Urns of Infinity

[Rhino]

Doggonit, extro..., you beat me to the chase.

The "empty-bucketers" are asking for the number of a ball that has not been removed.

My problem with that question is this: we supposedly have an infinite number of balls but each has a finite number on it. How can you name the last ball that was added? How can you name the last ball that was removed? You can't. There is no last ball for either. Sheesh, I feel like neo..."there is no last ball!"

Maybe these demons actually stopped time when they did this trick, and the sorceror was just messing with them so that he would never have to sign their time card.

Just a thought

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

[ZenBeam]

Wonko:

[dave10000]

extro writes: "No, there is no last number he would write down. There is no last step performed."

I don't understand this at all. The Demon is ordered to write down the number of balls in the bucket every time it changes. If the final number in the bucket is zero, then the final number he writes down must be zero -- otherwise he has violated his writing obligations. More specifically, the Demon is only allowed to do 3 things: (1) put balls in, (2) take one ball out, (3) write a number down. Since he (hypothetically) finished his task, he must have done one of these things last. Which one was it? If an "empty-bucketer" continues to maintain that "there was no last act," please explain how the Demon can possibly be finished?!

This, to me, is the problem with the "empty-bucket" position. It always has the Demon violating one of his instructions. THE ONLY WAY for the bucket to become empty is for the Demon to remove more than one ball without adding more. (If you disagree, would you care to explain how?) But removing 2 (or 3, or infinity) in a row without adding more VIOLATES the obligation that every removal of 1 be followed immediately by an addition of 9 or 10.

NOTE -- I'm not saying that infinity is the answer either, since you run into the "what ball is left in the bucket" problem.

The point is, it's no good to say the answer is zero, and then give an argument that has the Demon violate one of the rules. If we're going to do that, then we can come up with ANY answer (for example: 3 balls are left at midnight -- the Demon adds 3 balls to the bucket and then just disregards the rest of his task).

In sum, an answer of "zero balls" requires the Demon to disregard either the rule about "one ball at a time" or the rule requiring "add 9 or 10 after each removal". An answer of "infinity balls" requires the Demon to disregard the rule that ball N must be removed at 1/(N+1) minutes to midnight. This is because, as I have suggested in earlier posts, the premises of the puzzle are inconsistent.

ONE LAST TWO-PART QUESTION, an important one, for the empty-bucketers. Suppose there is another Demon -- #5 by now, I think. He has 2 buckets and one ball. One bucket is marked "IN" and one is marked "OUT". He is watching Demon #2, and is given the following, simple instruction: "Every time Demon 2 adds a ball to his bucket, put (or leave) your ball in the IN bucket. Every time Demon 2 takes a ball out of his bucket, put (or leave) your ball in the OUT bucket." Now, please tell me -- When Demon 2 has finished his task, is Demon 5's ball in the IN bucket or the OUT bucket? And, if you suggest that Demon 5 can never finish, or that his task or endpoint is ambiguous or undefined, then isn't that equally true of Demon 2's much more complicated task?

[Ghost Post]

ZenBeam:

[Rhino]

Yup, I just talked to the sorceror and we shared a good laugh. Those demons never finished...they're still playing with the balls and it's been 11:59:59.99999...9 in their little room for quite some time now. They'll never get to 11:59:59.999... er 12:00:00, though.

(just being silly here; please don't abuse me for it )

[Ghost Post]

dave10000: you ask many questions.

[dave10000]

quote:

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dave10000: you ask many questions.

quote:
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Since he (hypothetically) finished his task, he must have done one of these things last.
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No. The steps he performs can be numbered.
1 - Add balls 1-10
2 - write 10
3 - Remove ball 1
4 - write 9
5 - Add balls 11-20
6 - write 19
7 - Remove ball 2
8 - write 18
etc...

For ANY step, there is a next step. Whatever step you want to call his last step, it was followed by an infinite number of other steps.

---

Um, that's EXACTLY my point. If there is no last step, then the Demon has NOT finished his task. The empty-bucketers want there to be BOTH (1) no last step, and (2) a finished Demon. But those cannot simultaneously exist. As your responses to my questions establish. And so the premises of the problem contradict each other.

(I am still looking forward to any empty-bucketer answering my Demon # 5 question, above, which I think is key to resolving the problem/question/paradox.)

(And PS -- with respect to the A&B variant in your last post, I don't see that it changes the analysis. It's really the same problem. You can always find some time BEFORE midnight that will locate any required ball N in bucket B. But under the given circumstances, midnight can never be reached, AND the Demon's task is divergent, so the state of the buckets "at" midnight is undefined. The Demon cannot consistently both (1) comply with the requirements of the task, AND (2) finish the task.)

[This message has been edited by dave10000 (edited 05-19-2000).]

[Aarondalf]

You guys have been arguing for ages about pretty much nothing, as you actually CANT HAVE AN ANSWER, so just give up.

For Dave1000, you say that the people who think there is an empty bucket are wrong because the demon couldnt put down his last step, and has to be finished. Yes???

But did you ever think about your own solution of infinity balls??? You want there to be a finished demon and a last step, NEVER GOING TO HAPPEN.
Listen up, it is a question involving infinity so he can not ever write down his last step so please dont keep asking people to answer a question like that.

I also think there will be 0 balls left and the way he takes out his balls is completely different than Demon one.

First Demon one:

Suppose we split his task among two other demons, one demon put in balls numbered 1-OO
Then another Demon takes out all balls divisible by 10.

We have 9x-x as x--->OO = OO

There is a major difference between this and what Demon 2 does, and again we will split the tasks.

One Demon puts in balls 1-OO
Another takes out balls 1-OO

which is x-x=0

The fact that there is no last step is obvious, but how in hell does this prove that there are infinite balls in the Urn.

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You can't chainsaw a duck.

[Tom]

I've written far too much already, but seeing as there is something of a spirit of "us versus them" theme here, I thought I'd just say I agree with everything extro... has written so far .. I'm on his side. I think the "Was the demon ordered to instantly put an infinite number in at midnight?" is an absolutely perfect point, by the way.

A similar argument to the "9 more balls each step argument" might go .. At each step, the number of balls in the urn is finite. How can the number of balls suddenly become infite at 12:00.

But I'm just repeating extro..., really.

dave10000 - this is going back a bit, but I don't think the premises are inconsistent. I see what you're saying, but I think my line of argument is right, and the other line is wrong. I do think I've spent as much time trying to show what I think is wrong with other peoples reasoning, as supporting my own.
The "last number written down" problem is the same as your demon who stops when there is only one ball left, I think .. there is no last step. The balls really do all disappear at once at 12:00.

Wonko - I agree with most of what you're saying, too .. I had y ending up with any number of balls, because I didn't want to assume he took one out at random .. for all we know, he might take out the 1st, 2nd, etc he puts in. We don't know. I'm also inclined to side with extro... that the urn will not be empty, but that's not really the issue.

Just out of interest, is anyone actually going to believe Minotaur if he disagrees with what they already think anyway?

[Impartial Observer]

Yes, I agree Tom, that makes a great deal of sense. Those who disagree with you are obviously fools, aren't they?

[Tom]

I didn't write that! I did write this, though.

[Ghost Post]

dave10000:

[Tom]

dave10000, in response to your demon #5.

Suppose you have a demon (#6) that just adds a (numbered, in order) ball at each step, and ends up with an urn with an infinite number of balls in it. Do you allow this? I guess really, if you don't, then we have nothing to argue about .. you are just saying "infinities make no sense", we are saying they do, and that's it. We can't really argue against each other.

But, presuming demon #6 can do his task. Have demon #7 (with 1 ball and 2 buckets, EVEN and ODD) put the ball in EVEN when #6 puts in an even ball, ODD on an odd ball. Where is the ball at 12:00? Does this prevent #6 being able to complete his task?

#2's task is more complicated than #5's .. but it's convergent, and #5's isn't.

[This message has been edited by Tom (edited 05-19-2000).]

[dave10000]

quote:

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(Aarondalf)
For Dave1000, you say that the people who think there is an empty bucket are wrong because the demon couldnt put down his last step, and has to be finished. Yes???

But did you ever think about your own solution of infinity balls??? You want there to be a finished demon and a last step, NEVER GOING TO HAPPEN.

---

"Infinity balls" is NOT my solution. If you go back and read my posts, you'l see that I have the same type of problem with "infinity balls" as I have with "aero balls".

quote:

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(Tom)
The balls really do all disappear at once at 12:00.

---

Then the Demon has violated what the sorcerer has required him to do, namely (a) take out ONE BALL AT A TIME, and (b) follow every "take out" with an addition. I don't for a second (iotasecond?) deny that the bucket can be empty if the Demon violates his orders.

quote:

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(extro)
OK, forget the demons. Let's come up with a numbering for the decreasing length time intervals before midnight.
Interval 1 = 11:59 to 11:59.5
Interval 2 = 11:59.5 to 11:59.75
Interval 3 = 11:59.75 to 11:59.875
etc...

By your argument, since there is no last interval before midnight, midnight never arrives.

---

Nope. The sequence you just proposed is CONVERGENT, and thus has a well-defined sum, just like 1/2 + 1/4 + 1/8 + 1/16 ... = 1. This is also the basis for the "Achilles vs. the Tortoise" paradox that has been exhaustively examined in the literature (and in Doug Hofstadter's terrific book "Godel, Escher, Bach"). The Demon's task, however, is NOT convergent, and thus does not have a well-defined sum "at" midnight.

Mathematically, the sum of a non-convergent series is undefined. For instance, how much is:

X = 1 - 1 + 1 - 1 + 1 - 1 . . . ?

Looked at one way:

X = (1 - 1) + (1 - 1) + (1 - 1) ... = 0

Looked at another way:

X = 1 + (-1 + 1) + (-1 + 1) ... = 1

Neither way is necessarily a right or wrong way of grouping the numbers; the problem is the infinite sequence does not "approach" an answer, AND is not defined to have an answer "at" infinity, so X does not have an unambiguous value. Same with the Demons -- under one "grouping" (that is, bucket always has 9N balls in it), the number goes off to infinity. Under another "grouping" (each number N is removed, so bucket can't have anything left), it goes to zero. The end result of this process is no more defined than the hypothetical sum 1 - 1 + 1 - 1 ...

[Ghost Post]

dave10000: But you argued that the demon never finishes the task if, at midnight, there was no last step that he performed. That was the point I was addressing with the time intervals. If midnight arrives, he has finished, and there was no last time interval, so no last step.

[Tom]

you have a problem with aero balls?

I think (following your reply to Aarondalf) that if you have a problem with infinity balls too, then that's more of a philosophical problem .. you just don't think infinity makes sense in this context, do you?

Following your response to me, ok, all the balls don't disappear at 12:00 .. badly put by me. They all disappear before 12:00 (for any ball, you can even say when). But in some sense there is an infinity of balls removed at 12:00, because the no. of balls goes from steadily increasing to zero, discontinuously at 12:00.

To reply to your bit to extro..., suppose a demon has to write down the numbers of the intervals on a piece of paper .. you've used the arguments in the past similar to "because the demon can't write down a final number, the sequence can't converge". But it does, as you say. Similarly, I think demon #2s task "converges" (is completable). I don't agree that his task is similar to 1+1-1...

PS - what about demons #6 and #7? And do you believe #6s task is possible?

[dave10000]

Well, of course my "arguments" can be attacked when they are changed first to something that I did not say. You guys are like a moving target -- when argument A is shot down, you go to B, when B is shot down, you go to C, when C is shot down, you go back to A. I'm getting a little out of breath.

But anyway:

quote:

---

(extro)
dave10000: But you argued that the demon never finishes the task if, at midnight, there was no last step that he performed. That was the point I was addressing with the time intervals. If midnight arrives, he has finished, and there was no last time interval, so no last step.

---

The two situations are different. The Demon's task does NOT converge, the "approaching midnight by smaller step" does. By the "fundamental theorem of limits" (in pre-calculus), the content of the Demon's bucket is undefined at midnight, whereas the time-series approaching midnight IS defined at midnight (because it is convergent). If you want to use a special analysis that differs from the fundamental theorem of limits to support your position, then I can't argue against it, but that position would be strictly NON-STANDARD. Compare: if you want to argue that the three angles of a triangle add to more than 180 degrees, I can prove you are wrong in Euclidean geometry, but if you insist on using non-Euclidean geometry, then you may be right, but only in a non-standard system. Same here.

quote:

---

(Tom)
To reply to your bit to extro..., suppose a demon has to write down the numbers of the intervals on a piece of paper .. you've used the arguments in the past similar to "because the demon can't write down a final number, the sequence can't converge". But it does, as you say. Similarly, I think demon #2s task "converges" (is completable).

---

That's not even close to what I said. Whether the Demon can write down a final number has nothing to do with whether the sequence converges. For instance, there is no "final term" in the sequence 1/2 + 1/4 +1/8 . . ., but that sum certainly is CONVERGENT, and sums to 1. The sequence 1/2 + 1/3 + 1/4 + 1/5 . . ., however, does not converge -- by taking enough terms, the sum can exceed any positive number. The statement "Similarly, I think demon #2s task "converges" (is completable)" is a use of convergent that is different from standard. "Converges" does not mean "is completable" -- rather, in standard limit theory, it has a VERY SPECIFIC meaning (roughly, "gets closer and closer to a specific, finite number at each step"). Since this is NOT the case with the Demon's task, then the Demon's task (or rather, the number of balls in the bucket) does NOT converge as that term is used in standard limit theory. As I said above, if you want to say it "converges" under some non-standard definition, or want to apply non-standard limit theory to the task, then perhaps "zero balls" is a solution in under those rules. But not under the STANDARD fundamental theorem of limits.

[Kevin]

This discussion is getting way to deep. My mind is spinning in circles. But I need one of you smart guys to answer a question.

Earlier I stated that there are an infinit amount of numbers between 11:59 and 12:00. But, no one of you will dispute that 12:00 comes and goes each day. Well given that there are an infinite amount of numbers to keep us from reaching 12:00, How in the world does 12:00 come? Or even better,
HOW IN THE WORLD DO THE DEMONS FINISH THEIR JOB WITH AN IOTASECOND TO SPARE?

PLEASE HELP!

[ZenBeam]

[Green Dragon]

I think that I'll just stop arguing. None of us (on either side) are conceding our points, or being convinced by the others. We are just arguing the same points, over and over, and it's usless.

ZenBeam is right, that the probles in in some ways unsovleable. I believe, now, that this was a badly worded puzzle, and it is really quite usless to keep disscussing it. I am waiting (as we all must be) for the minatour's solutionn to be posted, but I will not continue disscussing it. There are diffrent ways to look at this puzzle, and until these are clarified, bye to all.

[Ghost Post]

ZenBeam:

[Borodog]

Somebody was getting at this before; I'm not sure who. Both of the following positions are defensible. You really can't argue with the logic of either one:

1) Every ball that gets put in gets taken out later. The urn ends up empty.

2) Every time you take out a ball, you put nine more in. The urn ends up infinitely full.

The problem is that these two propositions, are mutually exclusive, even though both follow the wording of the puzzle. The puzzle really and truly is NOT WELL FORMED. It has no "solution." You might as well say something like, "There's a ball in a bucket. Take the ball out, but when you do, leave it in." Now, is the ball in the bucket or not? Neither, because the problem has ceased to have any *meaning*.

Before I was firmly an "empty-bucketer" (good phrase, whoever that was). But now I'm
with Veridisaurus Rex. There REALLY IS NO SOLUTION, because there REALLY ISN'T A WELL FORMED PUZZLE.

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Insert humorous sig here.

[This message has been edited by Borodog (edited 05-19-2000).]

[This message has been edited by Borodog (edited 05-19-2000).]

[This message has been edited by Borodog (edited 05-19-2000).]

[Wonko the Sane]

I haven't had time to read through most of the argument, but I'd like to respond to extro.

You're still misinterpreting what I said. The probabilities don't add up. They change. After 3 iterations (well, 2.5, because the demon hasn't removed a ball yet), there will be 3n + 1 or 28 balls in the urn. EACH ONE has a 1/28 chance of being removed. After step 4 there are 4n+1 or 37 balls in the urn. EACH ONE has a 1/37 chance of being removed. By addding balls, the demon is changing the probabilities of EVERY ball being removed. If you take 1/37 and add it together 37 times you get 1, right? That's what I was trying to say.

[dave10000]

"Not What I Said" files, part 2:

quote:

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dave10000: You said the demon never finishes his task.

quote:
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If there is no last step, then the Demon has NOT finished his task.
--------------------------------------------------------------------------------

He does. Now you say it doesn't converge.

---

It's not just "now" that I say it doesn't converge, I've always said that. The response "He does [finish]" is hardly an argument. I guess my only response to that is "Does not!"

My point was, and is, that there is a difference between non-converging series and converging ones. With a converging series, there is a "value at convergence" so one can talk about what happens when you add up the infinte terms, or "finish". With a nonconverging series, there is no "value at convergence," so one cannot (at least in standard limit theory) talk about what happens "at" the end point. When the series converges, you can get to the finish; when it does not, you cannot. I have said this consistently since my first post.

quote:

---

quote:
--------------------------------------------------------------------------------
By the "fundamental theorem of limits" (in pre-calculus), the content of the Demon's bucket is undefined at midnight, whereas the time-series approaching midnight IS defined at midnight (because it is convergent). If you want to use a special analysis that differs from the fundamental theorem of limits to support your position, then I can't argue against it, but that position would be strictly NON-STANDARD.
--------------------------------------------------------------------------------

What is non-standard is attempting to use calculus to solve a simple problem dealing with discrete sets of integers.

---

WHAT WHAT WHAT??? I did not say calculus, I said (see your own quotation of my post) pre-calculus. Pre-calculus involves primarily the exploration of limits of series of discrete numbers. And that is EXACTLY what is involved in the problem. If you want to ignore the relevant mathematical field and apply your own personal logic, you can do so, but if your logic leads you to a result that conflicts with standard limit theory, then your result is what is non-standard. (Many problems in limits and probability started out with an answer that seemed "logical," only to be shown later by mathematics to be wrong.)

quote:

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(Borodog)
The problem is that these two propositions, are mutually exclusive, even though both follow the wording of the puzzle. The puzzle really and truly is NOT WELL FORMED. It has no "solution." You might as well say something like, "There's a ball in a bucket. Take the ball out, but when you do, leave it in." Now, is the ball in the bucket or not? Neither, because the problem has ceased to have any *meaning*.

Before I was firmly an "empty-bucketer" (good phrase, whoever that was). But now I'm
with Veridisaurus Rex. There REALLY IS NO SOLUTION, because there REALLY ISN'T A WELL FORMED PUZZLE.

---

Exactly! (Borodog's "in/out" question being analogous to my "red/blue" question several posts ago. If a question requires us to accept two material but contradictory facts, the answer is likely to be undefined. And that should not be at all surprising.)

[Ghost Post]

Wonko:

[ZenBeam]

[Ghost Post]

dave10000: OK, regarding the claim you made (again) that the demon does not finish his task(and ignoring the issue of convergence for the moment): You agree that midnight arrives. The demon performs each step during a specific (finite, non-zero length) time interval between 11:59 and 12:00. Every step performed by the demon is performed during a time interval that ends before 12:00. So every step is finished before 12:00. If each and every step is finished before 12:00, then at 12:00, all steps have been finished. That, again, is indisputable.

[Ghost Post]

Another PROOF:

There are two possibilities at midnight:
1) The urn is empty, or
2) The urn contains at least one ball.

It must be either 1) or 2), and not both.

I'll assume 2), derive a contradiction, therby proving 1).

If the urn contains at least one ball, let X be the lowest number labelling a ball in the urn (every ball is labelled with a positive integer, and every non-empty set of positive integers has a smallest one).

The ball labelled X was removed from the urn at 1/(2^(X-1)) minutes to midnight.

[dave10000]

extro:

You continue to attribute positions to me that I did not take, to deny math the way I learned it, at least, and to use math in ways that I learned, at least, were incorrect. I see no reason to continue to debate under such circumstances, and I choose not to. Life is too short. If you (or others) wish to consider this a victory for the "empty-bucketers," so be it. The half of this bulletin board that does in fact dispute what you proclaim to be "indisputable" (which shows that we are arguing from different premises), will draw their own conclusions.

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

[Wonko the Sane]

Extro, I see what you're saying, but I don't agree. But seing as neither of us are going to agree and it's not particularly important to the puzzle anyway, it's a moot point and I'm just going to drop it.

[Ghost Post]

I don't mean to be argumentative, but I can't see where I attributed something to you that you did not say. You said "If there is no last step, then the Demon has NOT finished his task." I addressed this, showing that he DOES finish his task, although there is no last step, just as midnight arrives, although there is no last interval before midnight.

You say I deny math the way you learned it, and I suspect that's also not the case. All you have said is that using limits, you can't conclude what's left in the urn. I agree. Using limits, you can't conclude what's left, just as you can't use limits to evaluate the infinite sum 1+1+1+.... I'm not using limits. The fact that use of limits does not provide an answer does not mean there is no correct and well defined answer.

You say I use math in ways that you learned were incorrect, but you haven't said where. You haven't ever objected to my arguments, except to imply that because mine arrive at a definite conclusion, whereas yours do not, mine must be wrong.

I doubt you were ever taught that anything I've used in my arguments is incorrect. It's just that you are using limits, which are undefined in this case, whereas I am using well defined finite and infinite sets, and operations on them that are equally well defined, and yeild well defined results.

In the proof I give above, it should be easy to show my error if there is one.
A) I assume a non-empty urn, and show a contradiction, so conclude an empty urn. You might dispute that that's a valid principal (proof by contradiction), but do you?
B) So, did I really show a contradiction? Well, I assume the labels on the balls in the non-empty urn are positive integers (you might dispute that), and that any set of positive integers has a least element (you might dispute that). So I take from the urn the ball with the least label, X, and tell you it was removed from the urn 1/(2^(X-1)) minutes before midnight (you might dispute that that's correct). And 1/(2^(X-1)) is greater than 0, so it was removed before midnight (you might dispute that). So there's a contradiction - the ball was still in the urn after it was removed and never replaced (you might dispute that that's a contradiction).

I've think I've pointed out every point that anyone might dispute (you might dispute that), just to make it easy. Perhaps I'm dense, but I can't see how any of those are disputable.

[stoatboy]

Well, this certainly became a heated discussion in the time between when I last posted and today.

does anyone doubt that when the daemon cheats, painting new numbers on old balls, that there are an infinite number of balls in the urn? I would assume not--he adds an infinite number of balls to the urn, and never removes a single one.

But don't the same laws of time and space, and number of steps he takes to complete his task also apply to him? Is anyone arguing that when the daemon cheats, there are a finite number of balls in the urn after midnight because there was never a last step? After all, there sure can't be none in there, because, for instance, at 11:59, he added nine balls which were never thereafter removed.

Similarly, does anyone argue that daemon #1 is not also left with an infinite number of balls after midnight? He puts in 1 through 10, then removes ball 10, adds 11 through 20, then removes ball 20, and so. At 12:01, you know for a fact that ball 1 is in the urn, because it was put in and never removed. Same with ball 2,3,4,5,6,7,8 and 9. And 11-19, and so on, to infinity. Again, the fact that there's no last step doesn't effect your argument, nor the fact that at any time before midnight that you check on him, there are still a finite number of balls in the urn.

And so it surely seems as though if he doesn't cheat, daemon #2 would, after midnight have the exact same number of balls as daemon #1--after all, every time daemon #1 adds 10 balls, #2 also adds ten balls. Every time daemon #1 removes one ball, #2 removes 1 ball. There's an exact mapping between the number of balls added and removed by daemon #1 and daemon #2, and daemon #1 is left with infinity balls in his urn when he's done. Oh sure, they have different numbers on them, but what difference could that possibly make?

I'm sure everybody has a different answer to that somewhat playful and rhetorical question, but instead of answering it, I'll ask another question: say there's a rope that stretches an infinitely long distance. At each minute from now 'til the end of time, I cut off the next foot of rope and toss it in a nearby black hole. Meanwhile, my friend also has an infinite length of rope, but every minute he walks nine feet, and cuts out a one foot section of rope, and tosses that in a black hole. When we're done, how much rope will I have left? How much rope will he have left? If you answer is "the same amount, because you both removed the same amount every time," can you demonstrate that we have the same amount of rope by somehow mapping his leftover rope to my leftover rope? If your answer is, "different amounts," can you apply this to the rhetorical question in the previous paragraph? If your answer is, "you will never be done, therefore the question is meaningless", reread this posting and try again.

[Ghost Post]

Just for the record:
Demon #1 is left with an infinite number of balls (labelled with all the integers except those divisible by 10).

Demon #2, without cheating, is left with 0 balls.

Demon #2, with cheating, is left with an infinite number of balls (labelled with infinitely long strings of digits, which don't represent integers since they are not finite).

Now, demon #1 and demon #2 (not cheating) do, at each step, add 10, then remove 1. And 10 minus 1 is always 9. And at any moment before midnight, they have the same finite number of balls in their urns (some multiple of 9).

The "trick" is that demon #1 and #2 each remove an infinite subset of the infinite set of balls, but demon #1 is removing a proper subset, whereas demon #2's subset is equal to the original infinite set.

Regarding the rope, one guy tosses all the rope (in 1 foot sections) into the black whole - an infinite amount. The other tosses 1/9th (or 1/10th) of his infinite rope into the black whole - also an infinite amount. But he is left with an infinite amount. And you can't map an infinite amount onto nothing.

The point being, you can start with an infinite amount, and remove an infinite amount, and, depending on how you did the removing (and not how much you removed - it's always equally infinite), you can end up with nothing, a finite amount, or an infinite amount.

[Wonko the Sane]

Thank you extro, that was very well put. I've been looking for a mathematically good way to put it (instead of the conceptual way of saying that it isn't the amount, but the order that matters).
What all of us have to realize is that we're dealing with two infinites here. Most people have a hard enough time dealing with one of them, but trying to jam two of them together is even harder.
We've got a combination of Zeno's Paradox and a relatively simple logic problem. Let's reduce the number of balls and restate the problem.
Two demon are each given 500 numbered balls. Demon 1 is told to add 10 balls to his urn and remove the 10th one until he runs out of balls to add AND runs out of balls to remove.
Demon 2 is told to add 10 balls to his urn and remove whatever ball in the urn has the lowest number until he runs out of balls to add and or runs out of balls to remove.

Now, what will their urns look like in the end? Demon #2's urn will be empty and he'll have 500 balls on the floor, demon #1's urn will have 450 balls in it and he'll have 50 of them outside the urn on the floor.
What about the cheater? The cheater will add 9 balls, hide the 10th, and paint a 0 on the first ball. He'll end up with 450 balls in the urn as well. However, he wont have any balls on the floor, the 50 that should have been there will be under the rug. He'll also have different balls in his urn than demon #1 did before.
Extend this to infinite and you can get a better look at the situation.
Demon 1 will have and infinite number of balls in the urn and an infinite number of balls on the floor
Demon 2's urn will be empty and there will be an infinite number of balls on the floor
The cheater's urn will have an infinite number of balls in it, but there will be none on the floor because those will be jammed under the rug.

At least it seems that way to me.

[Ghost Post]

I think it's about time Minotaur posted the REAL solution so that we can finally end these (essentially) pointless arguments, which don't seem to be convincing anybody of anything they don't already believe. And maybe a solution to Monkey #3 would be nice too?

btw... Sorry, I didn't mean to offend back on page 2!

[Tom]

I just wrote a big post, and just deleted it. It was more argumentative I'm right/you're wrong rubbish.

It seems this all got a bit heated, hopefully we can all stay friends . By the way dave10000, the reason you got so much argument back is, I believe, because you write nice, clear posts that are good to argue against (as did a lot of other people, but dave10000 started to sound a bit annoyed). I have a lot of respect for your intellect and ideas; you argue well, it's just that I happen to disagree with you here. Well, I would if extro.. didn't always disagree with you first.

I'm guessing there's a few maths degrees flying around here, and none of us are fools, obviously. There doesn't seem to be any point arguing any more (but not because the question badly defined .. that is what we're arguing about!) as no one is going to change anyones minds .. including Minotaur when he puts up his answer, I suspect.

Could anyone get in touch with a respested mathematician about this? That said, they all got Monty Hall wrong, so who can we trust?

[Tom]

Oh, and Mathieu, I think it was me who got stroppy on page 2 .. I was in the middle of arguing and took it personally. Don't worry about it.

[ZenBeam]

I shouldn't post this, because I, too, am getting tired of this. Nevertheless...

quote:

---

There are two possibilities at midnight:
1) The urn is empty, or
2) The urn contains at least one ball.
It must be either 1) or 2), and not both.

---

There is a third possibility, that the number of balls is undefined. Thus, a contradiction in 2) doesn't prove 1).

[Gandalf the White]

Mathmematicians got Monty hall wrong? Who? I'd never seen or heard of "Let's make a deal", but when someone explained the problem to me for the first time, thankfully explaining it right, I understood that I should be switching straight away and I'm terrible at mathematical reasoning.

------------------
How many Grey Labryntians does it take to change a lightbulb?

Precisely 0.99999...

[Tom]

I founf it, finally. Here you are. Read the "'So Easy' to Blunder" section.