The Ross-Littlewood Paradox

• We have a giant vase, which is infinitely big, that can hold an infinite amount of balls.
• We also have an infinite amount of balls that we can insert into the vase.
• The balls are numbered in order 1, 2, 3, …
• We are adding balls at a rate of 10 balls per each task.
• Tasks are performed at decreasing timing, with first task starting at 11:59:00 AM:
  • The first task would is completed in half a min (1/2 min).
  • The second task, would be completed in half the amount, so a quarter of a min (1/4 min).
  • The third one, in half the last one’s amount, so one eigth of a min (1/8 min).
  • and so on and so forth…until, the last task to be completed by noon.

So basically, it drills down to performing an infinite amount of tasks, in an finite amount of time, which is the definition of what we refer as a supertask.

We start with task 1, where we add balls 1 to 10, and remove the first, which is ball 1. At task 2, we add balls 11-20, and remove the remaining first, which is ball 2. At task 3, we add balls 21-30, and remove the remaining first, ball 3… and so on and so forth.

Now the question is: how many balls would be in the vase at noon?

Let us look into what happened on the actual the steps. At step 1, we removed ball 1, and then step 2, removed ball 2, step 3, removed ball 3,… so at step n, we would remove ball n, and hence all the balls will be removed from the vase by the end of the tasks. So, the answer to our problem is none.

Apparently, the order in which we add balls, and remove them, affects how many balls are in the vase at noon. Let us look at an alternative scenario.

• At task 1, we add balls 1 to 10, and remove the last, which is ball 10
• At task 2, we add balls 11-20, remove the last, which is ball 20
• At task 3, we add balls 21-30, remove the last, ball 30,
• And so on and so forth…

So again, how many balls would be in the vase at noon?
This time, the answer is different, and it is Infinity. This is actually due to the fact that only balls which are multiples of ten are the ones being removed, so the vase will not contain balls 10, 20, 30,… yet all the other balls will still be there, hence the infinite amount.