Are deterministic adversaries as powerful as probabilistic adversaries? SOURCE states the following in the proof of Theorem 2:


  Without loss of generality, I will assume that A is deterministic. If A is randomized, we can determinize it by fixing a sequence of coins that maximizes A’s conditional success probability; it is easy to see that this cannot reduce A’s advantage.


Is this correct? Does it mean that we do not need to consider probabilistic adversaries in cryptographic security games?
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.