Urns of Infinity

[hank]

Following up on kevin's recent question, I too considered that conundrum. What is an iota second anyway. According to the minotaur, that is the time interval remaining at which the demons are no longer able to complete their tasks.Thus the demons are no longer able to continue and must stop. I can not see any justification here to consider solutions base on the concept of infinity.We don't know how long an iotasecond is but we do know it exists and that it is capable to prevent the demons from continuing infinitely. Infinity and limits do not mix.

[Ghost Post]

My first post on this topic:

Yes, I have to agree. It says they finished just before midnight - with an iotasecond to spare. So an iotasecond is an interval of time, and a non-zero length interval (or they wouldn't have finished before midnight).

Then, on the other hand, it says that when it's over, there are an infinite number of lumps under the carpet.

Now, which demon is the lazy one. It say's #1, but also say's that what the lazy one is doing is "in the spirit" of the instructions given, so it must be #2. Also, if it were #1, it would be a very uninteresting puzzle. If it's #2, it get's interesting.

If #2 follows the instructions (and doesn't stop before midnight), at midnight there would be no balls in the urn. No paradox here. The argument that he left 9 times as many balls in the urn as he removed doesn't apply, since we're dealing with infinity here. The important point is that every ball that gets placed in the urn, later (but BEFORE midnight) get's removed. Name the number on the ball, and I can tell you when it was placed in the urn, and when it was removed.

Now, the interesting thing is if #2 is the lazy one. At any point before midnight, the balls in the urn are labeled with the same numbers as if he had followed the instructions he was given. But, he never removes any balls from the urn. At each step he puts 9 balls in the urn, and never removes any, so the urn can't possibly become empty.

It seems like a paradox. Each step is performed before midnight. And if two demons were working side by side - #2 and the lazy #2 - then at each step, the numbers on the balls in their urns (and hence the number of balls in their urns) would be identical. This is true for every step. But at midnight, one has an empty urn, the other an infinitely full one.

[hank]

My dad and I have gotten into a lot of arguements over the years, and many of these arguements were based on differing definitions of words. If the puzzle states that there are an infinite number of balls under the carpet does that mean "all" the balls are under the carpet? Or , is infinity a unique number so that there is only one infinity? I'm sensing that that is what some of you believe, or I am seriously misunderstanding some of the comments here. I have always defined infinity as an unrestrained unfathomable 'relative' number for myself which always seemed to have worked up until now. This definition allows for the concept of more than one infinity. Nine times infinity equals infinity is a valid equation I believe.

[Ghost Post]

An infinite number of balls under the rug doesn't mean all of them. The balls that go under the rug all have numbers divisible by 10, but there are an infinite number of balls divisible by 10.

Infinities can be equal in size, but given two infinite sets, equal in size, one may be "smaller", in the sense that it is a proper subset of the other. If I take the infinite set of integers (1,2,3,...), and remove the infinite set of integers that are divisible by 10 (10,20,30,...), I have an infinite set left. If I take the infinite set of integers (1,2,3,...) and remove the infinite set of integers (1,2,3,...) I have a finite (empty) set left. But the set's (1,2,3,...) and (10,20,30,...) are the same size. So I can start with two equal size sets, remove two equal size sets, and get sets of unequal size.

In any case, regarding the balls under the rug: One ball goes under the rug at each step, and if they stop before midnight, there will only be a finite number of steps, and a finite number of balls under the rug. An "iotasecond" may be intended to be infinitely small, whatever that means.

[Aarondalf]

Being such a powerfull sorceror I will put in my infinite number of cents.

What does the sorceror EXPECT to see when he comes back?

From Demon #1 he expects to see balls indivisible by 10 IN the urn and balls divisible by ten out of the urn. (ie: IN=1-9 OUT= 10, 20 , 30....)

From Demon #2 he expects to see all the balls out of the urn BECAUSE HE IS MAGICAL AND THE MINOTAUR SAID IT IS POSSIBLE FOR HIM TO DO IT!

Now if the Minotaurs wording is correct then Demon #1 is cheating and what he sees is NO BALL NUMBERED 1, because the Demon painted it. So this makes a pretty bad puzzle.

But if Demon #2 is cheating then he is really following the spirit of the task.(at least he is trying to)

But when he changes the balls to higher powers of ten, ie 1=10 which then equals 100=1000=100000=1.........) all he is doing is just painting an infinite number of zeros onto every ball --->infinity.


The whole paradox comes from the fact that we are not supposed to think that 10x-x=0 as x-->infinity, we are supposed to say that it equals 9x which would equal infinity as x approaches infinity.

THE DEMON THINKS THIS ALSO! HE THINKS HE CAN SKIMP ON ONE TENTH OF HIS WORK (assuming painting is less work than lifting) but really he should, as alot of people have said, just sit and laught at the other demon.


In Conclusion, DEMON #2 must be lazy, not Demon #1, because it isnt such a hard puzzle to work out otherwise.

The whole time Demon #2 is following the spirit of his task, untill that last iota second, when if he had done his work properly he would have none left, now the urn is filled-with each ball having a number from 1->infinity followed by infinity zeros.

I think my signature sums the whole situation up nicely.




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You can't chainsaw a duck.

[Green Dragon]

Yeah, I auctually hadn't thought about that part of the puzzle...

If either demon finishes at any point before midnight, then you cannot have infinity...

As somone said, infinity and limits don't go together. If he finishes even one iotasecond before midnight, there are nearly infinite balls used, (dosen't matter where they are)but not quite. This changes the puzzle quite a bit.

another incongruency of the puzzle... ah..


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What is the world coming to, when a reasonable group of maniacs can't even chainsaw a duck?

[Tom]

Oooh, good, lots of people who (seem) to be agreeing with me .. well, I agree with them anyway.

I must confess I wondered about the "iota-second", but write it off as just a turn of phrase, not intended to really mean anything.

I would claim, however, that limits and infinity do mix, in a sense .. getting rid of infinity problems was exactly why limits were invented. But here they are not the right thing to use.

PS - callypigian means "having shapely buttocks", Green Dragon.

Oh, and Kevin, the demons can't really do it .. that why the sorceror got demons to do it. But there are infinite numbers between 11:59 and 12:00. Imagine 1/2, 1/4, 1/8, 1/16 .. 1/2n .. this sequence goes on forever. Now imagine 12:00-(1/2 a second), 12:00 - (1/4 a second), ...

[Ghost Post]

Essentially, we have to answer a similar question to

"Which is longer: a line or a ray?"

This reminds me of another puzzle I solved in the past about a hotel with infinite rooms that was totally filled. An inifinite number of people came, wanting lodging, and they got it-how?

;-)

[Borodog]

Ok, the "iotasecond" bit means absolutely NOTHING. It is arbitrary. There's no difference between stopping at midnight or and "iotasecond" before midnight. It's just and offset. The puzzle states that the demons do in fact finish their tasks.

The interpretation that it has to be demon 2 that's cheating because there is no ball #1 in demon 1's urn if he's cheating doesn't wash if the sorceror checks what is outside the urn and not inside; he'll see balls numbered 10, 20, 30 . . . just like he should. I think maybe the solution would be this: If demon 1 leaves the real 10 under the rug (which he must,
the rug is lumpy at the end), then he can take it out, paint it, and turn it into a 100. Therefore the sorceror can tell the demon cheated because he has no 100 (etc) in his pile.



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Insert humorous sig here.

[Aarondalf]

Yes, saying that he could be seen cheating just by looking for ball #1 does work, because if he found ANY ball with a 1 on the end then the demon is cheating.

And why disagree with someone you essentially agree with?

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You can't chainsaw a duck.

[stoatboy]

to logically prove your assertion in a puzzle like this (i.e. the urn has an infinite number of balls or no balls at all) you'd generally play a little game like this. If daemon #2 doesn't cheat, the assertion is that there are no balls left in the urn, you assume someone disbelieves you, and that there's at least one. But if there's a ball in the urn, it must have a number on it, and you are able to disprove that person's claim by pointing out that ball #x was removed by the daemon when he added balls 10x + 1 through 10x + 10 to the urn.

The argument for when the daemon cheats (that there's an infinite number of balls in the urn) is pretty easy--it's enough to realize that the daemon adds an infinite number of balls to the urn, and never removes any of them, he simply paints new numbers on ones that are already there.

So here's a bonus puzzle: the person whom you thwarted in the first proof says, in response to your claim that there's an infinite number of balls in the jar when the daemon cheats, "Okay, if you're so smart, tell me the number on just one ball that's still in the jar." Of course, you can't. "Well, okay then," he says, "therefore by your logic, the jar must be empty."

What's the falacy of that argument?

[Sumudu]

Just a thought...I doubt that the sorcerer will be able to tell exactly which balls are in the urn and which are outside the urn, that is he won't be able to tell where ball number 1 is. However, I think that he will be able to compare the number of balls in the urn to the number outside (two infinities possibly, but he can tell if the urn is empty or not).

[Ghost Post]

Maybe I missed it, but no one has mentioned the relation of this problem to Zeno's arrow paradox. The connection can be made between an arrow traveling a finite distance and the elapsing of one minute of time. If the demon performs the task at one minute to midnite then half that time and half that time and so on; the demon could not possibly finish the task with one iotasecond remaining because he would have to continue the task at half an iotasecond and half that time and so on. Even if an iotasecond is just meant to be an infinitely small amount of time, you can still never reach midnite by performing a task at half the time until midnite repeatedly. If you understand the paradox involved, you will also understand that this is a very ill-conceived problem.

[This message has been edited by russel12_3 (edited 05-16-2000).]

[Aarondalf]

I thought that an iota second was an infintesimal, ie it is an infinitely small ammout of time, but i may be wrong.

[Tom]

Stoatboy - I do see your point, but I never mentioned it myself because it would just complicate things.

The fallacy is you are assuming balls with a certain number are removed from the urn the same way each time, which is not true. In the first case, they are removed by taking out. In the second case, they are removed by painting over.

In the first case, we have no number for a ball in the urn, and the only way to have a numbered ball not in the urn is to take it out, so all the balls must be taken out.

The second case (with painting) the only way to have a numbered ball not in the urn is by painting over the number an infinite number of times (nothing is taken out). So having no number for a ball in the urn tells us that all the balls have been painted an infinite number of times.

Ok?

As to the iota-second, I believe it was just a bit of blurb to round out a sentence, not meant to be used seriously in the puzzle. Anyway, as Aarondalf says, an iota-second is probably an infitesimal, so doesn't screw up anything.

[ZenBeam]

quote:

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So here's a bonus puzzle: the person whom you thwarted in the first proof says, in response to
your claim that there's an infinite number of balls in the jar when the daemon cheats, "Okay, if
you're so smart, tell me the number on just one ball that's still in the jar." Of course, you can't. "Well,
okay then," he says, "therefore by your logic, the jar must be empty."

---

I can imagine that the demon paints each succeeding zero half as big as the previous zero, so an infinity of zeros takes up a finite amount of space on the ball. In fact, this could be true for all the digits. The tens digit is half as wide as the ones digit, the hundreds digit is half as wide as the tens digit, etc. You pick up a ball, and see the first several digits, the rest trailing away beyond your limit of vision.

I don't think there is a valid answer to the question of whether the urn is empty or has an infinite number of balls. The situation isn't convergent, and that's why you get different answers depending on how you approach the problem. As I showed above, answering this question isn't necessary to solving the Minotaur's question.

quote:

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Just a thought...I doubt that the sorcerer will be able to tell exactly which balls are in the urn and
which are outside the urn, that is he won't be able to tell where ball number 1 is.

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This is the most important missing information in the puzzle statement: What is the sorceror capable of observing? Every solution proposed so far has at least implicitly made an assumption here, but the Minotaur never says. Perhaps the sorceror is blind, and can't read the numbers on any ball.

Which brings up another "bonus puzzle" for those of you who claim the urn is empty: If none of the balls had numbers painted on them, would the urn still be empty? If you answer "yes", why? If you answer "no", how does painting the balls change the number left in the urn?

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

[dave10000]

stoatboy and ZenBeam have hit the nail on the head by identifying that there's not an "answer" to the question "what's left in the bucket at the end?"

In my opinion, the best "answer" to the current puzzle is that the premises are self-contradictory and thus (1) since the tasks could not be carried out as stated, there is no answer, and (2) if at gunpoint I am required to give an answer, more than one "answer" would suit the given premises.

Consider the following 2 scenarios:

Scenario 1 -- Suppose that Demon 3 is given the following task: "There is just 1 ball and 1 bucket. At each of the (infinite) intervals, you are to put the ball in the bucket if it is out, or take the ball out of the bucket if it is in." Simple enough. Now, suppose I were to ask you where the ball is when the Demon is "finished." There's no answer -- the ball cannot be in the bucket, because at all times when the ball was in the bucket it was later taken out. The ball cannot be out of the bucket because at all times when the ball was out of the bucket it was later put back in. The "answer"? The premises of the problem are contradictory -- the Demon could not carry out the task as told AND THEN AT SOME POINT BE FINISHED. So trying to draw any conclusion from a situation that LOGICALLY could not exist is irrelevant.

Scenario 2 -- Is the argument that "there is no ball in the Demon #2 bucket because every ball would be removed" really valid? Consider this: Demon #4 is given the same task as #2, but with the following, additional instruction. "After you have put the first 10 balls in, if you ever get to the point where there is just one ball in the bucket, leave that ball in there and stop any further actions." QUESTION -- how many balls are in Demon # 4's bucket when he is done? It CAN'T be zero, because he was required to stop if he ever got down to one ball left. Can it be one or more? Well, would anyone care to tell me the number on that ball? Thus, the fact that you cannot identify a numbered ball left in the bucket does not "prove" that the answer must be zero, since with Demon 4 you can't identify a numbered ball but the result BY DEFINITION can't be zero. The only thing left is to conclude that the premises are self-contradictory, and that no valid conclusions can be drawn from them.

If the Minotaur wants to have a puzzle involving the performance of an infinite number of steps, then I have no inherent quarrel with that as long as the result of those steps is CONVERGENT. (Perhaps Demon 2 takes away 1/2 a ball, then 1/4 of a ball, then 1/8 of a ball, etc.) But to set a puzzle where the series is not convergent, and then to ask what happens "at the end," is, in my opinion, meaningless. If the point of the puzzle is to reflect upon the self-contradiction or non-convergence in this way, fine. But if it is to come up with a specific answer that arguably cannot be challenged, I do not agree that such can be done.

Thanks for listening.

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Eternity is a long time; especially the end.
-- Niels Bohr

[IrishJoe]

Given the fact that the demons completed their tasks, the number of balls could not be infinite.
Given a finite number of balls, Zeno's arrow paradox is not a valid analogy, so we discard it.
Given the fact that the puzzle is written as it is, Demon #1 WAS the one cheating, or trying to.
Assuming that the sorcerer was capable of discerning balls, numbers thereon, and the number of balls outside the earn, he would be able to calculate the number of balls which should be out of the urn, which will be precisely one greater than the number which are actually out of the urn, thus proving the cheating.

The only problem is that there actually would only be a near-infinite number of balls under the rug.

Smile, it's all in fun!

[Tom]

IrishJoe - they're demons, who're magical, so they can do an infinite amount of things in a finite time.

Dave1000 - I agree that your two examples do not work (the question "what happens at the end" makes no sense), but I don't think this is the case for the Minotaur's examples .. what happens at the end does make sense. I agree the function of "balls in the urn" to "balls handled by demon" is divergent .. it goes off to infinity. But we are not looking at this, we are looking at the function of "balls in the urn" to time .. this is just discontinuous at 12:00. So are loads of functions in maths, doesn't make them nonsensical to talk about.

But I'm fairly sure that won't convince you; all I'm really saying is I think it does make sense.

[dave10000]

TOM (and anyone else) --

It's not the discontinuity that makes the premises self-contradictory, it's the divergence. I'd have no problem (well, much less of a problem) with the premises if they stated "infinite actions up until midnight, then empty each bucket instantly at midnight." In that case, while the function of "balls in the urn" is clearly discontinuous, it is equally clear that is has a defined value at midnight (zero), so it makes sense to talk about the value at midnight. In the actual GL problems, and in my 2 alternatives, there simply is no defined value at midnight -- it is ambiguous. Thus, we can't talk about "what happens at midnight" in the same way.

If anyone continues to believe that a Demon who takes out balls 1, 2, 3 etc ends up with an empty bucket, how do you resolve the following inconsistency:

Suppose there are 3 Demons. X puts 10 balls in and removes the lowest ball (like Demon 2). Z puts 10 balls in and removes the lowest EVEN NUMBERED ball (so "all the odds" will remain at midnight). Y, in the middle, puts 10 balls in and pulls one out, but his balls do not have numbers on them. Now, it is pretty clear that X, Y, and Z have equal numbers of balls in their buckets at all times BEFORE midnight. And it is pretty clear that, if you believe that we can reach midnight and the Demons can complete their tasks, Z has an infinite number of balls (the odd numbers) in the bucket. What about X? Some people have argued that X will have zero balls at midnight. Well, the SMALL problem is how can the totals in the X bucket and the Z bucket ever differ, but OK, some will say that the discontinuity at midnight accounts for that. But now the BIG problem is how many balls does Y have in his bucket at midnight? We can "prove" that it must be the same as Demon X, and we can "prove" that it must be the same as Demon Z. (And I'm interested -- what do YOU think, Tom?) To me, this just reinforces that there is no unambiguous answer to examining the buckets AT midnight, and no unambiguous conclusions can be drawn from hypotheticaly doing so.

ALSO, I am curious as to why you agree that it does not make sense to ask what the bucket looks like at the end of the task of Demon 4 in my earlier hypothetical, who acts exactly like Demon 2 but stops if there is ever a single ball remaining in the bucket. For all those who believe that Demon 2 ends with an empty bucket, doesn't that NECESSARILY mean that there was a point that the bucket contained a single ball (since the Demon only takes out one ball at a time), and if so, doesn't that NECESSARILY mean that Demon 4 will end up with a bucket that has 1 ball. OR, if you believe that the function is truly discontinuous, then the bucket will go from gazzillions just before midnight to zero at midnight, and there will never be a time where the bucket had exactly 1 ball, so Demon 4's bucket will be empty. 3 questions -- (1) Which is it? (2) Why does Demon 2's task reach a defined result and Demon 4's task not? (3) If Demon 2 always takes out 1 ball at a time, how can the function be discontinuous (other than increasing or decreasing by 9 or less at any given time) -- doesn't THAT contradict the premise of the problem?

I do not believe that the questions posed in this post, and in my previous one, are capable of consistent, definite answers, and thus I continue to believe that the premises of the GL problem, as given, are self-contradictory.

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Eternity is a long time; especially the end.
-- Niels Bohr

[worm]

so, is this an "infinite" rug? if not would an infinite number of finite sized balls fit under it?

[Tom]

Btw, I think I should say here that though I obviously believe I'm right, I'm not convinced of it .. this to dave10000 especially.

Right .. here goes -

Well, as far as x,y and z go .. x ends up with an empty urn and all the balls outside; z ends up with all the odd numbers in the urn, all the evens outside. To my mind, the reason y can have the same number of balls as both x and z is because we don't actually have enough information to say how many balls are in y's urn - it depends how y goes about removing them. If he does it like x, he gets none, if he does it like z, he gets an infinite amount. y's task is under-specified. He must use some method to pick the ball to remove, we just don't know what it is.

Now to answer the 3 questions -
1) I guess I believe that #2's urn does go from gazillions just before midnight to zero at midnight. The function is truly discontinuous.
2) I suppose I believe that #4 would end up with an empty urn as well, really. The stopping when there's only 1 left just never happens.
3) Because he takes only 1 ball out at a time, but in the neighbourhood of 12:00, however small, he is always taking out an infinitte number of balls. To be more precise, you can have #2 taking out any amount of balls, however large, just by getting close enough (without touching) to 12:00.

I know the function you've got here has to be doing some weird stuff, but I don't see why you can't have this. As an example (not an analogy, really, as it is undefined at 0) sin(1/x) is continuous all the way down to zero, so only goes up or down by a measurable amount on its way to zero, but is discontinuous at zero.

[Ghost Post]

Yeah, I'm also less certain about the whole thing now. dave1000 makes a good point.

This is similar to the seeming paradox I mentioned between demon #2 and demon #3 (a cheating version of #2). They work side by side, in parallel. At each step of the way, they have the same number of balls in their urns, and with the same numbers on them. Yet it seems, when midnight strikes, demon #3 has balls in his urn with infinitely large numbers on them, while demon #2 has an empty urn, and never touched a ball with an infinite number.

[Ghost Post]

Infinity is strange. The order that you do something is really important in the outcome. Here is an example without the extra confusion of the discontinuity at midnight.

Al puts ball 1 in the urn after 1 minute, ball 2 after 2 minutes, ball 3 after 3, etc. (after n minutes he puts in ball n)

Bob puts ball 2 in his urn after 1 minute, ball 4 after 2 minutes, 6 after 3 minutes, etc. (after n minutes he puts in ball 2n) After he puts in the even numbers, he puts in the odds.

Now if you pick any integer I can tell you when Al will put the ball with that number in his urn. If the integer is, for example, 5834023 then Al will put that ball in after 5834023 minutes.

But when did Bob put in ball number 1? He never gets to it because there are an infinite number of even numbers to go through first.

The paradox is that Al will eventually put every nuber in the urn, but there are some numbers that Bob will never get to. An other way of saying the same thing is that there are the same number of even numbers as integers. It doesn't make sense, but that is what you get when dealing with infinity.

[Griffin]

I agree. Infinity is strange. Consider this:

Demon #1 and Demon #2 are standing in a room, 10 feet from the door. 5 feet from the door, there is a dot (dot 1) on the floor. There are also dots at 2.5 feet(dot 2), 1.25 feet(dot 3), 0.625 feet (dot 4)etc.

They both start walking towards the door. Demon #1 takes 1/2 minute to get to dot 1, 1/4 minute to get to dot 2, 1/8 to get to dot 3, etc. (Basically he's walking at a constant rate).

After the first half minute, Demon #2 is already at dot 10. It takes him 1/4 minute to get to dot 20, 1/8 minute to get to dot 30, etc.

At the end of one minute, how many dots are between the two demons?

[Ghost Post]

Interesting thought Griffin, but at the end of one minute both are at the door. At time t, in minutes, #1 is 10(1-t) from the door and #2 is 10*(1-t)^10 feet from the door.

[Green Dragon]

Dave10000, I think that I could just kiss you. I had no more to say without repeating myself, and you jumped in to support out position.

Anyway, I completly agree with what most of you have been saying. However, In whoevers problem, there will never be any odd numbers. There are an infinite number of them, and they have to run out for him the demon to switch.

I haven't gotten much sleep, so I don't really have anything more to say

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"They're all crazy! They're all crazy except you 'n me. And sometimes I have me dobuts about you."

[Wonko the Sane]

Dave1000's premise that Y's urn must look like both X and Z's urns is wrong. It must look like X's urn at midnight. Here's why.
X puts in balls and takes all of them out in order, leaving the urn empty
Z puts in balls and takes out all of the evens in order, leaving the odds behind
Y put in unmarked balls and takes them out in random order, leaving the urn empty.

The reason that Y's urn doesn't look like Z is because Y's urn has no order. What you're doing by saying Y's balls are unmarked is saying Y has balls 1-infinite in his urn and he takes them out in random order. He'll eventually remove them all because he can't remove any ball twice.
in demon Y's urn each ball has a 1 in 9n+1 chance of being removed where n is the number of times he's added or removed balls.
In X's urn, ball n has a 100% chance of being removed, where n is any integer
In Y's urn, ball 2n has a 100% chance of being, all other have a 0% chance of being removed where n is any integer

So, how many balls get removed? Well, the demon will have an empty urn at the end IF AND ONLY IF every ball has a better than 0 chance of being removed.
In X's urn, every ball has a 100% chance of being removed
in demon Y's urn, each ball has a greater than 0 chance of being removed at any given time (though the actual probability changes). After an infinite number of trials, every possible chance must be fulfilled, thus the urn will be empty.
Demon Z's urn will not be empty because every ball with an odd number has a 0% chance of being removed. Those balls with a 0% chance of being removed are the ones that will remain in the urn when the demon is done. Thus, he is left with the odds.

See why there's no incongruency here?

[Ghost Post]

Y's urn may end up with any number from 0 to infinitely many balls in it.
Also, assuming random selection of the ball to remove, the first ten balls placed in the urn each have about a .314 chance of being removed evntually. Balls placed in the urn later have a lesser chance. I have a hunch there is a 0 chance of a finite number being left (i.e. probability = 1 that there will be an infinite number left).

[ZenBeam]

[Ghost Post]

"The fact that you get different answers looking at the problem three different ways should be a sign that the arguments used are invalid."

I disagree.

Let's look at three sets:
A = {1,2,3,4,5,...}
B = {2,4,6,8,10,...}
C = {2,3,4,5,6,...}

These are all infinite sets, and all the same size.

Now A-A is empty, A-B if infinite, and A-C is finite, but non-empty. The point is, take an infinite set, remove an infinite number of elements from it, and you can get an empty set, an infinite set, or a finite, non-empty set. It just depends on which (infinite set of) elements you remove. Nothing invalid about that.

The empty urn answer (where 1-10 are added, 1 removed, 11-20 added, 2 removed, etc...) is based on a perfectly valid argument: For each and every ball that is placed in the urn, it is placed in the urn at some specific time before midnight, and is afterwards, but still before midnight, removed from the urn. You can't deny that much, can you?

So, at midnight, you can say without doubt that every ball that was previously added to the urn was, afterwards, removed. What then could possibly be in the urn?

The proof is simple: The urn starts out empty. Every ball placed in the urn is placed in before midnight, and is later, but still before midnight, removed. At midnight, the only balls that could be in the urn are ones that were placed in before midnight, but not removed before midnight. There are no such balls. So, at midnight, the urn is empty.

[This message has been edited by extro... (edited 05-18-2000).]

[ZenBeam]

quote:

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For each and every ball that is placed in the urn, it is placed in the urn at some specific time before midnight, and is afterwards, but still before midnight, removed from the urn. You can't deny that much, can you?
So, at midnight, you can say without doubt that every ball that was previously added to the urn was, afterwards, removed. What then could possibly be in the urn?

---

This is what I'm talking about. Always the same approach. I can take a different approach: After each step, the number of balls increases by 9. There is never a step where the number of balls decreases. There are an infinite number of steps. The number of balls grows without bound. You can't deny that much, can you? So at midnight, there are an infinite number of balls in the urn. How then could there possibly be zero balls in the urn?

Why is your approach more valid than asking how many balls are in the urn at any given time? Show where the approach of counting the number of balls fails.


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

[Ghost Post]

That's simple. Before midnight, at the n-th step, 10*n balls have been placed in, and n balls have been removed. 10*n - n is 9*n. But that only works before midnight, when n is some finite number. It fails at midnight.

At midnight, n is infinity. Then 10*n = n = infinity. An infinite number (10*n) have been added, and an equal infinite number (n) have been removed. And an infinite set minus an infinite set (i.e. removing an infinite number of elements from an infinite set) can validly be the empty set, or an infinite set, or a finite non-empty set, depending on which infinite set of elements is removed (as illustrated in previous post).

[Green Dragon]

And ZenBeam too. I'm so glad that you are coming over to my side.

The fact that there are no balls in the urn that you can name a number on does not mean that there are no balls in it. This is a restriction that is only in your mind. If you put in 9 balls in in each step, and you have an infinite number of steps, you end up with an infinite number of balls inside the urn. No one of us is denying the power of the demon, saying that it cannot empty the urn instantly, exactly at midnight. But that's just not a step that the wizard has ordered him to take. And the demon wants to do as little lifting as possible, right? So he dosen't empty it, and it still has an infinite number of marbles in it.

[Wonko the Sane]

ZenBeam, what you're failing to see here is it isn't the numbers that are important, but the order. The way you just described it (add nine balls and don't take any out) does not describe the puzzle. The puzzle says that demon 2 puts in 10 balls and takes out a different one (except for the first time). So he isn't just adding 9 balls.
Extro, you've misquoted me a bit. Although the first set of balls may have about a certain chance of getting taken out, that probability changes every time. That's why I used the equation 9n+1 to describe the probability of a ball being removed. Every ball in the urn has a 1 in 9n+1 chance of being removed. After 3 turns, each ball has a 1 in 28 chance, after 4 1 in 37 chance...and so on. Thus until midnight, any given ball has a finite chance of being removed. Thus after you do this an infinite number of times, all of them should be removed by midnight.

[dave10000]

(Briefly, because I am at work)

To all those who argue that the urn MUST be empty at midnight, because no one can name a ball that's in the urn, you are committing the fallacy of "avoiding the other" that is so common in paradoxes or semi-paradoxes. In (most) paradoxes, there are two ways of looking at a situation that lead to inconsistent answers -- let's say "A" and "B". It gets the analysis nowhere to argue that "B" must be wrong BECAUSE "A" is right -- the paradox is that there is a (seemingly) valid line of argument to both. To dispel the paradox, you must show what is wrong with the REASONING of the other line, not that there is a line of reasoning that leads to the other solution.

In the present case, there is not really even a paradox, but rather INCONSISTENT PREMISES. Let me give an example. Suppose this were the question:

"I have an object in this bag. It is completely blue (and no other color). It is round. It is completely red (and no other color). What color is it?"

Now, if I argue that the object is blue, it is no good for you to say "it can't be blue because the problem states that it is red." Similarly, if I argue it is red, it is no good for you to point out that I'm wrong because the problem states it's blue. In such a case, where there is INCONSISTENT information, either "red" or "blue" is equally right. Or, perhaps a recognition that the problem is internally inconsistent is the "most correct" answer.

Turning to a more famous question in philosophy, there is the debate over "What happens when an irresistible force meets an immovable object?" Again, although it SEEMS like it's a valid point to debate, the question is internally inconsistent. If an irresistible force exists, then it is impossible for an immovable object to exist. If an immovable object exists, then it is impossible for an irresistible force to exist. Because the question is self-contradictory, there is no "right" answer.

As I set forth in earlier posts, the situation is the same with respect to the Demons. A non-convergent, infinite-step task cannot be completed, and does not lead to a defined result at the end. Asking a question involving such a task, and then asking what happens "at the end," is self-contradictory, and does not have a single, defined answer. I think (hope) you'd agree with me that is is more or less silly to seriously ask the red-blue question set forth above. (If not, what do you think the answer is??) The Demon question is, ultimately, "silly" for the same reason, although the inconsistency is much harder to find. (NOTE -- It's "silly" if you are looking for a real "either-or" answer. But it's informative and interesting if you are looking to get at a deeper truth about mathematics.)

[Ghost Post]

Green Dragon : "The fact that there are no balls in the urn that you can name a number on does not mean that there are no balls in it."

That's not my argument. The point is, you start with an empty urn. It would stay empty, but balls get added. But each ball added, before midnight, later gets removed, before midnight. That's a fact I think you must agree with.

"If you put in 9 balls in in each step, and you have an infinite number of steps, you end up with an infinite number of balls inside the urn."

As long as you don't remove them later. But you do. Every ball has a number, and name any 9, and I'll tell you exactly when they get removed.

"No one of us is denying the power of the demon, saying that it cannot empty the urn instantly, exactly at midnight. But that's just not a step that the wizard has ordered him to take."

And no one claimed that. But that argument doesn't hold. At any moment before midnight, no matter how close to midnight, there are a finite number of balls in the urn. Was the demon ordered to instantly put an infinite number in at midnight? It is equally true that that is not a step the wizard ordered him to take. But the net effect of the steps he was told to take is to empty the urn. If he was, in effect, told to place an infinite number of balls in by midnight, then he was also, in effect, told to remove them all from the urn by midnight.

[Ghost Post]

Wonko: I understand what you are saying, but I think the conclusion is wrong. 1/10 + 1/19 + 1/28 + 1/37 + ... does not add up to 1, but something under .314

[dave10000]

extro --

Let's say that the Demon was further required to write down the number of balls in the bucket every time he took a ball out or added a ball. I assume you believe that the last number he writes down is "zero," because you believe (I think) that the bucket ends up with zero balls.

Then, some questions:

(1) What number did the Demon write down immediately before the final zero?

(2a) If the answer to Q1 is "1", then after removing a ball to get to 1, wasn't he then required to add 10 (or 9) more balls? If not, why not? (And, if the answer is "1", what number was on that ball?)

(2b) If the answer to Q1 is a number greater than "1" (or "infinity" or anything like that), then didn't the Demon violate the rules by either (a) removing more than one ball at a time, or (b) not writing down the number each time he took a ball out -- that is, if the numbers on the Demon's list ever decrease by more than 1 at a time (such as go from 100 to 98, or from 2 to zero), then didn't the Demon violate his marching orders?

I'm curious to see how an "empty-bucketer" answers these questions.

[Ghost Post]

dave10000 : "I assume you believe that the last number he writes down is zero"

No, there is no last number he would write down. There is no last step performed.