First, a comment states that $Z_1\ge Z_2$ means the inequality holds surely, i.e., that it holds for every $\omega$ in the probability space $\Omega$. (There must be such a space, or else the notions of probability makes no sense, though it is quite common not to specify or even mention it explicitly.) But in practice, one uses the weaker statement $Z_1\ge Z_2$ almost surely (abbreviated a.s.), which means $\Pr[Z_1. Quite frequently, I suspect this is the intended meaning even when the author omits the “a.s.” part.
You cannot write the condition $Z_1\ge Z_2$ using the quantities you list. For that, you need the combined probability distribution, i.e. you need to know the probabilities $P_{ij}=\Pr[Z_1=x_i \mathbin\& Z_2=y_j]$. In this case, $Z_1\ge Z_2$ a.s. if and only if $P_{ij}=0$ whenever $x_i. If you don't include the “a.s.”, then even the full probability distribution won't provide the answer. You need the underlying probability space in order to answer the question in that case.