That's what I thought.
But actually the 9ball (the ball closer to the CB) has signifcantly more margin of error than than the 8ball (the ball closer to the pocket). Here are the numbers I used...
CB distance to pocket opening = 10 feet (120'')
pocket size = 4.5''
9ball distance to front edge of pocket opening = 2.25'' (to the front edge of the 9ball)
8ball distance to CB = 2.25" (ball edge to ball edge)
Results...
8ball margin of error = +/- 0.356 degrees
9ball margin of error = +/- 0.558 degrees
So the 9ball has about 57% MORE margin of error than the 8ball. Those numbers don't sound very big. However, compare it to the worst case margin of error with the OB (as you would expect) being right at the midpoint between the pocket and CB. The margin of error in this case is a miniscule +/- 0.042 degrees.
I always thought that margin of error would be pretty much symmetric around the midpoint between the CB and pocket. After graphing the solution, it more or less does look symmetric until the OB gets a couple of ball's widths distance away from the CB and pocket. At this closer proximity, the margin of error on the CB side goes up much faster than the pocket side.
If you're not convinced, just think of the extreme case of the 8ball hanging right on the edge of the pocket compared to the 9ball being virtually frozen on the CB. For the 9ball, you can pretty much aim in any foward direction (again, neglect throw and friction effects) and the 9ball should head towards the pocket. Your margin of error is huge...pretty much the full 180 degrees. However, to pocket the 8ball, you still need to have the CB contact the 8ball, so your margin of error is still very limited.
So going back to the original example, you
should be able to pocket the 9ball about 50% more times than the 8ball...in theory. It's funny how the practical world doesn't exactly agree with the theory.
Neat, huh? If anyone is interested, I'll post my derivations and calculations...NOT!