Let $U$ be an $n$ qubit Haar random circuit applied to $|0^n \rangle$. Thereafter, the state is measured in the standard basis. Let $p_0$ be the probability of getting $0$ in the first qubit. We know that, \begin{equation} \mathbb{E}_U[p_0] = \mathbb{E}_U\big[\text{Tr}(|0\rangle \langle 0| \otimes I_{n-1} ~U|0^n\rangle \langle0^n| U^{*}\big] = \frac{1}{2}, \end{equation} where $I_{n-1}$ is the identity operator on the remaining $n-1$ qubits.
Let $p(x_0 = 0 | x_1 = b_1, x_2 = b_2, \ldots, x_n = b_n)$ be the conditional probability of getting $0$ on the first qubit, conditioned on the other qubits being $b_1, b_2, \ldots, b_n$ respectively, where each $b_i \in \{0, 1\}$.
Is $p(x_0 = 0 | x_1 = b_1, x_2 = b_2, \ldots, x_n = b_n)$ also $\frac{1}{2}$? How might we express and compute this quantity as a Haar integral?