Parallel Pollard's Rho: Number of distinguished points
The expected number of points $\sqrt(\frac{n\cdot \pi}{2})$ is calculated via the birthday paradox and it applies to the number of calculated (not just distinguished/reported) points. As you said, each distinguished point represents $2^{32}$ points. Even if a collision is not reported (because it is not on a distinguished point), the two walks which collided will continue on the same path and eventually find the same distinguished point. That is where the ...$+\frac{1}{\theta}$ comes from in the complexity in the original Oorschot et Wiener paper. $\theta$ is the probability that a point is distinguished. So you have, the number of computations until a collision occurs, + the number of computations before that collision is reported i.e. before you reach the next distinguished point.

In practice, I haven't had a problem with Robin hoods, but Oorschot et Wiener seem to suggest that to avoid Robin hoods you need to take $\theta$  small enough that trails are long.