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This is a classic puzzle. I first saw it about 1960. Do not consider physical restraints, your friend has a container of an infinite number of balls, each numbered sequentially. At 11:00 AM, he places balls 1-10 into your box and you throw out ball number 1. At 11:30 AM process is repeated for balls 11-20 and you throw out ball number 2. This continues by halving the time until noon each time and placing the next sequence of balls in your box and you throwing out the next in the sequence. Question: at noon, how many balls are in the box?

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This is a classic puzzle. I first saw it about 1960. Do not consider physical restraints, your friend has a container of an infinite number of balls, each numbered sequentially. At 11:00 AM, he places balls 1-10 into your box and you throw out ball number 1. At 11:30 AM process is repeated for balls 11-20 and you throw out ball number 2. This continues by halving the time until noon each time and placing the next sequence of balls in your box and you throwing out the next in the sequence. Question: at noon, how many balls are in the box?

Noon will never be reached since you keep splitting in 1/2

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"not reaching noon" , which Sindyin proposes, implies that the ball transfers occur at a fixed space and that time slows down. Rather, as time moves inexorably forward, the speed of the transfers (each of which adds 9 balls) increases hyperbolically towards infinity. At noon, there are an infintie number of balls in both boxes.

Noon will never be reached since you keep splitting in 1/2

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Since noon is reached after an infinite number of operations, the number of balls at the end of each operation increasing by 9, the number of balls in the box is 9 times infinity = infinity.

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There are two arguments for this puzzle. One is that each transaction nets a +9 addition to the box. The other is that an infinite number of balls were added, and an infinite number were removed and the answer would then be zero. But you can't try and say infinity-infinity=0. It doesn't work that way. If it did, we could divide by zero as well and there would be no math. If you were to graph this puzzle, you get a hyperbolic curve clearly approaching positive infinity as time approaches noon from the left. At noon there are an infinite number of balls in the box.

Edited by jjcanny1
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The other is that an infinite number of balls were added, and an infinite number were removed and the answer would then be zero.

The argument is not that infinity number of balls were removed, and it is not about 'infinity - infinity' tricks.

The argument is that each particular ball was removed at specific moment. You can provide a formula that takes the number of the ball and gives the exact time it is removed. Each particular ball is inserted once and then removed once before noon, thus no ball can remain inside the box at 12:00.

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Two correct answers! witzar and max0r4axor came to the right conclusion.

I think this paradox is a lot deeper than it seems. The mathematical community, at least, doesn't seem to agree on a single answer to this problem.

This is known as the ross-littlewood paradox. The categories of answers run the gamut from 'the box is infinitely full', to 'box is empty', to 'Problem is underspecified', to 'Problem is ill-formed'. For instance, here's an article by mathematician John Byl published in 2000 that states that the box is full (http://www.csc.twu.ca/byl/little.fin.doc )

Edited by bushindo
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... at the time any ball was removed, there remained in the box the following number of balls: 9 x No. of removed ball, so the process cannot be completed to result in no balls. Another way of concluding this is that for the process to be completed, the last ball to be removed would still leave behind it 9 x the number of that last ball, which is an absurdity.

For every ball dropped in the box, it is possible to calculate the time at which it was removed.

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This is a classic puzzle. I first saw it about 1960. Do not consider physical restraints, your friend has a container of an infinite number of balls, each numbered sequentially. At 11:00 AM, he places balls 1-10 into your box and you throw out ball number 1. At 11:30 AM process is repeated for balls 11-20 and you throw out ball number 2. This continues by halving the time until noon each time and placing the next sequence of balls in your box and you throwing out the next in the sequence. Question: at noon, how many balls are in the box?

Here is an interesting twist to this puzzle

Suppose that we have an infinite number of balls, and we label them with odd numbers 1, 3, 5, 7, 9, 11, and so on.

1) On turn i, for i = 1, 2, 3, ..., we put 10 balls with the smallest indexes into the box. So, for turn 1, we put in balls (1,3, 5, ..., 19); for turn 2, we put in balls (21, 23, ..., 39 ), and so on.

2) On turn i, after we are done with step 1), we 'remove' the ball with the index i from the box by relabeling it as ball 2*i. So, for turn 1, we put in balls (1,3, 5, ..., 19), and then we relabel ball 1 as ball 2. For turn 2, we relabel ball 2 as ball 4, and so on. Notice that we are not physically taking any ball from the box at all; we are simply relabeling balls i as ball 2*i during every turn.

Now, since we can argue that for any ball index i, we can calculate the time after which there is no such indexed ball in the box, then the box must be empty when the process is finished. For instance, after turn 179, there is no ball in the box that is numbered 179.

However, another argument is that since we never physically remove any ball from the box, the box must be infinitely full.

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I'll describe two other twists: but first let's establish a test for emptiness of the box.

To claim that a ball or balls remain in the box, their number must be specified.

Experiment #1: At turn n, place the two balls numbered 2n and (2n)-1 into the box, and then remove from the box the ball that has the lowest number.

Experiment #2: At turn n, place the two balls numbered 2n and (2n)-1 into the box, and then remove from the box the ball that has the highest number.

Experiment #3: At turn n, place the two balls numbered 2n and (2n)-1 into the box, and then remove from the box any ball, chosen blindfolded.

Perform an infinite number of steps [by infinite I mean Aleph null, the countable infinity]

Do this in finite time by performing the steps at times 1, 1/2 , 1/4 etc minutes, so we're done in 2 minutes.

If you're uncomfortable with breaking the speed of light with objects that have mass, then simply assign and remove mass-less integers to a set.

Can / do the numbers of balls left in the box [integers in the set] differ among these three experiments?

Putting it another way, can the number of remaining balls depend on their ink patterns?

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I think this paradox is a lot deeper than it seems. The mathematical community, at least, doesn't seem to agree on a single answer to this problem.

This is known as the ross-littlewood paradox. The categories of answers run the gamut from 'the box is infinitely full', to 'box is empty', to 'Problem is underspecified', to 'Problem is ill-formed'. For instance, here's an article by mathematician John Byl published in 2000 that states that the box is full (http://www.csc.twu.c.../little.fin.doc )

The wiki discussion of this paradox contains some loose arguments:

The impossibility of completion is compared to Zeno's paradox which asks whether a unit distance can be walked in an infinite geometric sequence of steps. But here the state of the walker at "completion" of this sequence is certain, so long as one is willing to consider completion of any infinite sequence.

Another comparison that would be invalid to make is that with Thompson's Lamp, where an infinite sequence of ON and OFF turns are taken, and a final state is asked. Here the outcome is undefined because a "last" or "highest" integer must exist to give the answer. No such integer exists.

The present paradox [and twists] differ from both.

I believe that if one admits to a completion of an infinite sequence of steps, each of the three experiments of the previous post has a defensible outcome.

<ahttp://brainden.com/forum/uploads/emoticons/default_wink.png' alt=';)'> and resembles, somewhat, Hilbert's hotel paradox.

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