Since the idea of a probability monad is not a fully formal concept, there can be no fully formal answer to this question. Nevertheless, I will argue below that *no nontrivial probability monad is a Hopf monad*.

But let's start with some background.

**Probability monads**: The idea behind this concept is that such a monad consists of the following:

- The underlying symmetric monoidal category is some kind of category of "spaces", where the tensor product is a version of the Cartesian product of spaces. (Which is not necessarily the categorical product; metric spaces with $\ell^1$-product are an example where it isn't.)
- For every space $X$, the object $TX$ is the space of probability measures on $X$, and the unit $\delta_X : X \to TX$ assigns to every point $x$ the Dirac delta $\delta_x$.
- The multiplication $T^2 X \to TX$ takes a probability measure on probability measures—often called a
*random measure*—and maps it to its expectation value.
- The monoidal structure maps $TX \otimes TY \to T(X \otimes Y)$ implement the formation of product measures.
- The opmonoidal structure maps $T(X \otimes Y) \to TX \otimes TY$ implement the formation of marginals by mapping every measure on a product space to the pair of its marginals.

**Hopf monads**: As per the nLab, an opmonoidal monad is a Hopf monad if the left fusion operator
$$T(X \otimes TY) \to TX \otimes T^2 Y \to TX \otimes TY$$
is invertible, and similarly for the right fusion operator.

**The actual answer**: Let's see now what the left fusion operator does in a probability context. An element of $T(X \otimes TY)$ is a joint distribution of an $X$-valued random variable and a random measure on $Y$. Here are two examples:
$$\frac{1}{2} \delta_{\left(x,\delta_y\right)} + \frac{1}{2} \delta_{\left(x',\delta_{y'}\right)}, \qquad \frac{1}{2} \delta_{\left(x,\delta_{y'}\right)} + \frac{1}{2} \delta_{\left(x',\delta_y\right)}$$
for points $x,x' \in X$ and $y, y' \in Y$. Both are measures which correlate a point in $X$ with a probability measure on $Y$. Now the first map above is $T(X \otimes TY) \to TX \otimes T^2Y$, which is one of the opmonoidal structure maps and therefore amounts to taking the pair of marginals. This gives the pair
$$
\left( \frac{1}{2} \delta_x + \frac{1}{2} \delta_{x'}, \frac{1}{2} \delta_{\delta_y} + \frac{1}{2} \delta_{\delta_{y'}} \right)
$$
for *both* of the elements of $T(X \otimes TY)$ above. Therefore our monad is not a Hopf monad, already because the first map is not injective. In fact, the second map is not injective either, and we could construct other pairs of elements of $T(X \otimes TY)$ where the invertibility fails at that step.

There are two more general and abstract ways to see what goes wrong:

In a probability monad, the monoidal structure maps $TX \otimes TY \to T(X \otimes Y)$ are right inverse to the opmonoidal structure maps $T(X \otimes Y) \to TX \otimes TY$. In probability terms, this expresses the fact that the two marginals of a product measure are exactly the original factors. (See our paper for more details.) Whenever you have this property, then the first map above is a split epimorphism; so if $T$ is a Hopf monad as well, then this opmonoidal structure map must be an isomorphism, and therefore $T$ is even strong monoidal, at least on all pairs of objects of the form $(X, TY)$. But strong monoidality trivializes the probabilistic interpretation, because it says that every probability measure is uniquely determined by its marginals. Since probability theory is all about how things are correlated, there is no nontrivial flavour of probability theory which satisfies this, and therefore no nontrivial probability monad that is a Hopf monad.

A probability monad is *affine*, meaning that $TI \cong I$ canonically, because there is exactly one probability measure on a one-point space. When you plug $X = I$ into the definition of Hopf monad above and use $TI \cong I$, you'll see that $T$ being a Hopf monad amounts to the multiplication $T^2 Y \to TY$ being invertible. Thus an affine monad is a Hopf monad if and only if it is idempotent. Once again this does not happen for any nontrivial probability monad, because it amounts to saying that the formation of expectation values is a bijection.