Two slits experiment with glitchy detector In the double slit experiment, when we place a detector on one of the slits, the electrons will start to act like particles. When we unplug the detector they return to acting like waves.
What if our detector was turning on and off with a high frequency in a way that we could not tell for all the electrons if during the time they were passing in the slits our detector was on or off, how would the interference pattern look then?
 A: You would not observe the interference pattern. In fact, when you consider 'interference pattern' and 'electron information', this situation would be a loss-loss, because you don't get both (on the contrary, plugging or unplugging completely would still get you an loss-win and win-loss).
So, why does this happen? Let's say you have a glitchy detector, which switches on and off every other second. And let's have a few electrons passing through the slit at a rate of say $h$ electrons per second. Now at $t=1s$, the first $h$ electrons pass through the slit, and the detector is off, so the electrons land according to the probabilities determined by the interference of their wavefunctions. Now, at $t=2s$, the detector turns on. But the corresponding batch of $h$ electrons have now been measured, so they hit the screen at random spots, adding a random pattern in front of the ordered interference pattern. 
And this goes on. The next batch forms an interference pattern, but now you wont be even able to tell it apart from the random arrangement. And finally, after the experiment, you will see random arrangements disrupting the interference pattern, which would be what you obtain had the detector remained switched off.
If you had a normal detector (one that stayed on), you would at least get the information of which slit the electron passed through. But, in this case, you can't even get that. This was the loss-loss situation I was referring to.
