It is well-known that the expected number of vertices on the convex hull of random set of points in the plane distributed uniformly within a $k$-gon is $O(k\log n)$ and within a smooth shape (e.g. a disk) is $O(n^{1/3})$. These bounds also extend to $\mathbb{R}^d$ with arbitrary $d$, by tacking on a $d-1$ exponent. But, I'm only interested in $\mathbb{R}^2$ for now.
My question is: What is wrong with the following argument for the number of layers in a $k$-gon to be $\Theta(n/\log n)$, taking $k$ as a constant:
In $\mathbb{R}^2$, the number of vertices of $conv(P)$ is $\Theta(\log n)$. After peeling off $i$ layers, the points $P_{i+1}$ enclosed by layer $L_i$ are also independently and uniformly distributed, so their convex hull $conv(P_{i+1})$ should also have $\Theta(\log |P_{i+1}|) = O(\log n)$.
I'm aware of the following result by Dalal:
In this paper, we show that the expected number of layers of a convex hull onion for n uniformly and independently distributed points in a disk is $\Theta(n^{2/3})$. Additionally, we show that in general the bound is $\Theta(n^{2/(d+1)})$ for points distributed in a $d$-dimensional ball. Further, we show that this bound holds more generally for any fixed, bounded, full-dimensional shape with a nonempty interior.
I guess my question could also be put more simply as this: Why is the expected number of layers for $k$-gons not $O(n/\log n)$? Why is there a difference in the size of the convex hull between the case when the random points are drawn from a polygon and smooth shapes, but there is no such difference in the expected number of layers, based on Dalal's result (unless I misunderstand his claim)? An intuitive reason should do, but a formal argument or a pointer to one will be great!