Everybody Loves Raymond...

I regularly re-run my favorite old Raymond Smullyan puzzle (that actually goes back to "Annals of the New York Academy of Sciences," 1979, Vol. 321, although my version is an adaptation from Martin Gardner's presentation in his Colossal Book of Mathematics), and recently realized I failed to do so in 2018, so may as well remedy that now. Skip over if you've seen it before, but down below I have newly-tacked on a video tribute to Raymond from a prior 'Gathering For Gardner' Celebration:

Imagine you have access to an infinite supply of ping pong balls, each of which bears a positive integer label on it, which is its 'rank.' And for EVERY integer there are an INFINITE number of such balls available; i.e. an infinite no. of "#1" balls, an infinite no. of "#523" balls, an infinite no. of "#1,356,729" balls, etc. etc. etc. You also have a box that contains some FINITE number of these very same-type balls. You have as a goal to empty out that box, given the following procedure:

You get to remove one ball at a time from the finite box, but once you remove it, you must replace it with any finite no. of your choice of balls of 'lesser' rank (from the infinite supply box). Thus you can take out a ball labelled (or ranked) #768, and you could replace it with 27 million balls labelled, say #563 or #767 or #5 if you so desired, just as a few examples. The sole exceptions are the #1 balls, because obviously there are no 'ranks' below one, so there are NO replacements for a #1 ball.

Is it possible to empty out the box in a finite no. of steps??? OR, posing the question in reverse, as Martin Gardner does: "Can you not prolong the emptying of the box forever?" And then his answer: "Incredible as it seems at first, there is NO WAY to avoid completing the task" (emptying out the box). [bold added]

Although completion of the task is "unbounded" (there is no way to predict the number of steps needed to complete it, and indeed it could be a VERRRRY large number), the box MUST empty out within a finite number of steps!

This amazing result only requires logical induction to see the general reasoning involved:

Once there are only #1 balls left in the box you simply discard them one by one (no replacement allowed) until the box is empty -- that's a given. In the simplest case we can start with only #2 and #1 balls in the box. Every time you remove a #2 ball, you can ONLY replace it with a #1s, thus at some point (it could take a long time, but it must come) ONLY #1 balls will remain, and then essentially the task is over.

S'pose we start with just #1, #2, and #3 balls in the box... Every time a #3 ball is tossed, it can only be replaced with  #1 or #2 balls. Eventually, inevitably, we will be back to the #1 and #2 only scenario (all #3 balls having been removed), and we already know that situation must then terminate.

The same logic applies no matter how high up you go (you will always at some point run out of the very 'highest-ranked' balls and then be working on the next rank until they run out, and then the next, and then the next...); eventually you will of necessity work your way back to the state of just #1 and #2 balls, which then convert to just #1 balls and game over (even if you remove ALL the #1 and #2 balls first, you will eventually work back and be using them as replacements).

Of course no human being could live long enough to actually carry out such a procedure, but the process must nonetheless, amazingly, conclude after some mathematically finite no. of steps. Incredible! (a pity Cantor isn't around to appreciate this intuition-defying, ‘see-it-but-can-hardly-believe-it’ puzzle).