Do probabilistic one-way functions imply deterministic one-way functions? Suppose $f$ is a probabilistic one-way function.  Then my question is, does there exist a construction of a deterministic one-way function $g$ based on $f$?
Or is it possible that probabilistic one-way functions exist but deterministic one-way functions do not?
EDIT: Probabilistic one-way functions are defined in definition 2.2.2 here.
Let's look at the definition in the linked thesis:

Definition 2.2.2 (probabilistic one-way function). A probabilistic function, $F$ (with randomness domain $R_n$), with a corresponding deterministic verifier, $V_F$ , is called one-way with respect to a well-spread distribution, $\mathbb{X}$, if for any PPT, $A$:
$$\Pr\bigl[x \gets X_n, r \gets R_n, V_F\bigl(A(F(x, r)), F(x, r)\bigr) = 1\bigr] &lt; \mu(n).¹$$
$F$ is called one-way if it is one-way with respect to the uniform distribution.

I will assume, since you did not specify anything about the distribution, that we are looking at the uniform distribution.
This definition is a bit problematic since it does not contain any kind of correctness guarantee for the verifier. In particular, take any function $F$ and define $V_F$ to be the constant $0$ function. According to the definition above the pair $(F,V_F)$ is one-way. Clearly this is not what was meant.
Definition 2.5.1 defines efficient verifiability, though this definition doesn't quite fit for a one-way function as defined above since it talks about an ensemble of keyed functions. However, in the same spirit, I will assume that Definition 2.2.2 meant to require the following from $F$:

Definition (Effcient Verification). A function, $F$ satisfies efficient verification if there exists a deterministic polynomial time algorithm, $V_F$, such that:
$$\forall x \in X_n, r \in R_n, V_F(x, F(x, r)) = 1.$$

If that is the case then the following holds.
Theorem Let $F : X_n \to Y_n$ be a probabilistic one-way function with randomness domain $R_n$. Then the deterministic function $G : X_n \times R_n \to Y_n$ defined as $G(x,r)=F(x,r)$ is a deterministic one-way function.
Proof. Let $A$ denote an arbitrary PPT algorithm such that $$\Pr[(x,r) \gets X_n\times R_n, G(A(G(x,r))) = G(x,r)] =\epsilon(n).$$
Note, that since the distribution over $X_n$ is uniform it holds that
\begin{align}
\epsilon(n) = &amp;\Pr[(x,r) \gets X_n\times R_n, G(A(G(x,r))) = G(x,r)]\\
=&amp;\Pr[x \gets X_n, r\gets R_n, G(A(G(x,r))) = G(x,r)]\\
=&amp;\Pr[x \gets X_n, r\gets R_n, F(A(F(x,r))) = F(x,r)]
\end{align}
where the last equality follows by the definition of $G$.
Now, consider the PPT algorithm $B$ that upon input $y$, executes $(x',r')\gets A(y)$ and outputs $x'$. By the definition of efficient verifiability, for all $x,x'\in X_n$, $r',r \in R_n$ it must hold that $$F(x',r') = F(x,r) \implies V_F(x',F(x,r))=1.$$
Thus,
\begin{align}
\epsilon(n) = &amp;\Pr[x \gets X_n, r\gets R_n, F(A(F(x,r))) = F(x,r)]\\
\leq &amp; \Pr[x \gets X_n, r\gets R_n, V_F(B(F(x,r)),F(x,r))=1] \leq \mu(n),
\end{align}
where the last inequality follows from the assumption that $F$ is a probabilistic one-way function.
Since the above holds for arbitrary PPT $A$, the theorem statement follows. $\quad\quad\Box$

¹Where $\mu$ denotes an unspecified negligible function.