Consider I have $S$ elements $1,\ldots,3000$.
And $30,000$ $3-sets$ in $C$.
(eg.$C = [[1,2,3],[4,5,6]\ldots]] 29,998 \text{ more to go}$)
This I find really strange. A solution has $\operatorname{length}S/3$ units. Our solution must have $1000$ 3-element units from $C$.
$1000/30,000$ odds is $1/30$ odds.
Since Exact Three Cover is NP-complete it isn't expected to take $1/\operatorname{length}(C)$ odds to find a solution assuming one exists. (That would be a very practical probabilistic algorithm)
Question
If my math is correct, that there are an essential $1/30$ odds of randomly selecting a correct solution, then why does it still take exponential time?
Is my math wrong?
You would be right if there were only 30 ways to choose 1,000 sets of 3 out of 30,000 items. But the real number of possible choices is computed by this formula
$$ \dfrac{n!}{(n-r)!} $$
where $n$ is 30,000 and $r$ is 1,000. That is 30,000 times 29,999 times 29,998 times ... times 29,001. I don't know the exact value of this number, but it can't be any smaller than $29,000^{1000}$ and that number has over 4500 digits. So the probability of guessing a correct solution is nowhere near 1/30.
A refresher on permutations and combinations is probably in order.