Probability of getting three zeros when sampling a set Say we have a set of $12$ numbers - $(0,1,2,0,3,0,4,0,5,6,0).$
We go thru this set one number at a time, and choose a number with a probability $1/3$; we sample. What is the probability we end up with $3$ $0$'s out of the $4$ numbers we will likely choose(on average)? 
Is there a closed-form solution for this?
If I understand the problem, you have a list of 12 objects, 5 of which are special (the 0s).  (I use the word "list" instead of "set" since the latter implies there is no repetition among the objects).  Each of the 12 objects will be independently considered, and "chosen" with probability 1/3.  You want to know the probability of 3 of the special objects being chosen.

In that case, the 7 non-special objects are irrelevant.  You just want to know the probability that exactly 3 out of 5 independent events occur, when each one occurs with probability 1/3.  This is the probability of a binomial random variable with $n=5$ and $p=1/3$ taking on the value $3$.  The linked Wikipedia page has a formula for computing this.
The probability that you want is $(1/3)^4(2/3)^8 N$, where $N$ is the number of subsets of length $4$ containing exactly $3$ zeroes.