After reviewing the more than half dozen questions on the DCQE here, this doesn't seem to have been asked or answered directly - if I've missed this piece in one of the other questions or answers please point me in the right direction and we can close this.
Using the detector notation from the DCQE Wikipedia article for brevity:
Let $D_0$ = idler photon collector plate
Let $D_1$ = no-path information detector 1
Let $D_2$ = no-path information detector 2
Let $D_1$ and $D_2$ be time-like separated from $D_0$ such that $D_0$ will register photon arrival hits some finite, measurable time prior to $D_1$ or $D_2$.
For the purposes of this question, we can ignore the cases where which-path info is collected. Frankly, we can ignore the "choice" to insert a beam splitter or not entirely. For the purposes of this question, a half-silvered mirror will remain in place (no which-path information is available).
After many runs, when I go back to correlate the arrival times of clicks at $D_0$ with clicks at $D_1$ and $D_2$, I note that 2 interference patterns are recoverable, with the location of $D_1$ idler entangled photons aligning with troughs of $D_2$, just as expected.
Since I now know the locations of the respective peaks and troughs on $D_0$ correlated with $D_1$ and $D_2$ detection, I wipe my detection plates clean and start again. I note that my first "click" at $D_0$ is in a location that corresponds to a peak in the $D_1$ fringes.
My question is simple:
Based on this knowledge, is the signal photon more likely to be detected at $D_1$? (my $D_1$ detector hasn't clicked yet due to the time-like separation).
It seems the answer must be yes.
If it IS yes, then why is there any discussion at all about backwards causation? If I can note the location of photon recording on $D_0$ and predict a higher or lower probability of recording an entangled sister photon on $D_1$ or $D_2$, then that's the ballgame. Where's the spookiness that has this experiment come up again and again?
If the answer is no in this setup, I struggle to see why, and so any clarification would be greatly appreciated.