Let $Z$ be uniform on $\{-1, 1\}$, $T$ uniform on $\{1, 2\}$, $Y = 0$ and $X = TZ$.

Then $\mathbb E(X | Y, Z) = 0$, but $P(X = 1 | Z = 1, Y = 0) = \frac{1}{2} \neq 0 = P(X = 1 | Z = -1, Y = 0)$.

Let $Z$ and $T$ be independent uniform on $\{-1, 1\}$, $Y = 0$ and $X = T(Z + 1)$.

Then $\mathbb E(X | Y, Z = 1) = \mathbb E(2T | Y, Z = 1) = 0$ and $\mathbb E(X | Y, Z = -1) = \mathbb E(0 | Y, Z = 1) = 0$, but $P(X = 2 | Z = 1, Y = 0) = \frac{1}{2} \neq 0 = P(X = 2 | Z = -1, Y = 0)$.