Confusion regarding a supertask problem in elementary probability The subject is example 6a from chapter 2 of "A First Course in Probability" by Sheldon Ross (9th edition):

Suppose that we possess an infinitely large urn and an infinite collection of balls labeled ball number 1, number 2, number 3, and so on. Consider an experiment performed as follows: At 1 minute to 12 P.M., balls numbered 1 through 10 are placed in the urn and ball number 10 is withdrawn. (Assume that the withdrawal takes no time.) At 1/2 minute to 12 P.M., balls numbered 11 through 20 are placed in the urn and ball number 20 is withdrawn. At 1/4 minute to 12 P.M., balls numbered 21 through 30 are placed in the urn and ball number 30 is withdrawn. At 1/8 minute to 12 P.M., and so on. The question of interest is, How many balls are in the urn at 12 P.M.?


The answer to this question is clearly that there is an infinite number of balls in the urn at 12 P.M., since any ball whose number is not of the form 10n, n >= 1, will have been placed in the urn and will not have been withdrawn before 12 P.M. Hence, the problem is solved when the experiment is performed as described.


However, let us now change the experiment and suppose that at 1 minute to 12 P.M., balls numbered 1 through 10 are placed in the urn and ball number 1 is withdrawn; at 1/2 minute to 12 P.M., balls numbered 11 through 20 are placed in the urn and ball number 2 is withdrawn; at 1/4 minute to 12 P.M., balls numbered 21 through 30 are placed in the urn and ball number 3 is withdrawn; at 1/8 minute to 12 P.M., balls numbered 31 through 40 are placed in the urn and ball number 4 is withdrawn, and so on. For this new experiment, how many balls are in the urn at 12 P.M.?


Surprisingly enough, the answer now is that the urn is empty at 12 P.M. For, consider any ball—say, ball number n. At some time prior to 12 P.M. [in particular, at (1/2)^(n-1) minutes to 12 P.M.], this ball would have been withdrawn from the urn. Hence, for each n, ball number n is not in the urn at 12 P.M.; therefore, the urn must be empty at that time.

My objection is as follows: How do the cases differ? All I see is that we are changing the order in which the balls are removed, not the quantity of balls removed in each time instant.
For every time instant, let us create a tuple of $(ball\  removed,balls\ added)$. Over the entire super task, the collection of our tuples will be:
$\left\{
(1,\left\{1,2,...10  \right\}),(2,\left\{11,12,...20  \right\}),(3,\left\{21,22,...30  \right\})\ ...  \right\}$
Now, the nth ball is removed at $(1/2)^{n-1}$ minutes to 12 PM. At the same instant, the set of balls $\left\{10n-1,...,10n  \right\}$ are added. Thus, for every time instant where a ball is removed, we are also adding a greater number of balls to the urn.
Ultimately, the number of resultant balls at $(1/2)^{n-1}$ minutes to 12 PM $= \ 9n$.
I can't seem to understand the reasoning put forth by the author.
 A: You're right that the number of balls after $n$ steps is $9n$ in either method. The question is about what happens as you take the limit $n\to\infty$, and as is often the case, infinities are weird.
In the first method, $\frac{9}{10}$ of the balls are put in, and stay in; the set of balls $k$ which are still in at $n=\infty$ is $\{k \in \mathbb N : 10 \not \mid k\}$. In the second method, every ball is inserted, and then later removed; the set of balls still in at $n=\infty$ is empty.
This may seem strange; if we have $9n$ balls after $n$ steps, shouldn't we have $\infty$ balls after $\infty$ steps? Well, sort of. The limit of the number of balls is $\infty$, yes, but the limit of the set of balls is empty. These limits don't match, because infinity is weird.
A: The urn will be empty if and only if given $n\in \mathbb N $, the $n$th ball is not in the urn. In the first case, ball $1$ is still in the urn. But in the second case, you can actually explicitly compute when that ball will be removed from the case. So if the case is not empty, then you can find some ball in it. But you can't because tell me any candidate, and I can tell you when I took it out.
A: This is known as the Ross-Littlewood Paradox.
And yeah, it's called a paradox not for nothing.
In fact, I personally think it's an excellent argument against the very idea of completing supertasks.
