(please excuse the top post, but with the HTML formatted original it's just easier than trying to add the tags to make my reply look right. Honestly I hate top posts. Just scroll down to see the rest of the converstation framed in html)
Nice job of the physics. The only thing you're missing is that on an angle the center of gravity will change, placing more weight on the rearmost wheels. On a car that was perfectly flat, say, a steel plate with tiny wheels, your math is perfect. On a car that was, say, 7 stories high, you can see that a small tilt would place *all* the weight on the rear wheels and the fronts would actually come up off the ground and it would tip over. Just prior to that the front wheel would have zero weight. At smaller angles, or a shorter vehicle, the shift would be someplace in between.
In a real car, much of the weight is low (drivetrain) and some of it is higher (greenhouse). The center of gravity is somewhere between the ground (steel plate) and 7 stories up (my tower car).
So a car on a slope will have some amount of increased weight distribution on the rearmost wheels, and therefore the /4 trick won't work even if the car is 50/50 on a level slope where /4 is correct.
I can't give you math for it, but I'd be the weight transfer on a moderate to steep driveway type hill would be on the order of 5% or so. Strictly a guess, but I suspect a generous one. BMW and other makers try to lower the center of gravity all the time for handling reasons, so it's probably not immense. And the number varies by the steepness of the hill of course. But I'm not up for a calculus function to describe the relationship, especially since I don't know what the center of gravity is to plug in.
So anyway most bimmers are close to 50/50 to start with. Let's go with that. Say going up the hill, it's now 45/55.
Front drivers are probably closer to 60/40 in general. Yes I know, I'm being very inexact here. But with the same transfer, you're now at 55/45. Amazing... the same weight on the drive axel in both cases.
My numbers are made up, poke at them all you want I don't mind. The concept is there though. Play with the numbers and get small variations.
But front drive still wins, no matter how you dice it up. I submit this. Back up you driveway in the FWD car. With math above, you have 65% of weight on the drive wheels, vs RWD's best of 55%. With your math, you have 60% vs 50%.
Now, your point about the direction of the vector of the force is valid. As the angle increases, the force decreases, until it reaches zero. Your car on the wall will indeed have zero force on the rear wheels, only because the vectors are straight down. At 89%, where there is still some amount of lateral force (not enough to produce enough friction to hold the car mind you) almost all the lateral force would be on the rear wheels, but the size of that horizontal vector has become very small. So yes, nearly 100% of the force is on the rear wheels and nearly zero on the fronts, but, the amount of this force in the useful direction is so little that it doesn't help the car to stay put and it slides down the hill. Consider a car on say a 70 degree angle. Add much more and the downward component of the vector will overcome the friction of the tires and the car will slide. Let's imagine that 70 degrees is very near this point. Now walk up to that car and lift the front bumper