Below I present an answer (too long to be a comment) in the same spirit as (the video linked in) the accepted answer by Roland Speicher. It is a synthesis of material presented in the tutorial (in French) *Probabilités libres* by Pierre Tarrago. I hope someone else finds this useful.

## Algebraic view of classical probability theory

Let's recall that a classical probability space is a triple $(\Omega,\mathcal F,\mathbb P)$, where $\Omega$ is an abstract set (called the sample space), $\mathcal F$ is a collection of subsets of $\Omega$, and $\mathbb P:\mathcal F \to [0,1]$, and these satisfy certain axioms (not recalled here). One then constructs an *expectation* operator $\mathbb E[f] := \int_\Omega f\mbox{d}P$, acting on $(\Omega,\mathcal F)$-measurable functions. When the dust has settled, this construction is in fact equivalent to the specification of an algebra $\mathcal A \subseteq \mathbb R^\Omega$, namely the set of $\mathcal F$-measurable functions on $\Omega$, and of a linear form $\varphi:\mathcal A \to \mathbb C$ such that

$$
\begin{split}
\varphi(a) &\ge 0\text{ if }a \in \mathcal A\text{ is "positive"},\\
\varphi[1_{\mathcal A}] &= 1.
\end{split}
$$

In the same way, one can abstractly define a *noncommutative* probability space from a von Neumann algebra. We recall that a von Neumann algebra is roughly an operator algebra on a Hilbert space, which contains the identity element (i.e is *unital*), and is closed in strong topology.

**Definition.** *A noncommutative probability space is a pair $(\mathcal A,\varphi)$, where $\mathcal A$ is a von Neumann algebra and $\varphi:\mathcal A \to \mathbb C$ is a ***state**, i.e a linear form such that
$$
\begin{split}
\varphi(aa^\star) &= 1\;\forall a \in \mathcal A,\\
\varphi(1_{\mathcal A}) &= 1.
\end{split}
$$

## "Random variables" and their law

Given an noncommutative probability space $(\mathcal A,\varphi)$, the state $\varphi$ plays an analogous role to the role of the expectation operator $\mathbb E$ in the classical theory of probability. A normal element $a \in \mathcal A$ (i.e $aa^\star = a^\star a$) will be called a *random variable*. The *law* of any such $a \in \mathcal A$ is the mapping $\mathbb C[X] \to \mathbb C$, $P \mapsto \varphi(P(a))$. Note that because $\mathcal A$ is an algebra $P(a) \in \mathcal A$ for every complex polynomial $P$, and so it actually makes sense to write $\varphi(P(a))$. One verifies (thanks to the *spectral theorem*) that such an application induces a measure $\mu_a$ on $\mathbb C$ with support contained in the centered ball of radius $\|a\|$. In this way, one says a sequence $(a_n)_n \subseteq \mathcal A$ of random variables converges to a random variable $a \in \mathcal A$ if $\mu_{a_n} \to \mu_{a}$, or equivalently, if
$$
\varphi(P(a_n)) \to \varphi(P(a))\;\forall P \in \mathbb C[X].
$$

## Examples of noncommutative probability spaces

We now provide some examples of noncommutative probability space $(\mathcal A,\varphi)$.

If $(\Omega,\mathcal F,\mathbb P)$ is a probability space in the usual / classical space and $\mathbb E$ is the induced expectation operator, then $(L^\infty(\Omega,\mathcal F),\mathbb E)$ is noncommutatve probability space which is ... *commutative*! Recall that $L^\infty(\Omega,\mathcal F)$ is standard notation for all bounded random variables on $(\Omega,\mathcal F)$, i.e bounded $\mathcal F$-measurable functions $X:\Omega \to \mathbb C$.

Let $L^{\infty-}(\Omega,\mathcal F)$ the random variables on $(\Omega,\mathcal F)$ which have all finite moments and let $M_n(\mathbb C)$ be the vector space of all $n \times n$ complex matrices. Then $(M_n(\mathbb C) \otimes L^{\infty-}(\Omega,\mathcal F),(1/n)\mathbb E \otimes \mbox{Tr})$ is a non-commutative probability space (of random matrices!), one of the most important in all of the theory. Note that in this example, every random variable is an $n\times n$ random complex matrix $M=(M_{ij})$ whose entries $M_{ij}$ are $L^{\infty-}$ random variables on $(\Omega,\mathcal F)$.