# Another lot of lots of balls

wolfgang59
Posers and Puzzles 08 Oct '10 21:20
1. wolfgang59
Mr. Wolf
08 Oct '10 21:20

nb. this is NOT meant to be frivolous, this is a SERIOUS discussion

thank you

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Suppose you and your friend come into possession of a bag of infinite capacity, along with an infinite supply of numbered balls and a stopwatch with an infinite number of gradations. After watching the universe explode over mimosas and eggs Benedict at the Restaurant at the End of the Universe, the two of you decide to pass the time by running the following experiment:

1. You begin by placing two balls (balls 1 and 2) into the bag, then set the stopwatch to count down from 60 seconds.

2. When the stopwatch reads 30 seconds, you place two more balls (balls 3 and 4) into the bag and your friend pulls out the last ball you placed in the bag (ball 4).

3. When the stopwatch reads 15 seconds, you place two more balls (balls 5 and 6) into the bag and your friend pulls out the last ball you placed in the bag (ball 6).

4. The two of you repeat this process, you placing the next two balls into the bag and your friend removing the newest ball, each time the remaining time on the stopwatch is halved (i.e. 7.5 seconds, 3.75 seconds, 1.825 seconds, etc...).

Question: How many balls will be left in the bag when the stopwatch reads 0 seconds?
2. Palynka
Upward Spiral
08 Oct '10 21:33
That's easy, infinite balls.
3. wolfgang59
Mr. Wolf
08 Oct '10 22:43
OK
Now you do exactly the same process but take different balls
(BUT THE SAME QUANTITY)
instead of the highest numbered ball you take the least numbered ball

... how does the result differ?
4. joe shmo
Strange Egg
09 Oct '10 00:40
Originally posted by wolfgang59
OK
Now you do [b]exactly
the same process but take different balls
(BUT THE SAME QUANTITY)
instead of the highest numbered ball you take the least numbered ball

... how does the result differ?[/b]
The least numbered taken in the pattern you mention excludes the odds, thus still an infinite number of ball left in the bag

balls in {1 2 3 4.....}
balls out {1 3 5......}
5. 09 Oct '10 02:181 edit
Originally posted by wolfgang59
OK
Now you do [b]exactly
the same process but take different balls
(BUT THE SAME QUANTITY)
instead of the highest numbered ball you take the least numbered ball

... how does the result differ?[/b]
That the results differ is reconciled through the fact that a countably infinite set can be placed in one-to-one correspondence with a subset of itself. There is no problem here.
6. wolfgang59
Mr. Wolf
09 Oct '10 08:18
Originally posted by LemonJello
That the results differ is reconciled through the fact that a countably infinite set can be placed in one-to-one correspondence with a subset of itself. There is no problem here.
I dont believe it is reconciled.
Lets say that LemonJello and Palynka come across the man with the infinitely large bag and stopwatch et cetera.

he whispers to Palynka that the ball he is removing is the latest to go in
he whispers to LemonJello that the ball he is removing is the earliest to go in

Both men observe him putting two balls in and removing one. (All the balls are identical)

Palynka confidently sees the bag growing as he has predicted that the bag will contain an infinite amount of balls at t=0

LemonJello is not concerned by this (though his chanting mantra of "one-to-one correspondance" has a tinge of doubt)

As the stopwatch reaches the moment of truth what happens?

Who is correct? Both men?
7. 10 Oct '10 00:081 edit
Originally posted by wolfgang59
I dont believe it is reconciled.
Lets say that LemonJello and Palynka come across the man with the infinitely large bag and stopwatch et cetera.

he whispers to Palynka that the ball he is removing is the latest to go in
he whispers to LemonJello that the ball he is removing is the earliest to go in

Both men observe him putting two balls in and rem bt)

As the stopwatch reaches the moment of truth what happens?

Who is correct? Both men?
Consider the following sets, which are directly at issue:

Set1 = {1,2,3,4,....}
Set2 = {4,6,8,10,....}

Set1 are the balls extracted when the oldest (in terms of bag age) is taken out at each specified time step. Set2 are the balls extracted when the youngest is taken out at each specified time step. Set2 is a proper subset of Set1. Regardless, there is one-to-one correspondence between Set1 and Set2 (consider, for example, that for any Set1 member, n, the associated Set2 member is 2n+2). These two sets have the same cardinality (the "same quantity" are extracted as you say by the "same process" as you say), and yet the balls in Set1 that are not members of Set2 constitute yet another countably infinite set:

Set3 = {1,2,3,5,7,9,....}

So what? You tell me: what -- exactly -- is the problem here? Please state precisely what is the problem. It is simply a property of the naturals that it can be placed in one-to-one correspondence with subsets of itself in such ways.

Who is correct?

That just depends on whom "the man" actually lied to.
8. wolfgang59
Mr. Wolf
10 Oct '10 07:46
Originally posted by LemonJello
Consider the following sets, which are directly at issue:

Set1 = {1,2,3,4,....}
Set2 = {4,6,8,10,....}

Set1 are the balls extracted when the oldest (in terms of bag age) is taken out at each specified time step. Set2 are the balls extracted when the youngest is taken out at each specified time step. Set2 is a proper subset of Set1. Regardless, t ...[text shortened]... h ways.

[b]Who is correct?

That just depends on whom "the man" actually lied to.[/b]
That just depends on whom "the man" actually lied to.

But the man is physically doing the same thing. 2 balls in 1 ball out. There cannot be two answers. THAT IS THE PROBLEM.

Personally I think the question is flawed/invalid
9. 10 Oct '10 16:11
Originally posted by wolfgang59
[b]That just depends on whom "the man" actually lied to.

But the man is physically doing the same thing. 2 balls in 1 ball out. There cannot be two answers. THAT IS THE PROBLEM.

Personally I think the question is flawed/invalid[/b]
I have a problem with any answer that varies according to a simple renumbering of the balls.
10. 10 Oct '10 22:142 edits
Originally posted by wolfgang59
[b]That just depends on whom "the man" actually lied to.

But the man is physically doing the same thing. 2 balls in 1 ball out. There cannot be two answers. THAT IS THE PROBLEM.

Personally I think the question is flawed/invalid[/b]
No, he is not physically doing the same thing in both cases: he is purposefully selecting balls with different history to extract in the two cases. If there's an actual problem you want me to take seriously, please state specifically what it is. This candidate of yours (that he is physically doing exactly the same thing in both cases) is false.

FYI, look at this example I found on the web, which is basically exactly the problem at issue:

http://www.math.hmc.edu/funfacts/ffiles/30001.4-6.shtml

So the good people at Harvey Mudd College Math Department are confused just like I am? You better inform them of their error. Their error is....what exactly?
11. 10 Oct '10 22:381 edit
Originally posted by iamatiger
I have a problem with any answer that varies according to a simple renumbering of the balls.
I don't understand why you think this is relevant to this problem. The answer here doesn't vary with "simple renumbering of the balls." The answers in these two cases vary because of a substantive difference in the two cases: balls with substantively different history in the bag are being targeted at each specified extraction point.
12. 11 Oct '10 07:31
Originally posted by LemonJello
I don't understand why you think this is relevant to this problem. The answer here doesn't vary with "simple renumbering of the balls." The answers in these two cases vary because of a substantive difference in the two cases: balls with substantively different history in the bag are being targeted at each specified extraction point.
"Take out the highest numbered ball each time" can be made to have exactly the same effect as "take out the lowest numbered ball at each time" if we change each step to:

"Take out the highest numbered ball each time and add one to each ball's number"

Similarly, we can do "Take out the lowest numbered ball and subtract one from each ball's number", and this is the same as taking out the highest ball.

Seemingly, by changing what we add or take away from the balls' numbers, we can change the actual number of balls from 0 to infinity!
13. wolfgang59
Mr. Wolf
11 Oct '10 12:26
Originally posted by iamatiger
"Take out the highest numbered ball each time" can be made to have exactly the same effect as "take out the lowest numbered ball at each time" if we change each step to:

"Take out the highest numbered ball each time and add one to each ball's number"

Similarly, we can do "Take out the lowest numbered ball and subtract one from each ball's number", an ...[text shortened]... from the balls' numbers, we can change the actual number of balls from 0 to infinity!
Very elegant. I'm glad I've got an ally!
14. 11 Oct '10 12:42
Which is why I disagree with all of you ðŸ™‚.
15. 11 Oct '10 16:23
Originally posted by iamatiger
"Take out the highest numbered ball each time" can be made to have exactly the same effect as "take out the lowest numbered ball at each time" if we change each step to:

"Take out the highest numbered ball each time and add one to each ball's number"

Similarly, we can do "Take out the lowest numbered ball and subtract one from each ball's number", an ...[text shortened]... from the balls' numbers, we can change the actual number of balls from 0 to infinity!
No. The numbering of balls is irrelevant. It doesn't matter if you simply don't number the balls at all. What is relevant is the relationship between the extraction command and the temporal indexing of the balls (in terms of bag age). In the actual problem statement, for the case presented here, the extraction command is stated clearly in reference to temporal relation -- what is extracted at each step is the last ball that was placed in. That is the actual substance of the command, and your re-writes simply do not respect this aspect of the problem. So, what you have done is just an unwarranted sleight of hand.

You are just ignoring what is intended to be the actual relevant difference between the two cases. The relevant difference, again, is the following: in one case, the youngest remaining ball (in terms of bag age) is extracted at each step; in the other case, the oldest remaining ball is extracted at each step. The results in the two cases will still differ (regardless of how you number the balls or re-number the balls or etc), as long as this difference is preserved.

This is just an exercise that exposes the counter-intuitive nature of the fact that a countably infinite set can be placed in one-to-one correspondence with a proper subset of itself.