we say 1/0 is undefined.
> I think you do not mean t->0- above. It should be t->0+, right? f(t)
> is undefined as t->0-.
> f(0+) = 9 + 9 log(1/0+)/log(2)
> = 9 + 9 log(OO)/log(2)
> = 9 + 9 * OO / log(2)
> = OO
I mean t->0-. I use t<0 for times before noon, and t>0 for times after
noon. The left-hand limit is +oo, but the f(0) and the right-hand limit
are both 0.
> f(0-) = 9 + 9 log(1/0-)/log(2)
> = 9 + 9 log(-OO)/log(2)
> Cannot proceed because log(-n) is not valid
That's why I pointed out that you had a sign error in your formula. It
should be log(-1/t), not log(1/t), in order to make sense for times
before noon.
>> > The statement: "All that is necessary to show that the vase is empty is
>> > to show that each ball is removed from the vase before noon and is not
>> > subsequently replaced." is a logical argument. Unfortunately, if
>> > infinity gets involved, this statement alone is not sufficient to claim
>> > the vase is empty before noon. The vase can only be claimed to be
>> > empty at a particular time t is when all the balls are removed at that
>> > time.
>> Incorrect. Change "at that time" to "at or before that time". Then the
>> time t=noon fills the bill.
>> > At t = 1, ball 1 is removed but ball 10 (the last ball) is added
>> > to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last
>> > ball) is in the vase. There is no such time that all balls are
>> > removed. Therefore, the vase cannot be empty.
>> All balls are removed before noon.
> Again this argument alone is not sufficient to claim that the vase is
> empty at noon. Do you have any other argument?
My other argument is that no ball is returned to the vase after it is
removed. Which ball do you think is in the vase at noon?
>> > If the problem is changed to remove the last ball each time then the
>> > argument would be much simplier - ball 1 is always in the base.
>> If the problem is changed so that all the balls are in the vase at 11:00
>> and from then on balls are only removed, never added, then do you agree
>> that it is possible for the vase to be empty at noon?
[... irrelevant finite case snipped ...]
> Unfortunately, 'all the balls' in this problem refers to an infinite
> number of balls.
> f(t) = OO - log(1/t)/log(2)
> f(1) = OO
> f(1/2) = OO - 1 = OO
> f(1/4) = OO - 2 = OO
> ...
> There is no such a time that the vase would be empty.
Wrong. Noon is such a time. If you disagree, name a ball that is in the
vase at noon.
> I hope you can see that the argument "All balls are removed before
> noon." alone is not sufficient to claim that the vase is empty at noon.
You keep asserting this, but you have never given any hint as to what it
is that makes you think such a thing. What is the reasoning behind your
outlandish claim?
You also keep ignoring my question. If you think the vase is not empty
at noon, then what is the number of a ball that remains in the vase?
As long as you keep ignoring my question and refuse to provide any
supporting argument of your own, we will never make any progress.
Herman Jurjus
Perhaps also a useful link:
http://www.arxiv.org/abs/math.CO/0401154
(Boaz Tsaban, Random strategies with memory for the Robin Hood game)
From the abstract:
The_Robin_Hood_ game is played as follows: On day i, the Sheriff puts
s(i) bags of gold in the cave. On night i, Robin removes r(i) bags from
the cave. The game is played for each natural nymber i. Robin wins if
each bag which was put in the cave is eventually removed from it;
otherwise the Sheriff wins. (...)
--
Cheers,
Herman Jurjus
William Hughes
> William Hughes wrote:
> > O.K, f(0) = 9 + 9*log(1/0)/log(2) which is undefined.
>
> Mathematically: f(0) = OO (infinity).
No, as David Ullrich has pointed out f is not defined
at 0 and in any case the value at 0 is irrellevant.
> In usual case we would say it is
> undefined but because we are dealing with infinities here so f(0) = OO
> is well defined.
> Physically: Infinity is undefined physically.
>
> > I guess we will have to use a different way to
> > find out how many balls are in the vase
> > at noon. If this were a physical problem we could use the limit
> > of f(t) as t approaches 0. However, this is not a physical problem.
> > >
> > > The statement: "All that is necessary to show that the vase is empty is
> > > to show that each ball is removed from the vase before noon and is not
> > > subsequently replaced." is a logical argument.
> >
> > And a sound one.
>
> Agree, a sound one but still not sufficient to make the claim.
>
> >
> > > Unfortunately, if
> > > infinity gets involved, this statement alone is not sufficient to claim
> > > the vase is empty before noon.
> >
> > True, but only because the vase is not empty at any time before
> > noon. The statement is sufficient to claim the vase is empty at noon.
>
> The operation of adding or removal of balls is undefined at noon. In
> order for the vase to be empty at noon, the vase has to be empty before
> noon. It is proved that the vase is not empty at any time before noon.
> How can the vase be empty at noon!
>
It is not empty after any finite number of steps, it is empty after
an infinite number of steps. As you point out, we can either
take an infinite time to do the steps in constant time or
we can take a finite time to do all the steps, by doing the steps
arbitrarially fast. The important fact is that we do an infinite
number of steps.
Consider the following:
A: Let the Vase start by containing all the balls.
At time noon - 1 second remove ball 1
. At time noon-1/2 seconds remove ball 2
At time noon-1/4 seconds remove ball 3
...
At noon the Vase is empty. At any time before noon the Vase
contains an infinite number of balls. There is no time
at which the last ball is removed.
B Let the Vase start empty
At time noon -(1+1) seconds add ball 1-10
At time noon -(1+1/2) seconds add balls 11-20
At time noon -(1+1/4) seconds add balls 21-30
...
The Vase starts empty. At noon - 1 second the vase contains
all the balls.
C Let the Vase start empty.
Apply B, then apply A
The Vase starts empty. At noon - 1 second the vase contains
all the balls. At noon the Vase is empty.
B' Let the Vase start empty
At time noon -1 seconds add ball 1-10
At time noon -1/2 seconds add balls 11-20
At time noon -1/4 seconds add balls 21-30
...
C' Let the Vase start empty.
Apply Both B' and A.
(Note we can apply A even though the Vase
starts empty, because applying B' puts the
balls we need into the Vase.)
The Vase starts empty. At noon the Vase is empty.
<snip>
-William Hughes
Kirby Cook
I ask, by what logic you make 9 + 9 + 9 +... =0?
Kirby
David Kastrup
You don't have 9+9+9+..., but instead you have something that appears
at first glance like:
1 + 10 - 1 + 10 - 1 + 10 - 1 + 10 - 1 + 10 - 1 ...
Now we all know that only absolutely convergent series may be
reordered at will. It turns out that the above _is_ already
reordered, since the first 10 terms of (-1) actually have to be
grouped with the first term of +10, being the same balls. If you
reinstate the proper order, then the sum becomes 0. The labels of the
balls establish the proper order.
If the balls are not labelled, there is nothing to order the terms by,
and the resulting sum is undefined: it can be positive, negative,
infinite in either direction or zero, depending on the order.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
Kirby Cook
action is to add 10 balls to the vase, then remove one. As to your
reordering the infinite sum to read, in effect, (10 - 10) + (10 - 10)
etc., my first response is that that is fallacious, a sophistry that
does not reflect what actually happened, which was that a net 9 balls
were added to the vase each time. My sum reflects what happened; yours
doesn't. If yours did, any and every infinite sum could be made to
equal anything you wished... OK, I'll have to think about that.
Kirby
David Kastrup
> David Kastrup wrote:
[...]
>>> Your challenge is persuasive; I cannot answer it. Can you
>>> answer when I ask, by what logic you make 9 + 9 + 9 +... =0?
>> You don't have 9+9+9+..., but instead you have something that
>> appears
>> at first glance like:
>> 1 + 10 - 1 + 10 - 1 + 10 - 1 + 10 - 1 + 10 - 1 ...
>> Now we all know that only absolutely convergent series may be
>> reordered at will. It turns out that the above _is_ already
>> reordered, since the first 10 terms of (-1) actually have to be
>> grouped with the first term of +10, being the same balls. If you
>> reinstate the proper order, then the sum becomes 0. The labels of the
>> balls establish the proper order.
>> If the balls are not labelled, there is nothing to order the terms
>> by,
>> and the resulting sum is undefined: it can be positive, negative,
>> infinite in either direction or zero, depending on the order.
>>
> First, to pick at a nit, your leading 1 doesn't belong there;
> the first action is to add 10 balls to the vase, then remove
> one. As to your reordering the infinite sum to read, in
> effect, (10 - 10) + (10 - 10) etc., my first response is that
> that is fallacious, a sophistry that does not reflect what
> actually happened, which was that a net 9 balls were added to
> the vase each time.
But you don't know which balls. And sums may be reordered only by a
finite number of term rearrangements. And for the net result it is
important to know the number of balls that have been put in and not
taken out. You want the sum of the total moves for every ball. And
that means that you must not move the corresponding +1 and -1 terms
for a particular ball arbitrarily far apart, or the end result will
not correspond to the number of labelled balls left at the end.
> My sum reflects what happened; yours doesn't.
But you don't want to know what happened. You want to know how many
balls there are _at_ _the_ _end_ that have been put in and not taken
out again. And the terms of this difference may not be moved around
in the sum without changing it.
> If yours did, any and every infinite sum could be made to
> equal anything you wished...
Uh, well, that is _exactly_ what reordering a not absolutely
convergent series _can_ do.
Dave Seaman
> Dave Seaman wrote:
> Your challenge is persuasive; I cannot answer it. Can you answer when
> I ask, by what logic you make 9 + 9 + 9 +... =0?
I don't. What made you think I did?