card riffle shuffling

The number of riffle shuffles with the cut in position $k$ is $\binom{N}{k}$. The number of riffle shuffles with the cut in position $k$, such that the card in position $N$ remains in position $N$ is the same as the number of riffle shuffles of a deck of $N-1$ cards where the cut is in position $k$ again, that is, $\binom{N-1}{k}$. We can think of this as removing the card from the bottom, shuffling, and putting it back in afterwards.

It follows that the probability for that card to remain in position $N$ is the ratio of these two numbers, that is, \[ \binom{N-1}{k} \binom{N}{k}^{-1} = \frac{N-1!}{(N-k-1)!k!} \frac{(N-k)!k!}{N!} = \frac{N-k}{N}.    \]

If I understand correctly, the total variation distance on a countable set is half of the $\ell^1$ norm, that is, half of the sum of the absolute values of the difference between the probabilities of single events. In our case, every shuffling has probability $1/N!$ for the uniform distribution, while for the compromised deck, it is $1/(N-1)!$ if $i$ is the card at the bottom, and $0$ otherwise. So we have that our distance is \[ \frac{1}{2} \sum_{\sigma \in S_N} \left\lvert \frac{\chi(\sigma(N)=i)}{(N-1)!} - \frac{1}{N!} \right\rvert \] which we can rewrite, separating the permutations with $\sigma(N)=i$ from the others, as \[ \frac{(N-1)!}{2} \left(\frac{1}{(N-1)!} - \frac{1}{N!} \right) + \frac{N! - (N-1)!}{2} \frac{1}{N!} \] and simplifying the factorials, as \[ \frac{1}{2} \left( 1 - \frac{1}{N} \right) + \frac{1}{2} \frac{N-1}{N} = \frac{N-1}{N}. \]
If $N=3$, a simple substitution tells us that the distance is $2/3$.

I hope I got all the definitions right (probability is not my area of expertise), if not, let me know with a comment and I will fix it.