Chris White wonderfully addresses this with some Statistics, but there's a less math-y way of looking at it, too. Firstly, to dispel this notion:
So my question is, since such an insanely huge number of photons are coming out of the sun constantly, why isn't any photon hitting a detector matched up with another photon that happens to be exactly out of phase with it?
There's an equal chance that a photon will be matched up with another photon of the same phase as it will be with an opposite phase. The phase of each photon entering is an independent variable. If we're talking about two photons, then there's an equal chance of constructive interference a there is of destructive interference. This holds even if you scale up. (See last section if you're not convinced of this)
There are basically three things that you must note here:
- The average value of a distribution is not always the most likely value. Indeed, it may not even be a possible value.
- Our eyes measure intensity, not amplitude. We do not distinguish between positive and negative amplitude. Rhodopsin works by absorbing energy, which does not distinguish between the sign of the phase
- Interference is local, not global. If one of your retinal rod cells receives positive-phase light and the other receives negative-phase light, there will be no cancelling out.
Energy conservation argument
Here's a very simple way of looking at it. Due to energy conservation, if there is destructive interference there must be constructive interference elsewhere. Otherwise one could cleverly place detectors and create/destroy energy at will.
Since the light from the sun is incoherent, at any given point in time, approximately half of the spots on a sphere drawn around it will have constructive interference, and half will have destructive (not necessarily full destructive, just that the net energy is less) interference. These spots will change at random -- if a spot had constructive interference at one moment, it could have destructive interference the next.
With this in mind, there will always be some significant fraction of your rod/cone cells (which take up a small sliver of this imaginary sphere) receiving constructively-interfered light. That's enough for you to be able to see.
Why it holds even when you scale up
I'm using + to mean positive phase, and - to mean negative phase. I'm neglecting the fact that phase is not just a binary value, as this involves calculations (see Chris White's answer). A number next to the sign is the new amplitude if it has changed.
The basic thing going on here is that the mean value is not always the most probable value. Take the case of three photons:
1 2 3 Amplitude Intensity
+ + + +3 9
+ + - +1 1
+ - + +1 1
+ - - -1 1
- + + +1 1
- + - -1 1
- - + -1 1
- - - -3 9
(Avg intensity is 3)
Note the lack of a 0 in the output column. 0 is the mean output amplitude, but it never is observed as a value of the output phase. In the case of a continuous set of phases, a case of total destructive interference is possible, and it is the mean phase, however, there are man, many other final phase values that are more probable.
If you make this chart for any odd value, you'll always have no total destructive interference. If you make it for any even value, half of the time you get destructive interference, however the other half you get constructive interference, so total destructive interference does not occur. In all cases, the average intensity will always be equal to the number of incident photons. You can scale this up as much as you want, it won't change.