Suppose there are $n$ people invited to a party. Seats are assigned and a name card is made for each guest. However, floral arrangements on the table unexpectedly obscure the name cards. When the $n$ guests arrive, they seat themselves randomly.
Show that the probability that no one seats at the assigned seats is $$P_n = \frac{n-1}{n} P_{n-1} + \frac1n P_{n-2}.$$
My work:
We consider a letter $i$, where $i \in {1, \dots, n}$.
Let $E$ denote the event that no one sits at the right spot.
Let $X_i$ be an indicator variable s.t. $X_i = 1$ indicates the ith person sits at the right spot.
Then $Pr(E) = Pr(E \cap X_i = 0) + Pr(E \cap X_i = 1)$.
Note that $Pr(E \cap X_i = 1) = 0$ because it's impossible.
Then only need to compute: \begin{align} Pr(E \cap X_i = 0) &= \sum_{k = 1, k \neq i}^{n}Pr(E \cap \text{person i in seat j}) \\ &= \sum_{k = 1, k \neq i}^{n} \left( Pr(E \vert S_{i,j} \cap S_{j,i}) \times Pr(S_{i,j}\cap S_{j,i}) \\ + Pr(E \vert S_{i,j} \cap S_{j,i}^C) \times Pr(S_{i,j}\cap S_{j,i}^C) \right) \end{align} where $S_{i,j}=\{\text{person }i\text{ in seat }j\}$ and $S_{i,j}^C$ denotes the complement, $S_{i,j}^C=\{\text{person }i\text{ not in seat }j\}$.
Then $$Pr(E \vert \text{person i in seat j} \cap \text{person j in seat i}) = P_{n-2},$$ $$Pr(E \vert \text{person i in seat j} \cap \text{person j not in seat i}) = P_{n-1},$$ $$Pr(\text{person i in seat j} \cap \text{person j in seat i}) = \frac{1}{n(n-1)},$$ $$Pr(\text{person i in seat j} \cap \text{person j not in seat i}) = \frac{n-2}{n(n-1)}.$$ And this sums up as $$\frac1n P_{n-2} + \frac{n-2}{n}P_{n-1}.$$
I couldn't figure out where I went wrong. Could somebody point me out? Or is my method completely wrong?