I've been reading this article on the overhand shuffle. In it the author uses a simplied mathematical model of the shuffle:

Pemantle’s model for the overhand shuffle is parameterized by a probability $p \in (0,1)$. The transition rule is the following: Each of the $n − 1$ slots between adjacent cards is, independently of the other slots, declared a cutpoint with probability $p$. This divides in a natural way the deck into subsequences, or packets, of cards. Reverse each of these packets without changing its position relative to the other packets.

(...)

To further simplify our calculations, we will to begin with assuming a circular deck convention, that is, regarding the top card and the bottom card as being next to each other. This means that the top packet and the bottom packet may be treated as a single packet (if the slot between that top and the bottom card happens to be a cutpoint).

The author then states without proof that in the case when $p=\frac 1 2$, the probability that any particular card moves $k$ spaces to the right at a single shuffle is $\frac 1 3 (\frac 1 2)^k$. This seems wrong to me. I must not understand the underlying model correctly, because for instance in the case of $3$ cards and $k=1$, I calculate the probability as $p^2(1-p)+p(1-p)^2=\frac 1 4$, and not $\frac 1 6$.