# The infinite bag of billiards balls

The following trio of puzzles comes from http://skepticsplay.blogspot.com/

My question is: What is the solution to variation 2?

The original author of these puzzles described them as being abstract. I understand this to mean that we don’t have to consider any real world physical constraints. For example, we don’t have to worry about the bag being big enough to hold all the balls OR whether it is possible to have an infinite number of balls OR any other considerations.

This is a classic puzzle that often provokes lively discussion among puzzling communities.

We start with an empty bag at 11:00. We put in 10 billiards balls, which are numbered 1 through 10. But then we take out the ball labeled #1. At 11:30, we add 10 more billiards balls, which are numbered 11 through 20, and remove the ball labeled #2. At 11:45, we add balls 21 through 30 and remove #3.

This process is repeated infinitely. At each step, the next ten billiards balls are put into the bag, and the lowest number in the bag is removed. There are an infinite number of steps before 12:00, since the time between each step decreases (halved) at each step.

Obviously this puzzle is meant to be abstract, as you cannot actually perform an infinite number of steps within an hour, nor can a bag actually hold that many billiards balls. But we can still ask a genuine mathematical question about the situation: what's left in the bag after 12:00?

Variation 1: Instead of removing the lowest numbered ball in the bag at each step, we instead remove the highest numbered ball. So on step 1, we remove #10, on step 2, we remove #20, and so forth. What's left in the bag after 12:00?

Variation 2: On the first step, we add only billiards balls labeled 1 through 9. Instead of adding #10, we just take #1 and paint an extra zero at the end. On the second step, we add numbers 11 through 19, and add an extra zero to the end of #2, resulting in #20. On the third step, we add numbers 21 through 29, and add an extra zero to the end of #3. This process is repeated infinitely before 12:00. What's left in the bag after 12:00?

• A black hole... Commented Oct 21, 2023 at 22:03
• Not a full answer, but changing the order of the terms in an infinite sum can change that sum. Commented Oct 21, 2023 at 23:41
• You never reach after 12.
– Moti
Commented Oct 22, 2023 at 5:49
• So the first two parts are trivial: rot13(ab onyyf ner yrsg va gur bevtvany chmmyr fvapr gur agu onyy vf erzbirq ng n gvzr 2^-a) and rot13(na vasvavgr ahzore bs onyyf ner yrsg nf ng rnpu fgrc 9 onyyf ner nqqrq gung nera'g erzbirq) but for last variation I am a little puzzled. It would seem we get rot13(na vasvavgr ahzore bs genafsvavgryl ynoryyrq onyyf fvapr rnpu onyy a vf hcqngrq na vasvavgr ahzore bs gvzrf) but that seems a bit too obscure to be the intended solution. Commented Oct 22, 2023 at 9:26
• @WillOctagonGibson: It was a joke, I forgot to add a smiley face :)! Just another comment: I know this puzzle under the name "Hilbert's subway/metro" (as an analogy to "Hilbert's hotel"). Thanks for the puzzles!
– Plop
Commented Oct 24, 2023 at 8:05

The balls which are added to the bag are exactly those balls where the numbers on it have no trailing zeros. During the infinite process the number $$X$$ on such a ball changes from $$X$$ to $$X*10$$, $$X*100$$, $$X*1000$$... ad infinitum but nervertheless the ball can be indentified by its original number $$X$$ on it.

• So at 12 o’clock, what are the labels on the balls? Commented Oct 22, 2023 at 19:35
• The labels are some finite number n with infinitely many zeros appended. If an infinite number of balls fit in the bag we can assume that also an infinite number of zeros fit on the ball. You can avoid problems here if you reduce the size of the following zero appended for example by a factor 2 compared to the previous zero appended. Commented Oct 22, 2023 at 23:46
• And if you remove all trailing zeros from the balls after 12 o’clock you reconstruct exactly the numbers on the balls you have added to the bag. Commented Oct 22, 2023 at 23:53
• I generally thought of this variant as an impossible supertask, since the individual balls don't converge to a single state at the end of the time period just like Thomson's lamp, but after your comment about the geometrically smaller zeroes, I think I'll have to reconsider. Commented Oct 23, 2023 at 8:44
• @JaapScherphuis are you saying the "it's the same ball" argument is valid if there is some concept of limit for the labels and invalid if not? What if you take instead the limit as 5-adic or 2-adic integers. All the labels will converge to 0 then. Does that mean the bag will eventually contain a single ball? Commented Oct 24, 2023 at 12:05

As OP acknowledges the whole question is an abstract one and as such it requires unambiguous definitions to be well posed. A bag that can hold infinitely many balls does not exist. Therefore it is mute to argue what properties such a thing may have.

Unless:

We clarify that our bag is metaphor for a set, more precisely, a family $$S_1,S_2,...$$ of sets indexed by time steps. The elements of set $$S_j$$ are the balls in the bag after $$j$$ steps of adding/removing balls. This allows to rigorously define what we mean by "bag after infinitely many steps". Namely the set $$\bigcup_{k=1}^\infty\bigcap_{j=k}^\infty S_j$$ If you do not read mathematical notation: It says that the set contains and only contains each element that is at some point added and never removed afterwards.

With that in mind the original question and variation 1 are easily answered.

Variation 2 introduces new ambiguities and unlike the "magic bag" one it does not have a generally agreed on resolution. Specifically, it is not clear what the "objects", i.e. things that are allowed to be in a set ("bag") are.

In the original problem and variation 1, balls are helpfully but redundantly labelled with unique numbers. It is an acceptable and reasonable mental economy to not distinguish between balls and labels. So, for example, $$S_1$$ is the set $$\{1,2,...,10\}$$ and not the set $$\{\text{ball labelled with }1,\text{ball labelled with }2,...\text{ball labelled with }10\}$$

Variation 2 introduces and leaves unanswered the question:

Is a relabelled ball 1) the same object as before? Or is it 2) a distinct object. If (2) then variation 2 is identical to the original problem. If (1) then no balls are ever removed.

Note (only applies to case (1)): Irrelevant for the question as asked but maybe interesting in its own right. What are the labels after infinitely many steps?

Answer: undefined. AFAIK, there is no "natural"/consensus/broadly accepted "limit" of the sequence of labels on a given ball.

Solution to Variation 2:

As an example, consider what happens to the ball initially labeled 5. It is put into the bag in the first step. In the 5th step, its label is changed to 50. In the 50th step, its label is changed to 500. In fact the label on this ball is changed an infinite number of times (on the 5th, 50th, 500th, ... steps).

So at 12:00 this ball will be labeled:

5000... where there is an infinite number of zeros.

All balls are relabeled in a similar fashion so at 12:00 the bag contains balls labeled:

1000...
2000...
3000...
$$\cdots$$
9000...

11000...
12000...
13000...
$$\cdots$$
19000...

$$\cdots$$

In summary, at 12:00 for every positive integer $$n$$ that is not a multiple of 10, the bag will contain a ball whose label is $$n$$ followed by an infinite number of zeros.