Following fkraiem's answer, I would share my thoughts.
Generally speaking, we do not know if randomness helps, i.e., P=BPP is an open question. So probabilistic-polynomial-time (PPT) adversaries may not equal to deterministic ones.
However, it seems that cryptography always (correct me if I am wrong) focuses on advantages (or something similar) related to PPT adversaries. In this case, it is sufficient to consider only deterministic adversaries. Here is a simple proof:
Suppose you have a PPT adversary A that has advantage $p$. Then you can fix all possible random coins and get a bunch of deterministic adversaries $A_1, A_2, ..., A_n$, where $A$ runs $A_i$ with probability $p_i$. Since $Adv(A) = p_1Adv(A_1) + \cdots + p_nAdv(A_n)$ and $1=p_1+\cdots+p_n$, there exists $k$ such that $Adv(A_k) \geq Adv(A) = p$. Note that $n$ may be exponential but each adversary $A_i$ runs in polynomial time. Also, knowing that $A_k$ exists does not imply that one can find it in polynomial time.
Update: The above argument can also be adapted to information-theoretic (i.e., computationally unbounded) adversaries.