This article is within the scope of WikiProject Philosophy, a collaborative effort to improve the coverage of content related to
philosophy on Wikipedia. If you would like to support the project, please visit the project page, where you can get more details on how you can help, and where you can join the general discussion about philosophy content on Wikipedia.PhilosophyWikipedia:WikiProject PhilosophyTemplate:WikiProject PhilosophyPhilosophy articles
This article needs a History section detailing who first proposed it, historical arguments, etc. I have not yet found any good references on the web, but I remember seeing two or three good articles in the past. —
Loadmaster22:14, 6 November 2006 (UTC)reply
The first reference now is to a different Ross (Alf Ross) and a different paradox (one from deontic logic), see e.g. Understanding Ross's Paradox by AZIZAH AL-HIBRI, The Southwestern Journal of Philosophy Vol. 10, No. 1 (SPRING, 1979), pp. 163-170 This needs to be fixed!
SnoTraveller (
talk)
10:55, 30 September 2021 (UTC)reply
On a sidenote, noon is expected to arrive at 2^(-(n-1)). According to the problem description, n is countably infinite.
Thus, n->infinite. Which means 2^(-(n-1))->0. That is also one point that illustrates that the problem's initial assumptions were problematic, a fact that is intuitively confirmed. (essentially - and if I'm mistaken please post - "how many steps of 9 does it take to count to infinity ?") —Preceding
unsigned comment added by
87.202.27.214 (
talk)
02:44, 7 October 2007 (UTC)reply
The "straightforward arrangement" is flawed. Even for a convergent series if you rearrange terms you can change the nature of convergence. It's even more the case for divergent series. 10-1+10-1+10-1+... diverges to positive infinity, whereas by contrast 1-1+1-1+1-1+... oscillates and is Cesaro summable to 1/2. See
Summation of Grandi's series for various summation methods and the differences between them and their results.
I went ahead and added those two references to the "References" section of the article. They need to be wikified into ref form, though. —
Loadmaster17:19, 22 February 2007 (UTC)reply
The Lamp Paradox
Shouldn't there be a mentioning of the Lamp Paradox? It is basicily the same as the Vase Paradox, it goes at follows: "Let there be a Lamp which can have two states: "on" or "off". At t=1 the lamp is off, t=1/2 the lamp switches on, t=1/4 the lamp switches off and so on. The question posed is: what will the state of the lamp be at t=0?"
Well, the two problems are similar, but they are not the same. The first asks for the final sum of balls based on adding and removing them from the vase, while the second asks what the "final" state of the lamp will be. The first question can be answered assuming that the time t=0 can be defined well enough. The second cannot be answered because even if time t=0 is defined, the answer relies on knowing the "last" step prior to t=0, when the lamp was switched on or off for the "last" time. The first problem needs no "last" step to be answered. It would be nice to have a "Lamp Paradox" article that could be added to the "See also" section, though. —
Loadmaster00:31, 6 March 2007 (UTC)reply
I found the article, named Thomson's lamp, for the paradox you describe. I've added it to the "See also" section appropriately. And I added a redirect for Lamp paradox as well. —
Loadmaster
I think this problem simply shows that you have to be careful about your terms - in particular, when talking about limits, you have to define your
topology first.
The problem can be phrased like this: We have a function f from [0,1] to some state space of the vase, let's call it X. We have defined f on a sequence converging to 1 and would like to know the continuous continuation.
If we let X be the
one-point compactification of the real numbers and f(1-2^-n)=9n, then the limit is infinity.
On the other hand, if we let X be (each component representing the presence or absence of an individual marble) with the
product topology or in other words the topology of pointwise convergence, and f(1)=(1,1,...,1,0,...) (10 ones, then zeroes) and f(2)=(0,1,1,...,1,0,...) (0, then 19 ones, then zeroes) and so on, then the limit is the zero sequence, since every component pointwisely converges to zero.
You see, it's just a matter of proper definitions. If there is a notable source for this argument, I strongly suggest adding it to the article.
Functor salad04:10, 16 October 2007 (UTC)reply
Benardete's Paradox
Another variation is
Benardete's paradox, said to be an extension of
Zeno's Paradox, which is described at
thinkquest.org. Briefly, the god
Zeus decides to kill
Prometheus by issuing an order to an infinitude of his demons, ordering the first demon to kill Prometheus if he is not dead by 2:00, the second to kill him if he is not dead at 1:30, the third to kill him if he is not dead at 1:15, etc., each demon killing Prometheus in half the time interval of the subsequent demon. The strangeness comes in the realization that Prometheus is indeed dead by 2:00, but yet no specific demon killed him. —
Loadmaster (
talk)
04:11, 30 March 2008 (UTC)reply
Fake question
Just like that infamous lamp, this question is a mathematical/philosophical/logical question masquerading as a physical one. Obviously when you're talking infinities, "balls" and "vases" are just placeholder concepts: the number of physical operations that can be performed in a finite time is finite. Casting this problem in physical terms is not valid. A more honest question is "what is the limit of the infinite sum "10 - 1 + 10 -1 ..."; this question can be mathematically discussed. A question of balls and vases falls at the basic hurdle of making physical sense. --
Slashme (
talk)
15:33, 27 May 2008 (UTC)reply
Actually, some proponents of
hypercomputation are hoping that at some point in the future there will be actual, physical computers performing infinitely many operations (in sequence) in a finite amount of time: a physical Zeno machine. —Preceding
unsigned comment added by
128.113.89.96 (
talk)
15:21, 28 May 2008 (UTC)reply
Well, when that time comes about, it'll be interesting to see what result comes out. From Wikiquote:
We all know Linux is great…it does infinite loops in 5 seconds.
-- Linus Torvalds about the superiority of Linux on the Amsterdam Linux Symposium
The Original Research tag is debatable. Set theorists are clearly in agreement that the first answer (zero balls left in the vase) is correct. The other solutions are variations on the conditions and "meaning" of the problem. —
Loadmaster (
talk)
02:25, 17 August 2009 (UTC)reply
The problem can be stated as <br\><br\>
<br\><br\>
The philosophers assuming zero as answer are getting that because they are implying that the two infinities are same. Their logic is: <br\>
<br\>
and <br\>
Hence the two infinities are assumed same! Which obviously aren't!!
People with other answers or variable answer based on which ball is removed are getting there because they are cancelling out unequal infinities somewhere in the process. — Preceding
unsigned comment added by
VinnieCool (
talk •
contribs)
01:34, 10 June 2012 (UTC)reply
So, VinnieCool, some infinities are more equal than other infinities?
I think it logically follows that IF they are UNequal, then one of them can not be INfinite.(edited from "finite"06:46)
Let adding and subtraction of balls take place simultaneus and assume end of operations IS possible (no supertask): then zero balls remain in vase because an infinite number has been removed
Sintermerte (
talk)
06:40, 22 February 2024 (UTC)reply
Section needs rewriting
I think the section on the "no balls left in vase" solution needs to be rewritten. It really doesn't make sense on face-- of course there is always a step n at which ball n is removed, but it's just as obvious that at at every step n, nine more balls are added. I'm not trying to argue against inclusion of this section; I just think it needs to be rewritten so that it at least makes some small amount of logical sense on face... I hope this makes sense. Thanks — Preceding
unsigned comment added by
70.36.226.136 (
talk)
21:50, 1 August 2013 (UTC)reply
Indistinguishability and probabalistic behavior
I did a quick google search, and I can find no mention of the distinguishability of balls. In physics, the statistics of particles depends on whether the particles are distinguishable or not, i.e. does swapping ball 1 and ball 2 change the state of the system in some way. It seems in this case that the balls are being counted, but with indistinguishable balls, there is no ordering which is unique. It makes no sense to say "I removed the first ball, then the second, then..." since there is no ball #1, 2 or 3. In an indistinguishable case, you can only specify the states of the balls probabalistically. After the first step, there are 10 balls in there, each with a 1/10 probability of being removed. After this, I believe the math gets hairy, since you can't remove a ball you already removed, and thus you need to calculate the probability of each ball being removed given that it was not removed in any previous step. Then you have to do a sum to find the number of balls left
47.21.153.202 (
talk)
15:31, 9 August 2013 (UTC)reply
I'm not sure, but you seem to be confusing physical reality with abstract mathematics. It's an abstract mathematical/logical gedankenexperiment, wherein we use an infinite number of abstract "balls". We assume each ball is uniquely labeled (e.g., with the natural number reflecting the order in which it is inserted into the vase). Such labeling allows us to keep track of which balls were inserted and which were removed at any given moment during the procedure. Perhaps we need to add some verbiage making it clear that there is a labeling employed, or some other means used to uniquely distinguish between different balls. —
Loadmaster (
talk)
23:01, 9 August 2013 (UTC)reply
I think my suggestion is that, when dealing with inifinites within the context of this problem, the ability to uniquely number the balls is broken. I suggest this due to what I percieve as a kind of "reductio ad absurdium", wherein changing which balls we remove, but not how many, changes how many balls are left in the vase. This challenges that we can uniquely number the balls we put into the vase, which seems to be a major premise of the proposed solutions to the paradox. I suppose my new premise, due to the statement of the problem not explicitly defining how we remove balls from this vase, is that they are removed at random, and that this may lead to a non-self-contradictory solution to the problem. Then again, it may just push the problem off on to how the probabilities are calculated, instead of which ball is removed. My parallel to physics is due to my personal training in physics and total lack of formal or significant informal education in the mathematics of infinites outside the scope of physics, which I can completely understand as significantly hampering my ability to comment on this problem.
47.21.153.202 (
talk)
18:15, 12 August 2013 (UTC)reply
Your suggestion sounds a lot like the assertion that the
Axiom of Choice is required for the infinite case, which is by no means a principle that all mathematicians take for granted. That's actually a very helpful observation for me, but seems a bit out of scope, as it is original research of the solution.
70.247.167.104 (
talk)
13:59, 11 February 2014 (UTC)reply
Paradox Solved
Really I think this paradox should be reported as "solved". I was working to a solution that I unfortunately found to be already published 13 years ago by John Byl
On Resolving The Littlewood-Ross Paradox on Missouri Journal of Mathematical Sciences Vol.12 (No. 1, Winter 2000): 42-47. At the end the urn contains an infinite number of balls (9ω), just numbering balls is some misleading.
Nerd4j (
talk)
09:49, 4 November 2013 (UTC)reply
Well, that's the whole problem with the paradox; there is no single solution accepted by everyone. Based on what I read, most mathematicians (logicians and number theorists) accept that, as the problem is usually stated, if each ball n is placed into the urn at some time tn and then removed at a later time t10n (or some similar moment), then after ω steps, every ball has been removed at some point after it was inserted, so there must be no balls left in the urn. Far from being misleading, the numbering of the balls, and more importantly the relative order in which the balls are inserted and removed from the vase, is the crucial point of the whole exercise.
If, however, we accept that there are an infinite number of balls left in the vase, we must ask which balls are left, i.e., what are their numeric labels? If we uniquely label each ball as it is inserted, and there are indeed 9ω balls left in the vase, how do we explain the labels on the balls remaining in the vase following the removals of the first 1ω balls?
In any case, before this turns into a lengthy and inappropriate discussion of the problem, be aware that the article does in fact explain the different interpretations of the problem and its possible resolutions. From that it should be obvious that we can't declare one solution as "the" solution here on WP. —
Loadmaster (
talk)
20:20, 4 November 2013 (UTC)reply
Which is reasonable. Just because a published article has no printed objections in no way implies that its premise is accepted by the mathematics community at large (it implies only that it was deemed suitable for publication at the time). I'll reiterate: the different interpretations of the problem are covered in the article. I'll also reiterate this: the order in which balls are inserted and removed really is the crux of the entire problem; simply declaring that it does not matter ignores the subtle nature of the problem entirely. —
Loadmaster (
talk)
21:55, 18 November 2013 (UTC)reply
You are both right about the pubblications, I don't think the Missouri Journal of Mathematical Sciences is peer-reviewed too, and actually I don't know which is the average competence level in the Central Missouri State University, but in my university each article is reviewed and discussed by doctoral candidates and professors before to be published and I assume that every university follows more or less the same behavior (I may be wrong). When I first read the Ross-Littlewood Paradox it was clear to me that the urn is always full because it follows trivially from the definitions of 'first countable' and 'limit point'. I was going to write an article to argue the overlapping of â„• as 'definition of countable set' and â„• as 'order topology' that leads to the misunderstanding. In fact saying that for each x in â„•, x was taken at instant x is not contradicting the fact that the urn is still full, therefore there is no paradox (that's my personal point of view). Before to start writing the article I searched for some previous work and I found Byl's article that argues the same point of view using the same arguments. So my work wasn't necessary and I didn't even start with it.
Nerd4j (
talk)
11:46, 19 November 2013 (UTC)reply
Responding to Loadmaster: the reductio ad absurdum is possible only if we define a labeling for the balls such that each ball has a unique label, so we are able to identify each ball in the urn, the problem is given by the assumption that all the values used to label the balls belong to â„•. My argument (and Byl's) is that this assumption is wrong because is contraddicting the definition of 'first countable' and 'limit point'. I know that it's a minority point of view but due to the fact that it leads to a consistent solution of the problem by simply applying well known definitions and concepts I think that it's deserving a further discussion.
Nerd4j (
talk)
11:46, 19 November 2013 (UTC)reply
Not here for the time being, I'm afraid. Please note that solving a paradox requires not only to propose a solution, but also showing how the contradicting approaches are wrong. You might want to find other places where this kind of discussion is appropriate. If you
WP:EMAIL me, I'll gladly explain the mistakes in Byl's paper. Regards,
Paradoctor (
talk)
15:01, 19 November 2013 (UTC)reply
In the introduction, it is never said how many balls are added or removed at each step. when reading the solution part, we have first "since at every step along the way more balls are being added than removed" but it was never assumed before. that is just a problem of formulation because after it seems obvious that 10 balls are added and 1 is removed at each step. but in the introduction, it seems that the problem is more general, and that any number of balls could be added and removed.
150.65.16.33 (
talk)
12:27, 25 June 2014 (UTC)reply
Yes, the article at present is a complete (erroneous) mess. The way Littlewood stated it, the answer is quite clear (not that this prevents crank nonsense, which WP seems to require equal attention to); the problem as stated in the article at present has no solution. (But I can't completely understand your comment either...) The article should be rewritten, starting by quoting the problem as posed by Littlewood... I have a copy of the Miscellany somewhere, and keep meaning to do it.
Imaginatorium (
talk)
15:31, 25 June 2014 (UTC)reply
Oh, I realized the text already contained this sentence in a different place. I removed the original sentence, so the effect of my edit is moving it to a place which makes the text more coherent.
MaigoAkisame (
talk)
15:23, 19 April 2017 (UTC)reply
The problem description adds the first balls at 11:59:30, while the figure adds the first balls at 11:59:00. The figure matches the description in Ross, so I think the problem description should be changed. Additionally, the problem structure is perhaps better revealed by this chart:
Illustration of the Ross-Littlewood problem description (following Ross).
The "Vase is empty" solution uses this logic: "This means that by noon, every ball labeled n that is inserted into the vase is eventually removed in a subsequent step."
Now consider the question, "at what time, or in which step, is the last ball removed?" The answer is there is no last ball. But I don't think this question can go in that section, because it's not part of the logic leading to this solution. Where could it go?
23.121.191.18 (
talk)
06:09, 28 January 2020 (UTC)reply
Ross deserves credit ... but NOT for coming up with this paradox
Ross, of course, deserves credit for whatever work he did on the problem.
But: The paradox was invented by Littlewood, not Ross. So Ross does not merit having his name attached to the paradox for developing it further, 35 years later!
I propose either moving "depends on the conditions" under "vase is empty" or to a new section about "variants of the problem", since it follows the same logic as the "vase is empty" solution except with similar problems
C7XWiki (
talk)
05:56, 27 March 2022 (UTC)reply