I am reading a paper from Yakir Aharonov and others revisiting Hardy's paradox and I have a question. Let me state the notation first as brief as possible so that we can use it in the discussion: * a positron e+ goes through either the overlapping arm o+ or the non overlapping arm no+ of a Mach-Zhender interferometer (MZI) * an electron e- goes through either the overlapping arm o- or the non overlapping arm no- of another MZI * the two detectors that e+ (resp. e-) may register in after passing the the MZI are C+ and D+ (resp. C- and D-) * when a single MZI is operated independently without which path measurements the particles always go into the C detector (destructive interference to D) -- C always "clicks" Now, when the arms of the MZI overlap to give Hardy's experiment, Ahranov et. al claim that, if detectors are placed within no+ and o-, then every situation in which both D+ and D- click, both of the which path detectors also click. I do not understand that. I do understand that, if a detector is in o- ONLY then every time D+ and D- click then o- must have clicked (otherwise e+ would have "interfered with itself" such that it would have to go into C+). However, if both the no+ and o- detectors are there, then I would expect 1/4 of the time there would be annihilation as usual (both e+ and e- enter overlapping arms), but the other 3/4 of the time the particles would collapse to the other 3 combinations no+o-, no+no-, and o+no-with equal probability. And, in each of the latter 3 cases, there would be a 1/4 chance of both particles being registered in their D detectors, right? So why would a measurement of, say no+no-, prevent the D detectors from clicking? Is it somehow impossible to register no+no-? Once the particles "collapse" into the non overlapping arms I'd think their should be no interference and they should independtly have 0.5 probability of going into D (.25 probability of D+D-). I guess I am missing something.