On a plane, there are $7$ seats. Each is assigned to a passenger. The passengers walk on the plane one at a time. The first passenger sits in the wrong seat (someone else's). For all the following people, they either sit in their assigned seat, or if it is full, randomly pick another. You are the last person to board the plane. What is the probability that you sit in your own seat?
1. Let's denote the seats as \( S_1, S_2, \ldots, S_7 \) and the passengers as \( P_1, P_2, \ldots, P_7 \). Each passenger \( P_i \) is assigned to seat \( S_i \).

2. The first passenger \( P_1 \) sits in a random seat that is not \( S_1 \). There are 6 other seats, so the probability that \( P_1 \) sits in any one of these seats is \( \frac{1}{6} \).

3. For the remaining passengers \( P_2, P_3, \ldots, P_6 \), they will either sit in their assigned seat if it is available or choose another seat randomly if their assigned seat is taken.

4. We need to determine the probability that the last passenger \( P_7 \) sits in their own seat \( S_7 \).

5. Consider the possible scenarios:
   - If \( P_1 \) sits in \( S_7 \), then \( P_7 \) will never sit in \( S_7 \) because \( P_7 \) will have to choose another seat.
   - If \( P_1 \) sits in any seat other than \( S_7 \), the problem reduces to a smaller version of the same problem with 6 seats and 6 passengers.

6. Let's denote \( E \) as the event that \( P_7 \) sits in \( S_7 \). We need to find \( P(E) \).

7. We can use induction to solve this problem. Let \( P_n \) be the probability that the \( n \)-th passenger sits in their own seat when there are \( n \) seats and \( n \) passengers.

8. Base case: For \( n = 1 \), \( P_1 = 1 \) because there is only one seat and one passenger.

9. Inductive step: Assume \( P_k = \frac{1}{k} \) for \( k \) seats and \( k \) passengers. We need to show that \( P_{k+1} = \frac{1}{k+1} \).

10. For \( k+1 \) seats and \( k+1 \) passengers:
    - The first passenger \( P_1 \) can sit in any of the \( k+1 \) seats.
    - If \( P_1 \) sits in their own seat, the problem reduces to \( k \) seats and \( k \) passengers, with probability \( P_k = \frac{1}{k} \).
    - If \( P_1 \) sits in any other seat, the problem reduces to \( k \) seats and \( k \) passengers, with probability \( P_k = \frac{1}{k} \).

11. Therefore, the probability that \( P_{k+1} \) sits in their own seat is:
    \[
    P_{k+1} = \frac{1}{k+1} \sum_{i=1}^{k+1} P_k = \frac{1}{k+1} \cdot k \cdot \frac{1}{k} = \frac{1}{k+1}
    \]

12. By induction, \( P_7 = \frac{1}{7} \).

The final answer is \( \boxed{\frac{1}{7}} \).