Do probabilities matter for resolving nondeterminism?

Consider a nondeterministic parity automaton $A = (Q, \Sigma, \Delta, q_0, \Omega),$ where: $Q$ is a finite set of states, $\Sigma$ is the input alphabet, $\Delta \subseteq Q \times \Sigma \times Q$ is the transition relation, $q_0 \in Q$ is the initial state, and $\Omega : Q \to \mathbb{N}$ is the parity acceptance condition. We say that a probabilistic automaton $P$ is obtained from a memoryless resolver for $A$ if, for each state $s \in Q$ and input symbol $a \in \Sigma$, the nondeterministic set of successors

$$\{ t \in Q \mid (s, a, t) \in \Delta \}$$

is replaced by a probability distribution $D_s^a$ over that set. The resulting probabilistic automaton $P$ retains the same state set $Q$, alphabet $\Sigma$, and initial state $q_0$, but transitions are decided by the probability distributions $\{ D_s^a \}_{s \in Q, a \in \Sigma},$ which define the probability of moving from state $s$ to state $t$ upon reading symbol $a$. We say that $A$ is memoryless stochastically resolvable if there exists a probabilistic automaton $P$, obtained from a memoryless resolver as described above, such that for every infinite word $w \in \Sigma^\omega$ accepted by $A$, the probabilistic automaton $P$ also accepts $w$ with probability 1. Moreover, we say that $P$ is obtained from uniform memoryless resolver if, for all $s \in Q$ and $a \in \Sigma$, the distribution $D_s^a$ is uniform over the set $\{ t \in Q \mid (s, a, t) \in \Delta \}.$ This leads to the question: If a nondeterministic parity automaton $A$ is stochastically resolvable using a memoryless resolver, is it always possible to construct such a resolver using uniform distributions?