Balls and gremlins puzzle

Factual Questions
1

I was given this puzzle today, and I’m baffled.

We start with an infinite supply of ping pong balls, helpfully numbered from 1 on up.

At minute 1, we begin. I put balls 1 and 2 into a large bucket.
A nearby gremlin immediately removes ball 1.

At minute 2, I put balls 3 and 4 into the bucket. The gremlin immediately removes ball 2.

At minute 3, I put balls 5 and 6 in the bucket. The gremlin immediately removes ball 3.

After an infinite time, how many ping-pong balls are in the bucket?

I say there are no balls left. After all, what ball would be in there? Number 13? No - the gremlin removed it at minute 13. Number 2398765? No - the gremlin removed it at minute 2398765. And so forth. There is no way any particular ball could be there.

But perhaps I’m not thinking of limits correctly.

2

“After an infinite time”?..no balls left. Infinity divided by two is still infinity.

After a finite time (2,398,765 minutes), there would be 2,398,765 balls in the bucket. 2n - 1n = number of balls where n = number of minutes

3

I doubt I’ve got the time to go into this again like last time, but you might be interested in reading this.

4

Well, hold on. The problem with your question is that the concept of “After an infinite time” is nonsensical. There can never be a time after an infinite period.

However, during the infinite period the number of balls in the bucket will equal the minutes you have been grabbing balls. The gremlin essentialy negates one of your ball grabs, so you are putting one ball in a bucket. As infinity goes on, you would eventually have an infinite number of balls in the bucket.

Asking which balls would be left in the bucket is pointless unless you choose a finite number. In that case, it would be every ball numbered past the number of minutes you have been going, for the number of minutes you have been going. (e.g. 2,398,765 minutes in you would have balls 2,398,765 to 4,797,530)

5

No, asking which balls are left makes perfect sense. Read the thread Cabbage linked to. Or are you saying it doesn’t make sense to ask which numbers are prime? That’s an infinite set, and could be envisioned as the result of a similar process.

Note particularly the similar scenario, where the gremlin removes the highest ball left in the bucket. In that case, all the odd-numbered balls are left. But in this case, the bucket is empty.

6

Actually, it is pointless to ask which balls are left in the OP’s scenario. The infinite period results in there being an infinite number of balls put in the bucket, and also an infinite amount removed from it. However, during no period of time will more balls be removed from the bucket than are being added. This results in a paradox; the bucket is both infinitely full and empty at the same time.

As for which particular balls would be in the empty bucket (yeah… that is odd), they would begin in a sequence infinitely long, stretching out past infinity. Yet another thing that cannot actually be.

7

Umm…

At minute four, you would have balls 5, 6,and 7 in the bucket.
At minute five, you have 6,7,8,and 9 in there…

Ball 13 gets grabbed at minute 13, yes. But, your bucket is starting to overflow at this point. Balls 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, and 25 are in there now.

At minute Twelve Gazillion, there would be Umpteen more balls in the bucket than the ones yoinked by yer gremlin. Not that you could take a snapshot of the process.

Or am I up too late again?

8

No, you’re absolutely right.

Bricker, it’s easier if you don’t think of the balls as being numbered. Just consider the number of balls in the pot at any particular time. This is a straightforward function of the number of elapsed minutes - which in the name of unoriginality I’ll call “x”.

f(x) = 2x - x = x

In other words, with each passing minute there’s an extra ping pong ball in the pot. After an “infinite” number of minutes there’s an infinite number of balls - both in the pot and under the gremlin.

I can see where you got the impression that there would be zero. That comes from thinking of “infinity” as a constant state or quantifiable number. Mathematically, we might say that the number of balls in the pot “tends to infinity” but that doesn’t mean that such a state actually exists. All that means is that the number of balls increases without limit.

So, it doesn’t make sense to ask what balls would be left “after an infinite number of minutes”. Think instead whether the gremlin, taking out one ball per minute, could ever catch up with the input of two balls per minute.

9

OK, people. Read the linked thread, where we debated this ad nauseam. It DOES matter that the balls are numbered, and it DOES matter which ones the gremlin takes out. Infinite sets don’t behave like your intuition says they should.

10

Since the previous thread, I’ve found some material on the web. Search for Littlewood or Littlewood-Ross paradox, or supertasks, using your favorite search engine.

Here is a usenet archive page which discusses it (about the fifth puzzle down). Also, John Byl published ON RESOLVING THE LITTLEWOOD-ROSS PARADOX (word document) where he takes the stance that the urn is full. He references several papers taking the opposing position, but I wasn’t able to find them online.

Finally, Here’s a page on supertasks in general.

11

Speaking ex cathedra I will insist that Bricker is right. No ball is left and that is just that. Another apparent paradox of the infinite. you just cannot expect rules of finite arithmetic to apply to the infinite. That doesn’t mean that it is ruleless, just that the rules are different.

I am reminded of a fatuous claim that Alan Guth made in his book on inflation, on Newton’s “error” in claiming that an infinite universe could not collapse because that would privilege a point, the point to which it would collapse, while the universe has no center. Guth justified this by saying that any sphere, no matter how large, would collapse and concluded the same for an infinite sphere. It is of course very dangerous to suppose that Newton erred and certainly was in this case. Again, you simply cannot use reasoning that is valid for finite things for infinite things without asking if the rules remain valid.

In this case, Bricker’s observation that no ball remains since each one eventually gets removed is irrefutable.

12

OK, since you asked so nicely, I’ll refute it. Zero balls in the urn leads to a contradiction.

I’ll need some additional equipment. First, paint a big, red “A” on the urn, to distinguish it from a second urn, with a big, blue “B”. The second urn differs from the first in that it can only hold two balls, no more.

Place both urns in a box. The urns completely fill the box, and any ball which is in the box must be in one or the other urn. Likewise, any ball within an urn is also in the box. We have an infinite supply of balls outside the box, and both urns are initially empty.

Prior to minute 1, place two balls in urn B, and number them 1 and 2. At the first step (minute 1), the two balls are moved into urn A, and ball 1 is removed from urn A, placed in urn B, and the 1 is erased and a 3 is written on it. A ball from the infinite supply is also placed into urn B and a 4 is written on it.

At minute 2, these balls are moved into urn A, and ball 2 is moved from urn A to urn B, the two is erased from it and a 5 is written on it. A ball from the infinite supply is also placed into urn B and a 6 is written on it.

At minute 3, these balls are moved into urn A, and ball 3 is moved from urn A to urn B, the 3 is erased from it and a 7 is written on it. A ball from the infinite supply is also placed into urn B and an 8 is written on it.

…and so forth…

No balls are ever removed from the box.

After an infinite number of steps, we clearly have an infinite number of balls in the box. Urn B can only hold two balls, so if urn A is empty, there can only be two balls in the box. What happened to the infinite number of balls? Where are they?

Note that from the point of view of urn A, this case is identical with the OP. Only balls numbered with consecutive integers are ever placed in urn A, and the numbers never change while the balls are in urn A. Also note that any argument like “no ball remains since each one eventually gets removed” or “any number you can name isn’t in the urn” is a local argument. It requries no information about what is going on outside the urn, so the extra urn and the box have no bearing on the validity of such an argument.

13

ZenBeam, your scenario may be different, because it involves relabeling balls. I can’t tell immediately, because I know better than to trust my intuition when it comes to infinite sets and infinite-length processes. Let me ask you: where is the ball that we labeled 1 at the end of the process?

Furthermore, in the original scenario, if there are an infinite number of balls left in the urn, name one. It does make sense to ask this; let me illustrate with an example. Suppose that instead of removing the ball with the smallest label each time, the gremlin removes the ball with the smallest composite label. I think we can both agree that at the end, you’ll have exactly those balls with prime labels in the urn. For any ball with a composite label, we can determine exactly when it is removed. What exactly is the difference between this scenario and the one in the OP?

Why is it so hard for you to accept that the infinite doesn’t always behave like the finite?

14

It’s pretty obvious that after n minutes, there will be n balls in the bucket, and they will be numbered n+1 through 2n. So, after 500 minutes, the bucket will contain 500 balls, number 501 to 1000. After 10000 minutes, the bucket will contain 10000 balls, numbered 10001 to 20000. After any finite number of minutes, you know exactly how many balls are in the bucket, and you even know specifically which balls they are.

So, after an infinite time, there should be an infinite number of balls in the bucket, and the lowest numbered one should be numbered infinity+1. But nobody was writing “infinity+1” on a ball; they are each labeled with a specific number. That appears to present a problem, but it doesn’t really because you never get to that infinitieth minute. The whole notion of checking the bucket after an infinite time has elapsed is nonsense, because an infinite time can’t elapse, or else it wouldn’t be infinite. “After an infinite time” is an impossible condition. That is the trick of the puzzle.

15

Zen Beam you have changed the problem with all that relabeling. As ultrafilter remarked, if at each stage you take out the highest numbered ball, you will end up with all the odd numbered balls in your urn. It doesn’t sound that different but it is. If you won’t see you won’t, but in the original scenario, every ball eventually leaves the urn, so none is there at the end.

Let me put it in other words. Every ball that goes into the urn has a number and that number never changes. So any ball found in the urn at the end has a number. So if the urn has even one, let alone infinitely many balls, what number does it bear?

16

What if each step occurs in 1/2[sup]n[/sup] seconds?

17

My response was to Manduck, although it’s a good question for anyone to ponder.

18

Since you add one ball at each step, the number of balls in the bucket has to be equal to the number of step. So, the bucket has an infinite number of balls if an infinite number of steps have been done. The question then is what is written on those balls?

You can figure out what the number is on the lowest-numbered ball by taking the number on the last ball you added, dividing it by two, then adding one. But, if there was an infinite number of steps, then there is no last-ball-added. So the question is nonsense and there is no answer.

19

It’s in the box. Whether it is in urn A or B may be undefined, but it surely is in the box.

Consider the box, ignoring the urns for a moment. Balls are placed in the box, relabled an infinite number of times, but they never leave the box. Do you disagree that there are an infinite number of balls in the box? I don’t think you do, but you can’t name one. I haven’t reviewed the previous thread recently, but my recollection is that all the “proofs” that the dish/urn is empty are proofs that there are no integer numbered disks/balls. But here we have an infinite number of balls in the box, yet none of them are numbered with integers. These 'proofs" don’t show that these non-integer-numbered disks/balls are not in the dish/urn.

I have no problem accepting this, and I’m sorry you feel this way, because I think it’s preventing you from really considering what I’m saying. I might ask in respose, why is it so hard for you to accept that your “proof” isn’t as air-tight as you claim?

I’ll have to think about the case where you are removing composite balls, but offhand I don’t see it making a difference. Unfortunately, I’m on travel next week, so I probably won’t be posting here again for a while.

All the relabeling takes place outside of urn A. As I pointed out, from the point of view of urn A, nothing has changed from the OP.

This is equally true for urn A in my scenario. There are clearly an infinite number of balls in the box. What numbers do they bear, and where are they? If you can’t name which balls are in the box, there is no reason to expect to be able to name which balls are in urn A, and likewise therefore in the urn from the OP.

20

Well, since I was trained as a physicist, the only way I interpret “after an infinite amount of time” is “in the limit as the length of time approaches infinity” , since that’s the only question that makes any kind of physical sense.

So the question is “Can we think of a number big enough that the amount of balls in the box will never exceed that number?” or, to put it another way, “for any number, no matter how big, is there a time after which the amount of balls in the box is always bigger than that number?” [the real mathematicians here might be able to more precisely define a test for whether limit f(x) [as x-> inifinity]= infinity, but this is close enough]

The answer is clearly, no there is no number big enough, and yes there is always some time where the box has more. Therefore the amount is infinite.

After all, if the gremlin grabs the highest numbered ball each time, then the box holds all the odd whole numbers, which I think everyone agrees is inifinite.
How can it make a difference what number is on the ball that the gremlin removes? What if there were no numbers at all on the balls, shouldn’t we get the same answer?