How many balls are in the box?

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December 22nd 2010, 03:26 AM
phillipshwong

How many balls are in the box?

One minute before 12:00, you place 10 balls labeled from 1 to 10 in the box. You randomly pick one ball out. In 1/2 min before 12:00, you put 10 ball in the box, and label them from 11 to 20. You then randomly pick one ball out. You continue in the manner until you reach 12:00. What is the number of balls in the box at 12:00?

In other worlds, for ( 1/2)^(n-1) before 12:00, we put and additional 10 balls in the box, and randomly pick one ball out. The question is how many balls in the box at 12:00 ?
December 22nd 2010, 05:23 AM
Soroban

Hello, phillipshwong!

This is classic problem . . .

Quote:

$\text{One minute before 12:00, you place 10 balls labeled from 1 to 10 in the box.}$
. . $\text{Then you randomly remove one ball.}$

$\frac{1}{2}\text{ min before 12:00, you add 10 ball to the box, labeled 11 to 20.}$
. . $\text{Then you randomly remove one ball.}$

$\frac{1}{4}\text{ min before 12:00, you add 10 balls to the box, labeled 21 to 30.}$
. . $\text{Then you randomly remove one ball.}$

$\text{You continue in this manner until 12:00.}$

$\text{What is the number of balls in the box at 12:00?}$

The surprising answer is: There are no balls in the box at 12:00!

At 1 minute before 12:00, we add balls #1 to #10 and remove one ball.
. . . . Suppose it is #1.

At $\frac{1}{2}$ minute to 12:00, we add #11 to #20 and remove one ball.
. . . . Suppose it is #2.

At $\frac{1}{4}$ minute to 12:00, we add #21 to #30 and remove one ball.
. . . . Suppose it is #3.

At $\frac{1}{8}$ minute to 12:00, we add #31 to #40 and remove one ball.
. . . . Suppose it is #4.

We continue this process until 12:00.
We can generalize the procedure:

At $\frac{1}{2^{n-1}}$ minute to 12:00, we add 10 balls and remove one ball.
. . . . Suppose it is number $n.$

At every fraction of a minute, we are adding 9 balls to the box.
Yet by 12:00, the box will be empty!

What about ball #35?
It was removed $\frac{1}{2^{34}}$ minute before 12:00.

What about ball #12,501
It was removed $\frac{1}{2^{12,500}}$ minute before 12:00.

Get it?
December 22nd 2010, 05:44 AM
phillipshwong

Quote:

Originally Posted by Soroban

Hello, phillipshwong!

This is classic problem . . .

The surprising answer is: There are no balls in the box at 12:00!

At 1 minute before 12:00, we add balls #1 to #10 and remove one ball.
. . . . Suppose it is #1.

At $\frac{1}{2}$ minute to 12:00, we add #11 to #20 and remove one ball.
. . . . Suppose it is #2.

At $\frac{1}{4}$ minute to 12:00, we add #21 to #30 and remove one ball.
. . . . Suppose it is #3.

At $\frac{1}{8}$ minute to 12:00, we add #31 to #40 and remove one ball.
. . . . Suppose it is #4.

We continue this process until 12:00.
We can generalize the procedure:

At $\frac{1}{2^{n-1}}$ minute to 12:00, we add 10 balls and remove one ball.
. . . . Suppose it is number $n.$

At every fraction of a minute, we are adding 9 balls to the box.
Yet by 12:00, the box will be empty!

What about ball #35?
It was removed $\frac{1}{2^{34}}$ minute before 12:00.

What about ball #12,501
It was removed $\frac{1}{2^{12,500}}$ minute before 12:00.

Get it?

There are different variation to the problem. In one case, with each addition of 10 balls, you always pick the largest labelled ball in the box. This means there are no balls for { 10*n, n>0}. At 12:00, there are thus, infinity many balls. This is obvious.

In your case, with each ( 1/2)^(n-1) before 12:00, you add 10 balls in the box, and pick the ball labeled n. So, at 12:00, all balls are gone, since what every ball label, q, q is out the box at ( 1/2)^(q-1) before 12:00. This means there are no balls at 12:00.

In my case, with each (1/2)^(n-1) before 12:00, in addition to adding 10 balls in the box, i randomly pick one ball out of all the balls in the box! So, what is the answer? The are no balls at 12:00, and an argument can be made that for each ball i in the box, and the number of trials n, the probability of i not being picked as n--> infinity is zero.

I want to know if there is another prove for the conclusion. I am very interested in the name of the problem if there is any.
December 22nd 2010, 05:04 PM
Archie Meade

Come on lads!

Granted, I'd be dropping the snookerballs all over the place with a few seconds left.

But the arithmetic sequence 9, 18, 27, 36, 45, 54, 63,....
"converges" on what exactly ?
December 22nd 2010, 05:33 PM
phillipshwong

why are you asking a question unrelated to anything said in this thread?
December 22nd 2010, 08:23 PM
matheagle

when in doubt I always go with 42
and I did meet Doug Adams a few years ago
He was a few blocks from where I live signing books with Terry Jones
December 23rd 2010, 02:00 AM
emakarov

Quote:

Originally Posted by phillipshwong

why are you asking a question unrelated to anything said in this thread?

Well, it's not unrelated to anything in this thread; the arithmetic sequence is the number of balls in the box after each placement. I also find each quite puzzling that the number can increase unboundedly and still turn into zero in the end. Even more puzzling is that the answer can depend on which ball is picked out each time. This is probably because I don't understand very well how the final set of balls is defined.
December 23rd 2010, 02:09 AM
CaptainBlack

Quote:

Originally Posted by emakarov

Well, it's not unrelated to anything in this thread; the arithmetic sequence is the number of balls in the box after each placement. I also find each quite puzzling that the number can increase unboundedly and still turn into zero in the end. Even more puzzling is that the answer can depend on which ball is picked out each time. This is probably because I don't understand very well how the final set of balls is defined.

It is a supertask that envisages the completion of an infinite number of steps (effectively you are dealing with a compete infinity rather than a potential infinity) this is what lets in the (apparent) paradoxes (the sequence of the number of balls in the box is increasing, but the final result is that every ball has been removed).

Some of us might conclude, with the ancients, that from this that such supertasks are illegitimate processes in some way.

CB
December 23rd 2010, 06:25 AM
Opalg

The "final" state, at 12:00, is only reached by means of a limiting process, because there are infinitely many steps at which balls are added and removed. The number of ball left in the box at 12:00 is therefore " $\infty-\infty$ ", and it is well known that this expression is undetermined. In some situations (such as Soroban's scenario above) the result can be 0. But any number, from 0 to infinity inclusive, is a plausible outcome.
December 23rd 2010, 06:31 AM
matheagle

I'll stick with 42
The mice told me so
December 23rd 2010, 06:55 AM
CaptainBlack

Quote:

Originally Posted by Opalg

The "final" state, at 12:00, is only reached by means of a limiting process, because there are infinitely many steps at which balls are added and removed. The number of ball left in the box at 12:00 is therefore " $\infty-\infty$ ", and it is well known that this expression is undetermined. In some situations (such as Soroban's scenario above) the result can be 0. But any number, from 0 to infinity inclusive, is a plausible outcome.

For random ball selection we can say:

If we calculate the probability p_n that ball #1 is removed on or before the n-th selection and the limit as n goes to infinity is less than 1 then we can say with probability 1 there are an infinite number of balls in the box at 12:00 (or rather that for any natural number M the probability that there ane more than M ball remaining is 1) (this would work since the probability that ball #1 is removed is an upper bound on the probability than any given ball is removed)

CB
December 23rd 2010, 09:36 AM
phillipshwong

One variation to this problem is that at ( 1/2)^(n-1) before 12:00 pm, the n labeled ball is taken out. The answer might be unintuitive, but you can still convince yourself this by picking any random labeled ball i, and you know that it is out at ( 1/2)^(i-1) before 12:00. The notion of infinity never comes into mind.

Let P(i) means " The i ball taken out at (1/2)^( i-1) before 12:00".

1. Any random i in N( natural numbers), P( i ) is true.

1 lead us to:

2. All i in N is such that P( i ) is true.

What is 2 mean? The cardinality of N is the number of balls in the box. P(i) is a mapping between time points, and N( the number of balls). What is the assumption? The only important assumption is that the size of N is infinite.
December 25th 2010, 11:18 PM
CaptainBlack

Quote:

Originally Posted by CaptainBlack

For random ball selection we can say:

If we calculate the probability p_n that ball #1 is removed on or before the n-th selection and the limit as n goes to infinity is less than 1 then we can say with probability 1 there are an infinite number of balls in the box at 12:00 (or rather that for any natural number M the probability that there ane more than M ball remaining is 1) (this would work since the probability that ball #1 is removed is an upper bound on the probability than any given ball is removed)

CB

On second thoughts we would have to do rather more than this since the removal or not of ball n is not independent of that of other balls.
December 25th 2010, 11:19 PM
CaptainBlack

An interesting discussion (with references) of this and similar supertasks (but not random ball removal!) can be found >>here<<

CB