Stochastic Supertasks

I really loved this illustration of the paradoxes of infinity from Infinite Series, so much so that I’ll sum it up here.

What if you could do one task an infinite number of times over a finite time span? The obvious thing to do, granted this superpower, is start depositing money into your bank account. Let’s say you decide on plunking in one dollar at a time, an infinite number of times. Not happy at having to process an infinite number of transactions, your bank decides to withdraw a 1 cent fee after every one. How much money do you have in your bank account, once the dust settles?

Zero bucks. Think about it: at some point during the task you’ll have deposited N dollars in the account. The total amount the bank takes, however, keeps growing over time and at some point it’s guaranteed to reach N dollars too. This works for any value of N, and so any amount of cash you deposit will get removed.

In contrast, if the bank had decided to knock off 1 cent of your deposit before it reached your bank account, you’d both have an infinite amount of cash! This time around, there is no explicit subtraction to balance against the deposits, so your funds grow without bounds.

Weird, right? Welcome to the Ross–Littlewood paradox. My formulation is a bit more fun than the original balls-and-urns approach, but does almost the same job (picture the balls as pennies). It does fail when it comes to removing a specific item from the container, though; in the Infinite Series video, the host challenges everyone to figure out what happens if you remove the median ball from the urn after adding nine, assuming each ball has a unique number assigned to it. Can’t do that with cash.

My attempt is below the fold.

Let’s start with the first step. We add nine balls, numbered 1 to 9, and take away the median one (5). What happens to ball #1 as we continue to add more balls to the urn? It has to stay in there, since it is less than the median and subsequent additions will not change that. So when finish the supertask, we know that at least one ball is still sitting in the urn.

Can we extend that? Easily. Suppose the urn contains 2N – 1 balls at some point in future. The median of that is the Nth ball, and by the same logic we know the first N-1 balls must stay in the urn. Since our choice of N was arbitrary, we can guarantee without limit and thus there must be an infinite number of balls remaining behind.

The amount of balls we add at each step only matters in one case, when that number is one. There, you just remove each ball you put in. Adding two balls can be a bit awkward, but after the second step you’ll either have balls #1 or #2 remaining, depending on how you did the median, plus ball #4.  From here on, that indeterminate ball is guaranteed to be less than the median. And things only get more sure from there.

As for the stochastic case… ah, why ruin it? Go check out the video to see the solution laid out in full.