Balls and vase $-$ A paradox? Question
I have infinity number of balls and a large enough vase. I define an action to be "put ten balls into the vase, and take one out". Now, I start from 11:59 and do one action, and after 30 seconds I do one action again, and 15 seconds later again, 7.5 seconds, 3.75 seconds...
What is the number of balls in the vase at 12:00?
My attempt
It seems like that it should be infinity (?), but if we consider the case:
Number each balls in an order of positive integers. During the first action, I put balls no. 1-10 in, and ball no.1 out, and during the $n^{\text{th}}$ action I take ball no. $n$ out. 
In this way, suppose it is at noon, every ball must have been taken out of the vase. So (?) the number of balls in the vase is
Zero???
My first question: if I take the ball randomly, what will be the result at noon? (I think it may need some probability method, which I'm not familiar enough with.)
Second one: is it actually a paradox?
Thanks in advance anyway.

This is essentially a pointwise convergence versus uniform convergence situation. For each ball separately, we have convergence to the state where that ball is not in the vase: as you say, every ball gets taken out before $12{:}00$. But if you look at the number of balls in the urn, that tends to infinity. So whether the process converges to the empty vase depends on what metric we use to define convergence.
A similar thing happens in real analysis. Define a sequence of functions $f_n:[0,1]\to\mathbb R$ as follows. $f_n(0)=0$, $f_n(1/2n)=2n$, $f_n(1/n)=0$, with $f_n(x)$ increasing linearly between $0$ and $1/2n$, decreasing linearly between $1/2n$ and $1/n$, and $0$ between $1/n$ and $1$.
Now for any $x$, $f_n(x)\to 0$. This means that $f_n$ converges pointwise to the zero function. But $\int_0^1f_n(x)\mathrm dx=1$ for every $n$. So pointwise convergence doesn't tell us how the integral behaves, and if that's what we care about then we need to define convergence differently.
