Snooker shot - does margin of error increase or decrease as the target angle increases? There is a perception (widely held) in snooker that a straight shot is more difficult than an angled shot. There are many forum discussion about this, and the reasons are usually accepted to be psychological. 
But I was wondering, is there a mathematical reason for it. Is the margin of error greater when the shot being taken is at an angle? 
Example - if the white ball was 1 degree off target on a straight shot, and one degree off target on an angled shot (same target for both shots), and each shot hit at the same speed, would the red ball travel off line to the same extent?

This may not answer your question but still worth a read.     
For straight shots, one needs to hit very close to Center of mass of the ball.Agreed? Now consider a standard sphere. I will take  projected area as $2\pi r^2$ ($r$ is radius)
To hit at the center of mass. Let's consider that hitting anywhere within $\frac{r}{2}$ of the projected area. Making it's favorable area will be $\frac{\pi r^2}{4}$ which happens to be $0.125$ of the total area. What I want to conclude by this is that the probability of a direct hit near center of mass is a tougher job Iff the probability of hitting anywhere on the ball is uniform.  
One more Physics related stuff.Hitting very close to the ball also does not guarantee you a head on collision. We simply can't ignore the Angular Impulse factor and friction offered by the table. That angular impulse will surely provide it a rotational motion along a axis which ain't passing through center of mass. This will surely affect it's trajectory (I can't calculate the error margin though).
Your question was interesting.  +1 for it.
