Classical observation: Given a "trinoulli" random variable ("trit") $X$ with possible states $0, 1, 2$, there always exist two Bernoulli random variables ("bits") $Y_1, Y_2$ which "simulate" $X$, namely the indicator random variables $$Y_1 = I_{\{X = 0\}}, \quad Y_2 = I_{\{X=1\}} $$ (Note that $I_{\{X = 2\}} = 1 - \max \{Y_1, Y_2 \}$, so $Y_1$ and $Y_2$ alone really are sufficient to completely determine $X$. Obviously $X$ determines $Y_1$ and $Y_2$.)
Question: Given a "qutrit" $Q$, which (if assumed to be in a pure state) I guess corresponds to a random variable taking values in $\mathbb{CP}^2$, can we always define two qubits $R_1, R_2$ whose values completely determine the value of $Q$ and vice versa?
And such that this is true for any choice of measurement basis for $Q$?
(I guess the measurement basises for $R_1$ and $R_2$ can be a function of the measurement basis for $Q$, although ideally they wouldn't have to be, because it seems like that might defeat the spirit of the question. I'm not really certain; it might not.)
Maybe for clarity: denote a choice of (computational) basis for $Q$ by $|0\rangle$, $|1 \rangle$, $|2 \rangle$, and (computational) basises for $R_1$ and $R_2$ respectively by $|0\rangle_{R1}, |1\rangle_{R2}$ and $|0\rangle_{R2}$, $|1 \rangle_{R2}$.
Observations: Obviously $Y_1$ and $Y_2$ are not independent, and equivalently the entropy of their joint distribution is less than the sum of their individual entropies. (I.e. they have non-zero mutual information.) Because they completely determine the value of $X$, and vice versa, their joint distribution has the same entropy as that of $X$.
So in the quantum case, I would guess that the two qubits $R_1$, $R_2$ would also have to have some statistical dependence in order to enable that the pair to completely determine $Q$ (which after all is only a $3 < 2 + 2$ level quantum system) and vice versa. Whether that has to manifest as quantum entanglement, I have no idea -- I am only familiar with the mathematics of the representation of a single qubit, not with systems of qubits or of individual qudits.
The classical observation clearly seems to imply that this is possible with sufficiently rigged choices of measurement basises for $Q$, $R_1$, and $R_2$, because for a fixed measurement basis $Q$ becomes a trinoulli random variable. But basically I'm wondering whether one can come up with a procedure which is "uniform" across all possible choices of measurement basises for $Q$.
Motivation: The motivation is that I am trying to test, and ideally extend, my understanding of the analogy between "classical" bits (Bernoulli RVs) and qubits. I thought I understood that fairly well, but although I understand how an arbitrary finite discrete RV can be simulated using Bernoulli RVs, I realized I have no idea how, or even whether, an arbitrary qudit can be simulated using qubits. This in spite of the fact that one frequently hears claims like "qubits are the fundamental building blocks of all discrete quantum systems". The Wikipedia article for qubits even says that only qudits with $2^N$ states can be simulated by qubits. Anyway the qutrit is the simplest possible non-qubit qudit.
This isn't necessarily an answer to your question, but my musings are much longer than a comment and I wanted to work some things out to understand your question more. And also to highlight some misunderstandings you seem to have about quantum information.
In your trinoulli example, you have a probability distribution $(p_0,p_1,p_2)$ and a random trinary variable $X$ having probability distribution given by $$ \operatorname{Pr}(X=0)=p_0, \quad\operatorname{Pr}(X=1)=p_1, \quad \text{and} \quad \operatorname{Pr}(X=2)=p_2. $$ You could "simulate" this random variable with a pair of correlated random binary variables $Y_0$ and $Y_1$ having joint probability distribution defined by the following \begin{align*} \operatorname{Pr}\big((Y_0,Y_1)=(1,0)\big) &= p_0\\ \operatorname{Pr}\big((Y_0,Y_1)=(0,1)\big) &= p_1\\ \operatorname{Pr}\big((Y_0,Y_1)=(0,0)\big) &= p_2\\ \operatorname{Pr}\big((Y_0,Y_1)=(1,1)\big) &= 0. \end{align*} Is this close to what you mean?
A state of a quantum system is not a random variable. Rather, the state of the system assigns a probability distribution to each possible measurement of that system. In finite dimensions, every orthonormal basis of a quantum system's corresponding state space defines a measurement. In your qutrit example, if the state of your quantum system is represented by a vector $|\psi\rangle\in\mathbb{C}^3$ of the form $$ |\psi\rangle = \alpha_0|0\rangle + \alpha_1|1\rangle + \alpha_2|2\rangle, $$ and you have some other orthonormal basis $\{|x_0\rangle,|x_1\rangle,|x_2\rangle\}$ of $\mathbb{C}^3$, then the probability distribution of the random variable that arises from performing this measurement is given by \begin{align*} \operatorname{Pr}(X=0)& = |\langle x_0|\psi\rangle|^2\\ \operatorname{Pr}(X=1) &= |\langle x_1|\psi\rangle|^2\\ \operatorname{Pr}(X=2) &= |\langle x_2|\psi\rangle|^2. \end{align*}
Analogous to your classical trinary example, you could "simulate" this qutrit system using two qubits as follows. Prepare the entangled state $|\phi\rangle\in\mathbb{C}^2\otimes\mathbb{C}^2$ represented by $$ |\phi\rangle = \alpha_0|10\rangle + \alpha_1|01\rangle + \alpha_2|00\rangle. $$ Then you could "simulate" measuring $X$ on the qutrit in state $|\psi\rangle$ by measuring your two-qubit system in state $|\phi\rangle$ according to the orthonormal basis $\{|y_0\rangle,|y_1\rangle,|y_2\rangle,|y_3\rangle\}$ defined by \begin{align*} |y_0\rangle &= \langle 0|x_0\rangle |10\rangle+ \langle 1|x_0\rangle |01\rangle + \langle 2|x_0\rangle |11\rangle\\ |y_1\rangle &= \langle 0|x_1\rangle |10\rangle+ \langle 1|x_1\rangle |01\rangle + \langle 2|x_1\rangle |11\rangle\\ |y_2\rangle &= \langle 0|x_2\rangle |10\rangle+ \langle 1|x_2\rangle |01\rangle + \langle 2|x_2\rangle |11\rangle \end{align*} and taking $|y_3\rangle$ to be orthogonal to those. Defining $Y$ to be the random variable produced by performing this measurement would yield the desired probability distribution that you want to simulate: \begin{align*} \operatorname{Pr}(Y=0) = |\langle y_0|\phi\rangle|^2= |\langle x_0|\psi\rangle|^2 = \operatorname{Pr}(X=0) \end{align*} and so on.