# Balls and gremlins puzzle **URL:** \ **Category:** Factual Questions\ **Created:** [May 15, 2003, 12:53am UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318 "2003-05-15T00:53:50Z")\ **Posts on this page:** 20\ **Page:** 3
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [May 23, 2003, 1:18am UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/41 "2003-05-23T01:18:50Z")
As to **Bill H.**'s question (“Why is this even debatable?”) – I’m not a math major, but here’s my take. A lot of mathematics is intuitive, and that’s no accident. There’s usually an attempt to match up with either a physical phenomena or at least to make things work on a simple and easily understood level. For example, it’s good to have the concept of addition actually correspond to what we perceive as addition, i.e., two + two = four lets us use arithmetic to know[sup]\*[/sup] that two monkeys + two monkeys = four monkeys. If instead, a+b = ‘one more than a’ it would not have much (if any) practical use. So we use the mathematics that fits with what we know. There’s often some point, though, at which things become less intuitive. Infinity is naturally going to be one of those; by its nature it doesn’t have a knowable correspondence with a physical situation. That’s exactly the sort of thing that confuses us in this case. We know that if anyone actually sat down and starting throwing the balls in the urn (and knowing that a person – or an imp – can only put in or remove balls in a finite amount of time) that we would never see the number of balls in the urn shrink. In order to find out the answer, we ask the mathematicians. And yes, they do come up with a definite answer – sort of. What they say is “starting with this, using these definitions of sets, and limits, etc. you get _this_” and that answer could not be different. That’s something everyone can agree on [sup]\*\*[/sup]. But in these cases, the answer goes against what we expect. While nobody disagrees with one particular conclusion, there is real debate. If you look at **ZenBeam** ’s first post in this thread, you’ll notice that there are academic papers written discussing this particular problem[sup]+[/sup]. There’s often papers that argue about a number of mathematical oddities (some of which arise as thread topics here). Not everyone agrees with the solution to the problem. There’s probably two main causes for debate. In reference to a particular problem, there may be the question of what I’ll call ‘the model’ of the problem. Often a basic question, of great interest when attempting to apply mathematics, is whether or not the mathematical tools used are actually modelling the ‘physical’ (or in this case, the ‘stated’) problem. This is what’s called into question with the second formulation of the problem (the two urns, and relabelling of the balls) – are they in fact equivalent; and why or why not? The other cause for debate is whether the math is good enough at all. In some way it’s related to physical correspondence but has to do with simplicity and ‘fitting in’ with the rest of mathematics. By good enough roughly what I mean is if despite its non-intuitive behavior in one case, does it follow intuitive behavior in other (simpler) cases? It’s a bit harder to define, and it can get a little wrapped up in technical details, but I suppose that’s why it becomes a topic for debate. This is the reason why **ultrafilter** had to have a second long post about his definitions of convergence. He had to show that these generally fit the way convergence ‘should’ behave – in his words “why I think this is a _good_ general notion of convergence for sequences of sets”. For example, I could, if I so desired, say that “all infinite sets converge to being empty”, and solve the problem - it’d look somewhat like **ultrafilter** ’s solution, and it could not be argued that it does not show the urn empty at the end. However, it’s clear that such a definition of convergence is nonsense. It not only doesn’t fit our conception of what should happen to a sequence of sets, but it doesn’t allow some of the basic operations to work that make it a useful tool for math. So despite the fact that mathematics proceeds in an orderly fashion to a single ‘right’ answer, it is the rigid definitions that shape the process that are called into question and the heart of many debates in mathematics; in more subtler cases, there are conflicting assumptions that both lead to ‘good’ results as we know now, but might be abandoned later. [sup]\*[/sup]According to common usage, but perhaps not by the strictest of epistemological standards. [sup]\*\*[/sup]Again, this raises a philosophical question (“why must there be universal agreement?”) that doubtless people are working on, but mostly we just accept that mathematics works the way it does because it does, or gives useful results, or whatever we like. [sup]+[/sup]This is completely unrelated to the rest of my post, but in the paper by John Byl linked to, I noticed what looks like an error when he is attempting to reconcile the continuity conditions : “… assuming that any ball coming to rest at a position which it does not leave before noon is still at that position at noon, then all balls in the urn just before noon should still be there at noon …” ---
**Author:** ![Bill\_H](https://avatars.discourse-cdn.com/v4/letter/b/a5b964/32.png) [@Bill\_H](https://boards.straightdope.com/u/Bill_H)\ **Post date:** [May 23, 2003, 1:19am UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/42 "2003-05-23T01:19:12Z")
**ultrafilter** wrote > [@](#): > > I didn’t think you meant anything by it. No hard feelings? of course not. ---
**Author:** ![missing\_link](https://avatars.discourse-cdn.com/v4/letter/m/d2c977/32.png) [@missing\_link](https://boards.straightdope.com/u/missing_link)\ **Post date:** [May 24, 2003, 11:38pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/43 "2003-05-24T23:38:46Z")
**Manduck** , my point was, that the infinite numbers of balls are labeled to begin with, so all of them already have a meaningful label. Imagine that the goblin doesnt discard/eat the ball, but puts it in another bucket. So if one says that the first bucket is empty after the infinite time, the second one has to have an infinite number of balls in it. What would be the labels on them then? In a purely mathematical way I’d say that both buckets will contain an Infinite number of balls. But as an answer to the riddle I’d say that the answer is “Who knows?” Because you never get around to check anyways because you are putting balls in the bucket till all eternity. ---
**Author:** ![Manduck](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/manduck/32/256_2.png) [@Manduck](https://boards.straightdope.com/u/Manduck)\ **Post date:** [May 25, 2003, 3:15am UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/44 "2003-05-25T03:15:47Z")
We are given as an initial condition that all the balls are numbered. The logic of the problem leads us to conclude that the gremlin takes out all the numbered balls, but we also have to conclude that an infinite number of balls remain, but they can’t be numbered. That violates one of our initial conditions, so that proves that one of the conditions must be wrong. Either there can’t be an infinite number of steps, or the balls can’t be numbered, or the imp or the player can’t behave as discribed. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [May 28, 2003, 6:13pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/45 "2003-05-28T18:13:27Z")
> [@](#): > > 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. If the answer with the smallest numbered ball removed is that there are an infinite number of non-integral balls remaining, then the consistent answer when removing the smallest composite number is that what is remaining is 1, all the primes, and an infinite number of non-integral balls. There’s no reason there can’t be both an infinite number of named balls, and an infinite number of non-named balls (if you accept non-named balls in the first place). > [@](#): > > Regarding **ZenBeam** ’s objection: I would contend that the act of relabelling the balls when they’re outside urn A does in fact change the problem, even though nothing has changed “from the persective of the urn”. Let’s suppose just for kicks that the balls have a second number written on them in invisible ink. Initially, these new numbers are sequential, in order, and in place before the experiment begins, but the new numbers don’t get changed when the balls are placed in urn B. > > Then the sequence of sets in urn A, _according to the second, fixed, labelling_, is {}, {2}, {1,3}, {3,2,4}, {2,4,1,5}, {4,1,5,3,6}, {1,5,3,6,2,7}, and so forth. In other words, according to the numbers written in invisible ink the balls in urn A at the end of minute n are the numbers from 1 to n except for f(n), where f is a function which assumes every positive value an infinite number of times. _This sequence of sets has no limit by the definition above_: the upper limit is the entire set of positive integers, the lower limit is empty. Every ball passes from urn A to urn B and back again an infinite number of times. First, not having a limit is not the same as having a limit of zero. Second, it doesn’t matter that this sequence has no limit, because the question is not _which_ balls are in the urn, but _how many_. Even though which balls are in urn B is not defined, is anyone here willing to say there is not an infinite number of balls in urn B? Both **Orbifold** ’s and **ultrafilter** definitions show that the numbers on the balls don’t converge, but we already know that. Again, the question is “how many?”, not “which?”. Their definitions can be used to deduce that an infinite number of balls remain. At each step, there can always be an enumeration of the balls in the urn, and there’s nothing which prevents us from using this enumeration, rather than the numbers printed on the balls. The simplest enumeration is to order them from lowest to highest. After the first step, there is always a first ball (the lowest numbered ball). After the second step, there is always a second ball, the second-lowest numbered ball. And so forth. For any _n_, there is always an _n_th ball, the _n_th-lowest numbered ball in the urn. Taking the limit as _n_ approaches infinity gives us an infinite number of balls. ---
**Author:** ![Bricker](https://avatars.discourse-cdn.com/v4/letter/b/977dab/32.png) [@Bricker](https://boards.straightdope.com/u/Bricker)\ **Post date:** [May 28, 2003, 6:28pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/46 "2003-05-28T18:28:17Z")
> [@](#): > > \*Originally posted by ZenBeam \* > > Both **Orbifold** ’s and **ultrafilter** definitions show that the numbers on the balls don’t converge, but we already know that. Again, the question is “how many?”, not “which?”. Their definitions can be used to deduce that an infinite number of balls remain. At each step, there can always be an enumeration of the balls in the urn, and there’s nothing which prevents us from using this enumeration, rather than the numbers printed on the balls. The simplest enumeration is to order them from lowest to highest. After the first step, there is always a first ball (the lowest numbered ball). After the second step, there is always a second ball, the second-lowest numbered ball. And so forth. For any _n_, there is always an _n_th ball, the _n_th-lowest numbered ball in the urn. Taking the limit as _n_ approaches infinity gives us an infinite number of balls. This is a very simple statement of the solution. I am sure it’s right on. Thanks to all, especially ZenBeam! ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 28, 2003, 7:41pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/47 "2003-05-28T19:41:54Z")
**ZenBeam** : my definition shows that the set of balls in the urn approaches the null set, and no other. If you believe otherwise, either point out the error in my argument, or prove what you’re claiming. The question is most definitely “which”, as the scenario in which the gremlin always removes the odd-numbered balls does result in an infinite number of balls left at the end (in the limit sense). In fact, I can come up with a scenario where the contents of the urn converge to **any** S [symbol]Í[/symbol] **N**. Here’s how it goes:[ol] [li]At each step, you put in the next two balls (starting with 0 and 1).[/li][li]The gremlin removes the lowest-numbered ball which is not in S.[/li][/ol]Using my definition, you can show that this sequence of sets converges to S, no matter what S is– **N** , [symbol]Æ[/symbol], {1}, or anything else. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [May 28, 2003, 9:50pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/48 "2003-05-28T21:50:14Z")
> [@](#): > > The question is most definitely “which” No, the question has always been “how many”. > [@](#): > > In fact, I can come up with a scenario where the contents of the urn converge to any S Í N. Your scenario removing composites is one case of this. I believe my answer there applies to the general case as well. That is, this doesn’t add anything new to the argument. > [@](#): > > my definition shows that the set of balls in the urn approaches the null set, and no other. Your definition shows that there is a first ball in the limit set, and a second ball, and so forth. It may be that your definition needs to be tightened up a bit. Having the number of elements in each set in the sequence of steps which aren’t in the limit set approaching infinity doesn’t give the warmest feeling that things are converging. **Bricker** , thanks for the thanks, but keep in mind there are three possible answers here: zero, infinity, or undefined. I believe zero is wrong, but undefined remains a possibility. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 28, 2003, 9:52pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/49 "2003-05-28T21:52:42Z")
> [@](#): > > \*Originally posted by ZenBeam \* > **Your definition shows that there is a first ball in the limit set, and a second ball, and so forth.** Please explain this to me in great detail. ---
**Author:** ![Bill\_H](https://avatars.discourse-cdn.com/v4/letter/b/a5b964/32.png) [@Bill\_H](https://boards.straightdope.com/u/Bill_H)\ **Post date:** [May 28, 2003, 10:08pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/50 "2003-05-28T22:08:24Z")
So, if removing the highest ball results in an infinite set of balls, and removing the lowest ball results in zero balls, how many balls would remain if a random ball is removed? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 28, 2003, 10:12pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/51 "2003-05-28T22:12:49Z")
All right, let me go through a detailed proof that which balls we take out are important. Assume we’ve got the setup in the OP, save that the gremlin is always removing the lowest-numbered ball whose label is in some set S. As always, every ball is labeled with an element of **N** , and there’s a ball corresponding to every natural number. We’ll also assume that there is no ball labeled 0, as in the OP. I’m also going to be a little sloppy about the distinction between a ball and its label, although I can correct that for interested parties. So we start dropping balls in two at a time, and the gremlin is removing them one at a time. The question is, which balls are left after step n? Well, at most there are the first n elements of S (if S is **N** ), and at worst, there are no elements left (if S has no elements less than n + 1, for instance). The key observation is that once the gremlin removes a ball, it never gets put back in. We can all agree on that, right? There are two cases we need to consider, the case where S is infinite, and the case where S is finite. In the case where S is infinite, let’s look at two sequences, \ and \. \ is the sequence of elements of S, and \ is the sequence of elements of **N** - S. When is s[sub]k[/sub] placed in the jar? Well, at the nth step, we’re putting in the balls 2n - 1 and 2n. If s[sub]k[/sub] is odd, we’re putting it in at step (s[sub]k[/sub] + 1)/2, and if it’s even, we’re putting it in at step s[sub]k[/sub]/2. As the gremlin only removes balls whose labels do not fall in S, s[sub]k[/sub] is never removed. When is t[sub]k[/sub] removed from the jar? This is a little trickier, as that depends on S. So we’ll come up with an upper bound instead. At step t[sub]k[/sub], we have removed at most t[sub]k[/sub] - 1 balls from the jar. Every lower ball whose label is not in S has been removed by now. So t[sub]k[/sub] will be removed on step t[sub]k[/sub], if not earlier. It could be earlier, if S has a member lower than t[sub]k[/sub]. And as we noted before, every ball taken out is never put back in. So by my definition, the sequence of sets of balls in the urn converges to S. If S is finite, things are a little easier. S has a maximum element, [symbol]s[/symbol], which will be put in on step ([symbol]s[/symbol] + 1)/2 if [symbol]s[/symbol] is odd, and step [symbol]s[/symbol]/2 otherwise. After that, all elements of S are in every iteration. As before, t[sub]k[/sub] is removed by step t[sub]k[/sub], and never placed back in. Again, by my definition, the sequence of sets of balls in the urn converges to S. That’s why you can’t just count the number of balls left. You have to know which ones are there. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 28, 2003, 10:52pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/52 "2003-05-28T22:52:25Z")
> [@](#): > > \*Originally posted by Bill H. \* > \*\*So, if removing the highest ball results in an infinite set of balls, and removing the lowest ball results in zero balls, how many balls would remain if a random ball is removed? \*\* Depends on which ones got pulled out. However, the probability that a finite number of balls is left works out to be zero. Why? Well, every probability is a measure, and the measure of a countable set is always zero, and there are a countable number of finite subsets of **N**. ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [May 28, 2003, 10:56pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/53 "2003-05-28T22:56:24Z")
**ultrafilter** - I could be wrong, but based on your conclusion, seems like it should be : “the gremlin is always removing the lowest-numbered ball whose label is [not] in some set S.” (So if S = empty, the urn contains S (i.e. is empty) at the end.) The conclusion still makes sense. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 28, 2003, 10:58pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/54 "2003-05-28T22:58:50Z")
> [@](#): > > \*Originally posted by panamajack \* > \*\ ***ultrafilter** - I could be wrong, but based on your conclusion, seems like it should be : > “the gremlin is always removing the lowest-numbered ball whose label is [not] in some set S.” > > (So if S = empty, the urn contains S (i.e. is empty) at the end.) > > The conclusion still makes sense. \*\* Oops. Good catch. ---
**Author:** ![Bill\_H](https://avatars.discourse-cdn.com/v4/letter/b/a5b964/32.png) [@Bill\_H](https://boards.straightdope.com/u/Bill_H)\ **Post date:** [May 28, 2003, 11:20pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/55 "2003-05-28T23:20:53Z")
**ultrafilter** wrote > [@](#): > > However, the probability that a finite number of balls is left works out to be zero. Why? Well, every probability is a measure, and the measure of a countable set is always zero, and there are a countable number of finite subsets of N. Yes, but.l. What you’re saying is that the probability of pulling out a given ball, say ball #1403 approaches zero. If this is true (and I assume it is), then isn’t that logic true even if you’re pulling out the lower balls? I.e. if you have a big urn with every odd number in it, the odds of actually pulling out a given numbered ball is zero. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 29, 2003, 12:33am UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/56 "2003-05-29T00:33:43Z")
> [@](#): > > \*Originally posted by Bill H. \* > \*\ ***ultrafilter** wrote > > Yes, but.l. What you’re saying is that the probability of pulling out a given ball, say ball #1403 approaches zero. If this is true (and I assume it is), then isn’t that logic true even if you’re pulling out the lower balls? I.e. if you have a big urn with every odd number in it, the odds of actually pulling out a given numbered ball is zero. \*\* That is correct, but it’s not quite relevant. Measure theory gets to be pretty tricky, and you have to rely on proofs rather than intuition. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 29, 2003, 12:36am UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/57 "2003-05-29T00:36:51Z")
Actually, if you have a big urn with all the odd-numbered balls in it, the probability of pulling out a given ball is _undefined_. This stuff gets really nasty, really quickly. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [May 29, 2003, 5:22pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/58 "2003-05-29T17:22:07Z")
> [@](#): > > Originally posted by ZenBeam > > > [@](#): > > > > Your definition shows that there is a first ball in the limit set, and a second ball, and so forth. > > Please explain this to me in great detail. While the sequence of steps is being executed, there is a lowest ball, which is taken to be the “first” ball, there is (after the second step) a second-lowest ball, the “second ball”, and so on. I didn’t want to say that in the limit set, the balls will necessarily be orderable, so I didn’t want to call them “lowest”, “second-lowest”, but they are the balls corresponding to the lowest, second-lowest, etc. balls in the sequence of sets of balls. Taking the balls in order in the sequence of steps isn’t important, it’s just an easy way of enumerating them. **ultrafilter** , I said “It may be that your definition needs to be tightened up a bit.” Here’s a clearer example of why I think this: Suppose I have a sequence of sets, S[sub]n[/sub], n = 1,2,…, each of which consists of two numbers: 1/(n+2) and 1 - 1/(n+2). Your definition of the limit of this sequence of sets gives me the null set, since no number ever appears more than one time. Therefore the number of elements in the limit set is zero. This is highly counterintuitive since both the number of elements and what they should be seems to be very well defined. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 29, 2003, 5:42pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/59 "2003-05-29T17:42:47Z")
> [@](#): > > \*Originally posted by ZenBeam \* > \*\*Please explain this to me in great detail. While the sequence of steps is being executed, there is a lowest ball, which is taken to be the “first” ball, there is (after the second step) a second-lowest ball, the “second ball”, and so on. I didn’t want to say that in the limit set, the balls will necessarily be orderable, so I didn’t want to call them “lowest”, “second-lowest”, but they are the balls corresponding to the lowest, second-lowest, etc. balls in the sequence of sets of balls. Taking the balls in order in the sequence of steps isn’t important, it’s just an easy way of enumerating them.\*\* [/quote] Every ball in the limit set has to be put in at some point. We’re only putting balls in whose labels have natural numbers. Therefore, the limit set is a subset of the natural numbers, which is a well-ordered set (meaning that every non-empty subset has a least member). So if the limit set is non-empty, there definitely is a ball with lowest label. \*\* > [@](#): > > **ultrafilter** , I said “It may be that your definition needs to be tightened up a bit.” Here’s a clearer example of why I think this: Suppose I have a sequence of sets, S[sub]n[/sub], n = 1,2,…, each of which consists of two numbers: 1/(n+2) and 1 - 1/(n+2). Your definition of the limit of this sequence of sets gives me the null set, since no number ever appears more than one time. Therefore the number of elements in the limit set is zero. This is highly counterintuitive since both the number of elements and what they should be seems to be very well defined. \*\* If no number ever appears more than once, you have convergence to the null set. It may not be desirable or intuitive, but that’s what you get. My definition is well-defined and preserves important operations, which is good. Not only that, but **Orbifold** ’s definition, which is _standard_, gives the same result. Remember, when you’re talking to a mathematician, you’re not generally talking to someone who cares whether something behaves intuitively. They only care that its nature is well-defined. Ever heard of the Banach-Tarski paradox? ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [June 5, 2003, 9:45pm UTC](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318/60 "2003-06-05T21:45:07Z")
Sorry for the delay; real life has been intruding. (Although I see you’ve got the 0.999… != 1 folks to keep you busy. 😉 ) Anyway… > [@](#): > > Ever heard of the Banach-Tarski paradox? Yeah, it’s been covered here. I’m not sure of the relevance here, though. I can qualitatively see how that could be done. > [@](#): > > If no number ever appears more than once, you have convergence to the null set. It may not be desirable or intuitive, but that’s what you get. My definition is well-defined and preserves important operations, which is good. Not only that, but Orbifold’s definition, which is standard, gives the same result. To put it in a “balls and Gremlins” form, assume the Gremlin starts with a stick of length unity. At the first step, he divides it into two sticks of lengths 1/3 and 2/3. At the next step he cuts a bit off the shorter piece and adds it to the longer piece so he has two sticks of length 1/4 and 3/4. At the next step, the sticks are 1/5th and 4/5ths, and so forth. If I ask how many sticks he has after an infinite number of steps, and what the total length is, are you saying the answers are no sticks (since we’ve converged to the null set), and zero total length (since there are no sticks)? [Previous page](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318.md?page=2) [Next page](https://boards.straightdope.com/t/balls-and-gremlins-puzzle/175318.md?page=4)