A lot of balls

"If you've done six impossible things this morning, why not round if off with breakfast at Millyways?" - Douglas Adams

Suppose you and your friend come into possession of a bag of infinite capacity, along with an infinite supply of numbered balls and a stopwatch with an infinite number of gradations. After watching the universe explode over mimosas and eggs Benedict at the Restaurant at the End of the Universe, the two of you decide to pass the time by running the following experiment:

1. You begin by placing two balls (balls 1 and 2) into the bag, then set the stopwatch to count down from 60 seconds.

2. When the stopwatch reads 30 seconds, you place two more balls (balls 3 and 4) into the bag and your friend pulls out the first ball you placed in the bag (ball 1).

3. When the stopwatch reads 15 seconds, you place two more balls (balls 5 and 6) into the bag and your friend pulls out the second ball you placed in the bag (ball 2).

4. The two of you repeat this process, you placing the next two balls into the bag and your friend removing the oldest ball, each time the remaining time on the stopwatch is halved (i.e. 7.5 seconds, 3.75 seconds, 1.825 seconds, etc...).

Question: How many balls will be left in the bag when the stopwatch reads 0 seconds?

Infinite.

EDIT:
I just realise there isn't a ball with the smallest number on it. Hmmm.

Originally posted by Thomaster
Infinite.

EDIT:
I just realise there isn't a ball with the smallest number on it. Hmmm.

The balls are numbered in order, starting with 1. That's the smallest number in the bunch.

Originally posted by PBE6
The balls are numbered in order, starting with 1. That's the smallest number in the bunch.

But when the stopwatch reads 0, infinite balls have been put in the bag. Half of that number has been picked out of it, but it still is infinite. In the bag, there isn't a ball with the lowest number.

I don't believe the stopwatch will ever reach exactly zero if the time is only ever halved and not decremented.

Originally posted by Tatanka Yotanka
I don't believe the stopwatch will ever reach exactly zero if the time is only ever halved and not decremented.

You don't think the stopwatch will ever count down from 60 seconds to 0 seconds? How do you boil an egg then? 😕

Originally posted by Thomaster
But when the stopwatch reads 0, infinite balls have been put in the bag. Half of that number has been picked out of it, but it still is infinite. In the bag, there isn't a ball with the lowest number.

Ah, I see what you mean. I don't think that matters much, but I've been wrong on this kind of thing before.

Originally posted by PBE6
Ah, I see what you mean. I don't think that matters much, but I've been wrong on this kind of thing before.

To answer your question, there are infinite balls in the bag.

Originally posted by Thomaster
To answer your question, there are infinite balls in the bag.

I have a different answer. Hopefully a few others will weigh in soon so we can discuss the merits of various solutions.

Originally posted by PBE6
You don't think the stopwatch will ever count down from 60 seconds to 0 seconds? How do you boil an egg then? 😕

"4. The two of you repeat this process, you placing the next two balls into the bag and your friend removing the oldest ball, each time the remaining time on the stopwatch is halved (i.e. 7.5 seconds, 3.75 seconds, 1.825 seconds, etc...). "

7.5; 3.75; 1.825 ...

If you would be so kind, continue that pattern to exactly zero kthx.

Originally posted by Tatanka Yotanka
"4. The two of you repeat this process, you placing the next two balls into the bag and your friend removing the oldest ball, each time the remaining time on the stopwatch is halved (i.e. 7.5 seconds, 3.75 seconds, 1.825 seconds, etc...). "

7.5; 3.75; 1.825 ...

If you would be so kind, continue that pattern to exactly zero kthx.

Let's represent the sampling times by the following formula:

T(n) = 60/2^n

Therefore, we have:

T(0) = 60/2^0 = 60
T(1) = 60/2^1 = 30
T(2) = 60/2^2 = 15
...
T(m) = 60/2^m

This sequences converges to the value 0 as m->inf, as can be proved using the methods of calculus (delta-epsilon proofs, specifically). Since the sequence converges, it is held that the stopwatch will indeed reach 0 seconds after an infinite number of sampling times have taken place. This method has been accepted as a solution to the various forms of Zeno's paradox since the 19th century. If you'd like to argue that the stopwatch will never reach 0 seconds, all I can do is offer to sit down with you and boil an egg. I've never had an philosophical argument regarding the reality of a boiled egg before, so it would be an interesting activity for all involved.

Originally posted by PBE6This sequences converges to the value 0 as m->inf, as can be proved using the methods of calculus (delta-epsilon proofs, specifically). Since the sequence converges, it is held that the stopwatch will indeed reach 0 seconds after an infinite number of sampling times have taken place.

I see. So after something (in this case; balls dropping) has happened an infinite number of times, the stopwatch will reach zero. If only I possessed a proper (read: different) understanding of the word "infinite" since my current understanding is presenting me with a paradox that you assert is not there.

I know: could you give me your definition of infinite?

There will be no balls left in the bag. 😲

But if the friend only removes one of the balls I've just put in, then there will be infinite balls in the bag. 😲😲

Originally posted by Thomaster
In the bag, there isn't a ball with the lowest number.

If there are any balls in the bag, I take a random one from that set. It has number X. If X is not the lowest number in that set, then the ball number X-1 is in that set. Repeat, until you reach number 1. No ball 0 exists, hence 1 must be the lowest number ball in your set. Contradiction.

But here's another argument. Each time we add two balls, and remove one. The net effect is to add one.

If we start at time T, then at time T(1/2)^n, there are n + 2 balls in the bag.

Let n -> infinity

Have we just proved that the operation is impossible, then?