QFT Causality: Real Scalar Field & Probability

[Aleolomorfo]

Hello everyone!
I have a question regarding the causality in QFT.
If I take into consideration a real scalar field and I calculate:
$$[\phi(x),\phi(y)] = 0 \space \space \space \space \space \text{if (x-y)}^2 < 0$$
Thanks to this relation we state that causality in QFT is preserved: a measurement in $x$ cannot talk to another one made in $y$.
However, this is not the same of saying that the probability of going from $x$ to $y$ is $0$, as a matter of fact it is $D(x-y) \neq 0$. So the probability is not zero. I do not understand how we can link these two results coherently. From my perspective they are in contradiction (I know they are not, I'd like to understand why): the latter is what we call strictly causality and so it is broken.
Then usually books make another step forward. If I calculate the probability of going from $y$ to $x$ I found $D(y-x)$. As long as $x-y$ is spacelike there is a continuos Lorentz transformation between $x-y$ and $y-x$ and since $D(x)$ is Lorentz invariant :
$$P(x \rightarrow y) = P(y \rightarrow y)$$
I understand the single steps, but I do not find the link between them. Thanks in advance!

 

[DarMM]

It's best to think of $D(x - y)$ not as the probability to go from x to y, but as a measure of how correlated the value of $\phi(x)$ is with the value $\phi(y)$.