A problem about random permutations

Define indicator random variables $X_1,\ldots, X_n$ where $X_i = 1$ if $i$ appears to the left of $i+1,\ldots, n$ in the random permutation. Consider the random variables $$ Y_n = (-1)^{X_1+ X_3 + \cdots} $$ and $$ Z_n = (-1)^{X_2 + X_4 + \cdots}. $$ Note that $\mathbb{E}Y_n$ will give us the answer to the question since it yields the difference between the probability of the sum being even and the sum being odd.

We will try to compute $\mathbb{E}Y_n$ and $\mathbb{E}Z_n$ inductively on $n$. Conditioned on the choice of any ordering $\pi$ of $2,\ldots, n$, the probability that $X_1 = 1$ is exactly $1/n,$ which implies that $$ \mathbb{E}\left[ (-1)^{X_1}\ \right|\left. \pi\right] = \left(1-\frac{2}{n}\right). $$ To compute $\mathbb{E}Y_n$, we note that $$ \mathbb{E} Y_n = \left(1-\frac{2}{n}\right)\cdot \mathbb{E}_\pi (-1)^{X_3 + X_5 + \cdots} = \left(1-\frac{2}{n}\right) \cdot\mathbb{E} Z_{n-1} $$ where we switched to $Z_{n-1}$ since we are now summing the $X_i$ variables corresponding to $i$ of even rank among $\{2,\ldots, n\}$. Similarly, we can also show $$ \mathbb{E}Z_n = \mathbb{E}Y_{n-1}. $$

This implies that $\mathbb{E}Y_n = (1-\frac{2}{n}) \mathbb{E} Y_{n-2}$ for $n > 2$ and hence $$ \mathbb{E}Y_n = \frac{n-2}{n}\cdot \frac{n-4}{n-2}\cdots $$

For the base cases $n = 1,2$, we can explicitly compute $$ \mathbb{E}Y_1 = -1, \ \ \ \mathbb{E}Y_2 = 0. $$

This shows that if $n$ is even, $\mathbb{E} Y_n = 0$, implying that the probability of the sum being odd and even are both $1/2$.

When $n$ is odd, $\mathbb{E}Y_n = -\frac{1}{n}$, implying that the probability of the sum being even is $\frac{1}{2} - \frac{1}{2n}$.

Here's a quick proof that the probability is $1/2$ when $n$ is even, but I couldn't figure out a similarly quick proof (i.e. one that avoids induction) when $n$ is odd.

Notice that the following procedures are equivalent:

1. Pick a uniformly random permutation.
2. Pick a uniformly random permutation, then with probability $1/2$, flip the positions of $n - 1$ and $n$.

But if $n$ is even, then flipping the positions of $n - 1$ and $n$ flips the parity of the desired sum, because it toggles whether $n - 1$, which is odd, is included in the sum. So if $n$ is even, procedure 2 makes the desired sum even with probability $1/2$, and thus so does procedure 1.