Urns of Infinity

[Griffin]

Urns of Infinity

First, I'd like to complement Minotaur on the nice looking graphic that he made. But I also want to point out a mistake. I think "But Demon #1 was lazy" was meant to be "But Demon #2 was lazy."

Anyway, here's my take on it:

Demon #1's urn has a infinite number of balls. Had Demon #2 followed directions, his urn would have been empty by 12:00. Because he did not, Demon #2 by 12:00 has a urn containing an infinite amount of balls labeled 1000..., 2000..., 3000..., etc., each with an infinite amount of zeros (perhaps each zero is half the size of the previous one).

Though any time before 12:00, there is no way to tell that Demon #2 cheated. I think the question is how and when did the two scenarios become different.

[Wonko the Sane]

The wording on this puzzle is ambiguous at best. I would request clarification on a couple passages from the Minotaur.

"Instead, at one minute to midnight, he placed balls 1-9 into the urn. Then, he quickly painted a zero on the first ball, making it look like ball ten..."
Okay, problem, ball 1 is already in the urn. So why'd he put it in in the first place?
"At a half-minute to midnight, he placed balls 11-19 into the urn, and painted a zero on the ball numbered "2" turning it into a 20."
Great...but ball 2 is in the urn

So let's assume that he takes ball 1 and ball 2 out of the urn when he converts them. Okay, that works, but yes, the sorcerer will have no problem exposing him.
If the demon did this an infinite amount of times, theoretically, by the time he used up the infinite number of balls, the urn would be empty because he pulled out all of the balls in order.
However, if he had followed the sorcerers orders, at midnight, the urn would have an infinite number of balls in it. Therfore the sorcerer simply has to walk in and look at the empty urn to know that demon #1 cheated.

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It's not the size of the spork, it's whether or not it's made of #7 recyclable plastic.

[worm]

man, that makes me feel a lot better that you think there is a mistake, too. i was spinning with the wording the way it is, thinking:

"if #1 really wants to cheat to prevent extra lifting, why the hell doesn't he just put 1-9 in and leave 10 out?" (i.e. don't put 10 in to begin with)

i haven't considered the problem much in the new context you've provided, but it sounds like the problem that was intended.

[Mercuria]

i'm not understanding the cheating part at all. according to the description, the cheating demon didn't put the balls divisible by ten into its urn in the first place, then converted the balls, in order, to balls ten times their value. if it was demon 1, then the first half would have finished his cheating for him. if it was the second demon, then there would be no reason for him to add the 0 at the end of the number. i think it is the first demon who cheated... though i don't understand why he had two of each ball out of the urn...

btw, if this doesn't make any sense, i'm sorry... lack of sleep...

[hank]

I'm puzzled by the logic of the cheating method for demon #1 (I agree with Mercuria, that demon #1 was the intended cheater.). It seems to me that the easier method is to skip balls 10, 20, 30, etc. from the operation altogether and leave them on the table. In this case however, at any given moment at the end of each cycle, including midnight, demon #1's urn will be 11.111% lighter. The sorcerer should be able to pick up on that. Is 89% of infinity still infinity? Actually, there can't be an infinite number of balls as there was a limitation imposed. What is an iotasecond anyway?

[This message has been edited by hank (edited 05-07-2000).]

[shoobbie]

just one simple question:

quote from the puzzle: "Then, he quickly painted a zero on the first ball, making it look like ball ten, WHICH he shoved under the rug. "

what does the "which" refer to (ball 10 or the first ball)?

[hank]

Ya, I pondered that one too, but decided that the purpose of the painting was deception which would have been irrelevant had the painted ball been hidden under the carpet. Perhaps there is a grammartician in the crowd?

[Wonko the Sane]

Here's what I'm getting out of this.

He puts 1-9 in the urn and ditches the real 10 under the rug, then he adds ball 1 to his pile of things he pulled out (it now looks like a 10). He adds 11-19 then pulls ball 2 out of the urn, hides the real 20, then puts the fake in the pile of removed balls. He continues until he reaches 90. Now we have a problem. He sticks 91-99 in to the urn, but then he needs the real 10 for the next part, right? So what's he do? Add the real 100 to his pile, or pull 10 out from under the rug and add a 0 to the real 10? I'm guessing the latter of those options. Therefore, at the end of it, he'll end up actually having no balls in the urn (for 110, he'll pull 11 out of the urn and everything progresses regularly. This problem i pointed out is only a problem at factors of 10). So, an empty urn is a good indication that he cheated since in the real case, he should have been left with an infinite number of balls. That's my spin.

[ZenBeam]

I think there needs to clarification on what the sorceror can observe to see if the demon has cheated. Does he look at which balls are left outside the urn (apart from the ones under the rug), or does he look at the ones in the urn. If it's really demon #1 who is cheating, and the sorceror looks into the urn he would see there is no #1 ball. If instead, the sorceror looks at the ones left outside the urn (but not under the rug), there is no #10 ball, because it gets turned into #100 after the demon puts in #s 91-99.

If instead, it is demon #2 who cheats, and the sorceror looks at the balls in the urn, there won't be a #10 in the urn, because it got turned into #100 after the demon put in balls 91-99. Finally, if it's demon #2 who cheats, and the sorceror looks at the balls outside of the urn (but not under the rug), the only balls out of the run are the ones under the rug, so this case doesn't make sense (the sorceror could tell because there are no balls out of the urn, but there should be).

Well, regardless, the sorceror can tell the demon cheated.


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It is so clear, and so it is hard to see.

[hank]

I find it interesting that seveal of you project empty urns, based on cheating. I just can't see it. Every ball removal in any case, demon #1 or #2 whether cheating or not, is only warranted following the introduction of 9 or 10 balls in each cycle. There may be an infinite number of balls under the carpet, which would require an infinitely large carpet in an infinitely large room in an infinitely large sorcerer's den etc. etc etc. But there would still be 8.1 times the number of balls in the urn, which would have a density greater than a black hole, i.e. , an infinite density. As I am rambling on here, I realize the sorcerer won't be able to tell anything, as the entire village including the sorcerer will implode on the urn and destroy civilization in the process.

[Wonko the Sane]

Although he's adding balls at a faster rate than he's taking them out, it doesn't matter at infinite, does it? Because he adds and removes an infinte number of balls. The balls that he removes correspond to the set of natural numbers, both which are of size aleph-null. Therefore the demon will remove every ball he adds by the time he hits the limit of infinite.
However, I think this is what the puzzle is about. Yeah, now I've got it. Okay, listen up. The question is this. If he both removes and adds an infinite number of balls, but he adds balls at a faster rate than he removes them, when you hit the limit, are there 0 balls left in the urn, or infinite? That's the puzzle.

[Wonko the Sane]

Oh yeah, one more problem in the wording of the puzzle, though it doesn't really effect the puzzle much. Since demon #1 has to pull all of the balls that he hid under the carpet out to paint another zero on them in the future, when the sorcerer gets there, there wont be any balls under the rug. At each factor of 10, he hides ball n, then he pulls the ball n/10 out from under the rug and paints an extra 0 on it, then puts it in the pile of balls that were removed. So in the end, at least in my interpretation, there will be no balls under the rug.
Actually, even under the other interpretation there shouldn't be. Here's the two sides of the puzzle, at least from what I can gather:
1) Even though he adds balls at a faster rate than he pulls them out, since the two infinites are the same size, he'll remove all the balls from the urn, and remove all of the fakes (since he has to remove ball n from under the rug when he reaches ball 10n.
2) Since he adds balls at a faster rate than he adds them, there will still be an infinite number of balls in the urn at the end. Also, balls n*10^aleph sub-null (for all n from 1 to 9) will still be under the rug (but not an infinite number as the puzzle states) because there corresponding balls at n*10^alpeh sub-null * 10 - k for all n and k from 1 to 9 are still in the urn. If that last bit didn't make sense, don't worry about it, because the actual mathematics determining the values of the balls is meaningless in terms of the infinite. It's just to clarify why there are still 9 balls under the rug at the end under this interpretation.

Personally, I have to agree with interpretation #1. If there are more interpretations that I missed (and I'm sure there are) please post them.

[ZenBeam]

Apparently there is a meta-puzzle, which is "What, exactly, is the puzzle?" Perhaps we need to solve this before we can solve it.

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It is so clear, and so it is hard to see.

[worm]

here's why i think demon #2 is supposed to be the cheat and not #1

the cheating demon puts 1-9 in and paints 1 to look like 10. so it looks like balls 2-10 are in the urn after the first step, which is exactly what the urn should look like after demon #2 is done with the first step. after the second step of the cheat, 3-20 are in the urn, which is exactly the way the urn should look after demon #2 is done with the second step. the two methods (the cheat's method and what demon #2 is supposed to do) give the same balls in the urn: 4-30 (3rd step), 5-40 (4th), 6-50 (5th), 7-60 (6th), 8-70 (7th), 9-80 (8th), 10-90 (9th), 11-100 (10th), etc.

if the question is written correctly, then yes, the sorceror will be able to see that he's been cheated.

if it's not written correctly, i don't think there's any way he can tell.

[Rhino]

I think you're right and wrong, worm. I agree that the puzzle should have demon #2 as the cheat, but that the sorceror might be able to tell. I don't mean to criticize, but you should have thought more about Griffin's original answer.

The way demon #2 is supposed to do the task is add 10 balls, remove a ball, add 10 balls, remove a ball, ad infinitum. So he is adding an infinite number of balls and removing an infinite number of balls. When demon #2 cheats, he adds 9 balls, adds 9 balls, ad infinitum. So he adding an infinite number without removing any.

The paradox is how can "adding 10 balls and removing 1" and "adding 9 and leaving the 10th out to begin with" not give you the same end result?

Hank has a valid gripe. Infinitely adding 10 and removing 1 seems like it would leave the urn with an infinite number of balls. It seems like the removing step would never catch up. But is there a ball number that would never be removed? I can't come up with one. Is the urn empty after midnight? Somebody who has a better grasp of infinity can try answering that.

But infinitely adding 9 without removing any would give you an infinite number of balls in the urn.

[shoobie]

i don't understand the idea about the urn being left empty at the end. I just don't see how that could happen. the fact that the deamon adds 10 and removes one, or adds 9 doesn't matter, even in infinitum:

let x be number of repetitions:

lim (10x-x) = lim (9x) = infinite
when x->infinite


what bugs me, is the question: how could the deamon take something out of the urn anyway? if the balls are of finite size, and the urn is infinite, then how could you find the balls in the urn once you put them in there? (could the answer be of course that they are deamons? but if so than the sorcerer must also be able to see inside the urn)

[HyToFry]

This puzzle also fails to mention what the demons were supposed to do with the balls that they removed. If they were supposed to stack them in a neat little infinate pile, then yes D2 has some splaining to do, because he will have no pile.

I agree too that D2 would have to be the cheater, this must be an oversite of minotaur???

[HyToFry]

One thing I like about this puzzle it that it shows that there have to be different levels of infinaty, or maybe not.

Let x=infinity
Demon 1
10x-1x
Demon 2
9x

Now 10x = x (if x = inf) and x = x

so would not 10x-x=0?

so demon one's urn would be empty, while demon 2 has a full urn.

this is not possible though becuase D1 would never remove numbers 1-9 or 11-19 etc...

so this leads me to belive that there are in fact different amounts of infinity.

hehe, i had 9x-1, had to change it


[This message has been edited by HyToFry (edited 05-08-2000).]

[Wonko the Sane]

But demon 1 does remove balls 2,3,4,etc. That's the point of this puzzle

You see, if demon 1 followed his orders, then 1-9,11-19 and so on would be in the urn at the end. The point is, he never adds ball 10 or ball 20. He paints a 0 on ball one and sets it in the pile, then hides ball 10. He pulls out ball 2, then paints a zero on it, buts it in a pile, then hides ball 20. When he gets to ball 100, he pulls 10 out of it's hiding place, adds a 0, puts it in the pile, then hides 100. You see what I'm saying? He removes all of the balls, in order.
The only problem with this puzzle is that even mentioning that there's a demon #2 just mucks up the puzzle. The only reason to mention demon #2 is to illustrate that essentially, demon #1 is doing the same thing as demon #2 in the long run. The difference will be instead of having balls 1-infinite in his pile at the end as demon #2 will, demon #1 will have balls 10-infinite in his pile of discarded balls.
Now, that isn't what allows the sorcere to know about the cheating, because that's what demon #1 would have ended up with anyway. The puzzle, as I stated before, is this. Will demon #1's urn be empty, or have an infinite number of balls in it at midnight?
I say it'll be empty. He adds and infinite number of balls, and he removes and infinite number of balls. Both infinites are the same size (aleph-null), therefore even though he adds at a faster rate than he removes the balls, he'll remove them all by midnight. However, we can't prove that. That's why this is a paradox. If you had read what the Minotaur said, this puzzle is a semi-paradox. Mathamatics denotes that the urn will be empty. However intuition says that the urn will have an infinite number of balls in it. That's why it's a paradox...sort of. Anyone getting this yet, or do you want me to try and explain it more simply?

Also, HyToFry, these couple messages are for you. By your logic, demon #2's urn will be empty. Read over what you wrote, you should see why. Also, there are different sizes of infinite, however, the one's that we deal with here are the same size.
The two infinites that are simplest to explain are aleph-null and aleph-c.
Aleph-null is the size of the infinite of natural numbers, integers, and rational numbers. The reason is because there is a one to one correspondance between them. Let me show you. (N = natural numbers, Z = integers, Q = rational numbers for those who don't know)
N 0 1 2 3 4
Z 0 -1 1 -2 2
Q 0 1/2 -1/2 1/3 -1/3

Although Z grows faster than N and R grows faster than both Z and N, you can line up the numbers 1/1. The size of these infinites is called aleph-null.
You cannot line up the real numbers like this because every decimal place can contain all of the real numbers. Then when you put that whole set together, you have another number that contains all of the first digits. Then another with the diagonal line. And so on. Theoretically, the set of real numbers is larger than the set of naturals, integers, or rationals. It's size is called aleph-c.
Anyway, not exactly going to help you answer the puzzle, but I hope that helps you understand the infinite a bit better HyToFry.

[This message has been edited by Wonko the Sane (edited 05-08-2000).]

[Wonko the Sane]

I also want to respond to Rhino's post, but I'm going to do it in a new sub-post for emphasis.
You ask how adding 10 balls and removing 1 give a different result from adding 9 and leaving one out?
The reason this is different is because demon #1 is removing a ball each time. It's only the first time that he adds 9 and leaves one out. After that, he adds 9, then takes one out. Here's what demon #1's urn will look like for the first few seconds. Balls such as 1(0) and 2(0) are balls that he painted zeros on.

Balls in the Urn | Balls on the floor | Balls under the rug
-----------------------|-------------------------|---------------------------|
2-9 | 1(0) | 10 |
3-9, 11-19 | 1(0),2(0) | 10,20 |
4-9,11-19,21-30 | 1(0),2(0),3(0) | 10,20,30 |

Now, here's when we hit factors of 100, things get a bit goofy, but it's not too bad. Here it is.

11-99 excluding | 1(0),2(0),3(0),4(0) | 20,30,40,50,60,70 |
multiples of 10 | 5(0),6(0),7(0),8(0) | 80,90 |
| 9(0),10(0) | |

See? At 100, he has to pull ball 10 out from under the rug, paint a 0 on it, then add it to his pile of balls he removed. You noticed though that at 100, every ball 1-10 has been removed from the urn? If demon 1 had what the sorcerer said, it would have looked like this.

1-99 excluding | 10,20,30,40,50,60 | none
multples of 10 | 70,80,90,100 |

So, instead of removing just multiples of 10, the demon is removing every ball, then hiding multiples of 10. Essentially, he takes ball n, paints a 0 on it, and adds it to the pile when he hides ball n*10. If ball n is in hiding, then he simply takes it out of hiding.

This thing is pretty much just a big punchline. See, demon #1 wasn't being bright, because when he finishes what he's doing the way he's cheating, he'll end up exactly where he started (all balls out of the urn). The wizard can tell he cheated because he was supposed to have an infinite number of balls in the urn and an infinite number of balls out of the urn. Demon #2 gains a lot more by cheating. If he wanted to cheat effectively, he should have just walked away, because he'll be right back where he started at midnight anyway.

[CrystyB]

WONKO, will you understand that the 2nd daemon is cheating? He paints 1 to 1(0), then 2 to 2(0), ..., then 9 to 9(0) and THEN 1(0) to 1(00) !!!

[Tom]

I'm glad you wrote that, CrystyB, I didn't want to have to check Wonko was wrong. The demon takes 10 from the urn (ie 1 with a 0 painted on, not the one under the carpet), and paints on a 0, etc. etc.

We are all agreed that the Minotaur made a mistake, yeah?

So .. I (as usual) agree with Griffin completely .. though isn't the point that the sorcerer knows after that #2 was cheating, as he has stuff in his urn after 12:00, when he shouldn't. Though what number is painted on the balls is a bit of a hard question .. I think that the demon in the second case is doing something ``impossible'' (ie painting an infinitely big number on a small ball) .. but the question states he is successful, so I guess it's ok.

HyToFry - this question has nothing to do with different sizes of infinity (we only use 1 here) but just demonstrates subtraction with infinities isn't defined .. not a great deal different than the fact that division by zero is not defined.

x + infinty = infinity, so what is x? Dunno.
x * 0 = 0, so what is x? Dunno.

[This message has been edited by Tom (edited 05-08-2000).]

[Ghost Post]

I was relieved to see other people also concluded that it must be Demon 2 that was cheating. I couldn't figure out any way that what Demon 1 could do what was described and have it even superficially mimic what he's supposed to be doing. Anyway, I'm assuming that D2 is the cheater.

I agree with the conclusion that D2's urn should be empty at the end. Any ball you pick will have a finite (although usually very large) number written on it. If D2 follows the rules this ball will get removed at step n, (n being the number on the ball). So any ball you care to name will be removed after a finite number of steps, and they all will be removed eventually.

Since D2 doesn't remove balls but instead changes ball n to ball 10*n at step n, what is left in his jar is ... very strange. Any finite numbered ball should have been removed, so no ball, in the end, can have a finite number written on it. There must be a countable infinity of balls each with a number followed by a countable infinity of zeroes. Does anybody know whether this can actually be achieved in a countably infinite number of steps, or has D2's attempt to save labor actually resulted in an un-completable task?

My brain hurts.

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"Space is blue, and birds fly through it."
Werner Heisenberg

[Ghost Post]

Gahh. I take so long to marshall my thoughts that in the meantime Tom says everything I plan to, only more clearly.

However, my brain still hurts.

[notlob]

I just want to see if I understood correctly:
is Wonko the Sane arguing that

lim(10x - x)=0 ?
x->infinite

[HyToFry]

Makes sense to me..

but the x->infinate is confuseing... i would have just put x=infinate.

maybe i don't know what i'm talking about though

[Mercuria]

acutally it has to be x->* and not x=* because that's how limits are written. (i'm hoping that * shows up in the forum, but, if it doesn't, then it says infinity)

[Mercuria]

well, that's disappointing...
btw, it's not that
lim (10x-x) = 0
x->infinity

it's more of infinity - infinity = 0

[Wonko the Sane]

I'm not saying that the limit of 10x-x is 0. However I can see that nobody is understanding what I'm saying. I'm sticking by my solution until the Minotaur makes an official call. However, I still say he's right. If it is demon two, then the sorcerer will be unable to tell that he cheated, because his urn will be empty in the end as well.

[Ghost Post]

I think that the order that things happen is important.

If you put balls 1-10 in the urn and then take out ball 1, then put 11-20 and take out 2, etc you wind up with an empty urn. At time n, you have put balls 1-10n in and removed 1-n. For any given ball you can find a time when that ball was removed. Thus the urn must be empty at the end.

If you put all the balls in one at a time first and then take out 1, then 2, then 3. It is different because you will never finish putting the balls in the urn in the first place. At time n, you have put in balls 1-n and not yet removed any.

So when he just renumbers a ball already in the urn, he is never removing any balls at all and the urn is full.


I think an easier way for the sorcerer to notice the cheating is to see that his infinite bucket of paint is empty.


[This message has been edited by Drew (edited 05-08-2000).]

[HappyMutant]

I may be wrong, but I still don't get the argument for the urn being empty at the end.
I understand that (10*inf)-(inf) = 0, but couldn't you say 9*inf = inf and have the same thing? Feel free to correct me if I'm wrong.
Yeah, right. "if." lol

to compare the instructions of devil 1 and devil 2...
D1 has infinitely many balls left in his urn. (1-9, 11-19, etc.) But D1 and D2 add the same # of balls and take out the same # balls per step.



[This message has been edited by HappyMutant (edited 05-08-2000).]

[Rhino]

Yes, I think that part of the paradox is that D1 and D2 are supposed to add and remove the same number of balls at each step and yet come up with different results.

But when D1 removes a ball, it only affects the set he's working with. For the first step, he removes the 10 ball, which is a part of 1-10, second step 20 ball, which is part of 11-20. i.e. he never touches 1-9 or 11-19

The method D2 is supposed to use removes balls from previous sets. 1 through infinity should be removed when he's done.

[Griffin]

How about Demon #2 does this instead:

One minute before 12:00, he puts in 1-10. Then 1/2 minute before, he puts in 11-20, then 21-30, etc. Then, after he's done, at one minute before 12:30, he removes ball 1, the 1/2 minute before, he removes ball 2, then ball 3, etc.

It's easier to see that the urn will eventually be empty in this scenario. But it is basically the same thing as what Demon #2 was suppose to do, the tasks are just separated.

[This message has been edited by Griffin (edited 05-09-2000).]

[Tom]

Erm .. I don't know if this helps, but the limit problem comes up because everyone is assuming that the no. of balls as a function of time is continuous, which it isn't. The no of balls goes up and up and up, then pings down to zero at 12:00. This may seem ridiculous, but around 12:00 the demon is doing an infinite number of step in any finite amount of time, so all the balls can be removed in "no time", as it were.

The graph looks a bit like this (great drawing )



the blob is balls=0 at 12:00. So the limit as x->infinty has nothng to do with the limit as t->12:00 (well, what they have in common is neither exist)

[Rhino]

I love that graph!!! Your and Grif's explanations have been very helpful in understanding the problem.

[Tom]

Hey, y'know, you do your best (blush)

[Trenin]

I think that the problem is slightly misleading, but the gist is correct.

If demon #1 (hereafter refered to as demon) had followed instructions, then the set of balls in the urn would be;
U = { 1-9, 11-19, 21-29, ...} = { (i), (i+1), ..., (i+8) for all i= 1, 2, ... }

The set of balls not in the urn would be the complement of U or;
F = { 10, 20, 30, ...} = { (10*i) for all i=1, 2, ... }

However, the demon did not follow instructions. Instead, at each iteration the demon will add 9 of the 10 balls under consideration to the urn and leave the 10th one out. Approximately 90% of the time, the demon will also remove a ball from the urn and paint it.

Lets say that the balls not added to the urn are placed under the rug and the painted balls are placed on the rug. I assume this is what is happening because the demon is trying to hide the fact that he is cheating, so he hides the real balls under the rug. (If he hid the painted ones under the rug, why did he have to paint them in the first place?)

For example, on the first iteration, balls 1-9 are added, ball 10 is left out (under the rug), and ball 1 is removed and painted (on the rug).
On the second iteration, balls 11-19 are added, ball 20 is left out (under the rug), and ball 2 is removed and painted (on the rug).
...
On the tenth iteration, balls 91-99 are added, ball 100 is left out (under the rug), and ball 10 - which is under the rug - is taken out, painted, and placed on the floor.
...
On the hundredth iteration, balls 991-999 are added, ball 1000 is left out (under the rug), and ball 100 - which is under the rug - is taken out, painted, and placed on the floor.

So, the set balls on the rug is the following; (the numbers are the original numbers, not the painted numbers)
F = {1, 2, 3, ...} = { (i) for all i = 1, 2, ... }

But this is exactly the same as the original set of balls!!!
B = {1, 2, 3, ...} = { (i) for all i = 1, 2, ... }

Therefore, there all the balls are on the floor!

To see this a different way, try to pick a ball in the urn. Any numbered ball you pick from the urn will _eventually_ be taken out, painted, and placed on the floor, or it was left out to begin with and placed under the rug. If it is placed under the rug, then, eventually, it will be taken out, painted, and placed on the floor.

I think most people are having trouble with the infinity - infinity = 0 statement made earlier. It does not make sense to subtract infinity from infinity. Think rather of the set of balls in the urn, the set of balls on the floor, and the set of balls under the rug. As the number of iterations approach infinity, the set of balls on the floor approachs the set of original balls.

Yet another way to look at it is this.
At iteration x, there are balls numbered 1-x on the floor (taken out of the urn and painted), and balls numbered (x+1) - (10*x - 1) in the urn (excluding all multiples of 10). As x approaches infinity, the lowest numbered ball in the urn approaches infinity. At infinity, the lowest numbered ball _is_ infinity. Since infinity is not a real number (i.e. the set of real numbers - or natural numbers or integers, for that matter - does not include infinity), there cannot be any balls in the urn.

[Wonko the Sane]

Thank you Trenin...nice to know that I'm not completely alone in my view that there isn't anything wrong with the wording of the puzzle, even if it is a bit ambiguous.

[Green Dragon]

I have two issues with this puzzle, questions that I have.

#1) why do we have two demons? Other than to confuse us as to which demon really cheated, and as a useless distraction, Why? I think that the solution must invove both demons, or at leat that it may. What would the sorceror notice as a diffrence between the two piles that the demons made, or something along that line.

#2) Slightly unrealeated to the puzzle, I really don't see why you believe that the urns will be empty. Maybe it is so, but I must be a dunce, because I really don't completly understand an validity in your statments as to why that is.

See, I understand that right before midnight, the demon is preforming an infinate number of operations in a finite amount of time. So I can see why it is POSSIBLE to empty the urn at midnight. But mr. magick didn't order either demon to completly empty the urn. From whatI remember of the semester I took on advanced probibility and infinities, (I still don't get why that was one class) I do remember that there are diffrent infinities. If you put an infinate number of marbles in in steps of ten, and take out a single marble each step, that is the exact same as putting in nine marbles an infinate number of times. You still get marbles in the urn, in fact, still an infinate number of them. Do I remember wrong? just think about it for a while (Your Glers- should I really need to tell this to you? probably not, since I'm guessing that you've dreampt dreams of urns and marbles since this puzzle came out)

Feel free to correct me, I just want really good proof along with the correction, proof that an idiot like me will understand.

[Green Dragon]

sorry, I really have three

#3) I agree with those of you that demon #1 is the one cheating. I haven't known the GL for more than a few months, if that, but how often does the minotaur make a mistake this crutial to the puzzle and then not correct it for so long? No mathematical basis, just a profound devotion and worship for his almighty lordship.