That's the same question that comes up in the original transparent-boxes problem, of course. We probably shouldn't try to recap that whole debate in the middle of this thread. :)
Don't worry; I don't want to do that :). If I recall the original transparent-boxes problem correctly, I agree with you on what to do in that case.
Just to check my memory, in the original problem, there are two transparent boxes, A and B. You see that A contains $1M and B contains $1000. You know that B necessarily contains $1000, but A would have contained $1M iff it were the case that you will decide to take only A. Otherwise, A would have been empty. The conclusion (with which I agree) is that you should take only A. Is that right? (If I'm misremembering something crucial, is there a link to the full description online?) [ETA: I see that you added a description to your post. My recollection above seems to be consistent with your description.]
In the original problem, if we use the "many choosers" heuristic, there are no choosers who two-box and yet who get the $1M. Therefore, you cannot "choose to be" one of them. This is why two-boxing should have no appeal to you.
In contrast, in your new problem, there are two-boxers who get the $1M and who get their E module to output True. So you can "choose to be" one of them, no? And since they're the biggest winners, that's what you should do, isn't it?
According to Ingredients of Timeless Decision Theory, when you set up a factored causal graph for TDT, "You treat your choice as determining the result of the logical computation, and hence all instantiations of that computation, and all instantiations of other computations dependent on that logical computation", where "the logical computation" refers to the TDT-prescribed argmax computation (call it C) that takes all your observations of the world (from which you can construct the factored causal graph) as input, and outputs an action in the present situation.
I asked Eliezer to clarify what it means for another logical computation D to be either the same as C, or "dependent on" C, for purposes of the TDT algorithm. Eliezer answered:
I replied as follows (which Eliezer suggested I post here).
If that's what TDT means by the logical dependency between Platonic computations, then TDT may have a serious flaw.
Consider the following version of the transparent-boxes scenario. The predictor has an infallible simulator D that predicts whether I one-box here [EDIT: if I see $1M]. The predictor also has a module E that computes whether the ith digit of pi is zero, for some ridiculously large value of i that the predictor randomly selects. I'll be told the value of i, but the best I can do is assign an a priori probability of .1 that the specified digit is zero.