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Imagine a game where at each step, a person is given 10 dollar bills (numbered in the order they are given, starting at 1), and then one bill is removed from the bills that the person has received during the course of the game. Suppose the game goes on for a countably infinite number of steps. How many bills are left afterwards? As you may have already surmised, the answer depends on how the bills are removed.

If the lowest numbered bill is removed each time, then no bills are left. The removal of every bill can be traced to a specific step, so no bill can be identified which could remain in the person's collection. There are other removal patterns that result in total loss as well, but this one is simple.

If the highest numbered bill is removed each time, then there will be an infinite number of bills left, because each of the infinite steps gives the person nine bills which will never be removed. Again, there are other methods to achieve this, but this one is simple.

There are also removal patterns one could specify to end up with any finite number of bills, but it's basically just waiting until the player has the right number of bills and then implementing a pattern of total removal starting at a bill of some specific number. I can give examples later, if they are needed.

What if the removal process is randomized? At step 1, bills 1-10 are given and one of those 10 bills is randomly selected for removal. At step two, bills 11-20 are given, and one of the 19 remaining bills is randomly selected for removal. And so on.

What I'm after is how to figure out what the probability of either no or only a finite number of bills remaining after an infinite number of steps. I know about the Infinite Monkeys Theorem, but that is about a finite string being contained in a countably infinite string and this issue is about a countably infinite string being part of the set of patterns that remove all bills, and that difference seems significant. It seems like I can't just compare the set of possibilities that do and don't have this result because they're both aleph-1 sets, and I have a proof that equinumerous infinite sets of possibilities can't be assumed to just balance out in that way (I don't know if one exists elsewhere, but I got myself to that point).

TBH, I feel like I should be able to work this out, but I'd really appreciate some help here because I'm writing my philosophy MA thesis and this is a minor point, but one I'd really like to leave in if possible. Of course, any assistance given will be cited (we can discuss how you want this to be done). If I work it out on my own (or think I have done so), I'll post a follow up and see what y'all think.

Thanks in advance!

John
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  • You're not framing this precisely. In particular, I'm not certain what you mean by "end up with" or "remaining." I think what you mean to ask is the probability that all but (perhaps) finitely many specific bills will be removed by the game conclusion with probability 1. – Robert Shore Jun 15 '19 at 03:18
  • I suppose I was being a bit loose with language there, thanks. "End up with" simply means the bills that the player still possesses after the game is complete (that is, after a countably infinite number of steps have iterated). Bills that are "remaining" are those that are still in the player's possession at whatever time is indicated in the sentence. I will think about ways I can make that more explicit, but I think you answered my question already =) – John Jun 15 '19 at 04:38
  • The reason your phrasing is a little confusing is that obviously the number of bills that you have at each step is always increasing, so in one sense you "end up" with infinitely many bills. But you're focusing on the fate of each individual bill, not the total number of bills. – Robert Shore Jun 15 '19 at 04:53
  • I'm interested in both the individual bills and the total number because the fate of individual bills can determine the number of bills the player has after all the steps are complete. That said, I'll think about what you have to say about the phrasing and make sure I'm avoiding that potential confusing in the thesis. Thanks for clarifying – John Jun 15 '19 at 05:46
  • Related: https://math.stackexchange.com/questions/869916/a-strange-puzzle-having-two-possible-solutions/ – Asaf Karagila Jun 15 '19 at 06:00

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The probability that Bill number 1 persists through to the end is:

$$\prod_{k=1}^\infty (1-\frac{1}{9k+1}).$$

This infinite product converges (to a number greater than $0$) if and only if:

$$\sum_{k=1}^\infty\ln (1-\frac{1}{9k+1}) \text{ converges.}$$

Taking the Taylor series of $\ln(1-x) = -\sum \frac {x^n}{n}$, we find that this infinite sum (all terms of which are negative) is larger (in absolute value) than a harmonic series, so the infinite product diverges to $0$.

For bills entering at step $k$, you're simply removing the first $k-1$ terms from the product, so the infinite product still diverges to $0$. In short, for any specific bill you care to identify, the probability that it will eventually be removed is $1$, so there is no bill with a non-zero probability of persisting until the end of the game.

Robert Shore
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  • Thank you for the formula! That tracks perfectly with my intuitions about the case. That P=0 would still be possible, as per the Infinite Monkeys Theorem, yes? – John Jun 15 '19 at 04:42
  • I don't think the Infinite Monkeys Theorem has any application to your problem. If you haven't already done so, upvotes as well as acceptances are always appreciated. – Robert Shore Jun 15 '19 at 04:51
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    Fair enough, I'll have to mill this over for a bit to make sure I'm really getting it. I did hit the accept (assuming that's the check mark), and I upvoted, but since I'm brand new, the vote doesn't display publicly. Thanks again! – John Jun 15 '19 at 05:41