**A recursive solution:** Once a passenger who has been displaced from their assigned seat sits in the seat assigned to passenger $1$, no more passengers will be displaced. But if the displaced passenger takes some other seat, a passenger still standing in line will become displaced

Let the plane have $N$ seats and let $r_n$, $1\le n\le N-1$, be the probability that with $n$ passengers standing in line none of the standing passengers is displaced from their assigned seat. We have $r_{N-1}=\frac{1}{N}$ because that is the probability the first passenger happens to choose their own seat. If $n$ passengers are standing in line and somebody in line is displaced—it will always be exactly one passenger—the probability that the displaced passenger is at the head of the line is, by symmetry among the standing passengers, $\frac{1}{n}$. And if the displaced passenger is at the head of the line, the probability that they take the seat assigned to passenger $1$ is also $\frac{1}{n}$. It follows that
$$
r_{n-1}=r_n+\frac{1-r_n}{n^2}.
$$
By induction, we find that $r_n=\frac{1}{n+1}$ and hence $r_1=\frac{1}{2}$.

**Another recursive solution:** This expands on Shashank's answer, emphasizing a key point that caused me some confusion at first.

When a non-displaced passenger gets to the head of the line, they simply take their assigned seat, but when a displaced passenger gets to the head of the line, they choose uniformly at random from the set
$$
\{\text{passenger 1's seat}\}\cup\{\text{seats assigned to passengers standing behind them in line}\}.
$$
Although technically passenger $1$ is not displaced, they follow the same procedure as the displaced passengers do, so we lump them in with that group. As a consequence, if passenger $k$ is displaced, then when that passenger gets to the head of the line, the situation is just like the original problem, but for a plane of $N-k+1$ passengers.

Define $p_n$ to be the probability that, in the setup described, the last passenger to board an $n$-seat airplane gets their assigned seat. Then for $n\ge2$,
$$
p_n=\frac{1}{n}\cdot1+\left[\frac{1}{n}\sum_{k=2}^{n-1}p_{n-k+1}\right]+\frac{1}{n}\cdot 0.
$$
Each term in this expression corresponds to a seat choice of passenger $1$: if the choice is seat $1$, passenger $n$ gets their assigned seat with probability $1$; if it's seat $k$, for $2\le k\le n-1$, then passengers $2$ through $k-1$ take their assigned seats and, when passenger $k$ comes to choose a seat, it is like the original problem but with $n-k+1$ seats; if the choice is seat $n$, then passenger $n$ gets their assigned seat with probability $0$.

Multiplying by $n$ and reversing the order of summation gives
$$
np_n=1+\sum_{k=2}^{n-1}p_k.
$$
Subtracting this from the same recurrence with $n$ replaced by $n+1$ gives
$$
(n+1)p_{n+1}-np_n=p_n,
$$
for $n\ge2$. Hence $p_2=p_3=p_4=\ldots$. One easily sees $p_2=\frac{1}{2}$.

The stumbling block for me was in seeing the similarity between passenger $1$ and the displaced passengers. Passenger $1$ chooses randomly among the available seats, which include their assigned seat, and I got hung up on the possibility that passenger $1$ might get their own seat, which can't happen for other displaced passengers. That, however, is not the correct parallel to draw. The right parallel is that both passenger $1$ and the displaced passengers can get passenger $1$'s seat, which ends the cycle of displacement.

Several other answers on this page, like (this, this, and this) —jump straight to an even simpler recurrence. The recurrence does hold and can be derived from this one, but haven't been able to understand how to get that recurrence directly. This issue was raised by Hans on this page in a number of his comments. If I stumble upon any insight that might help other readers with this point, I will add it here.

**Solution using conditional probability:** we have seen that if seat $1$ is chosen before the last passenger is seated, then that passenger sits in their assigned seat; if seat $N$ is chosen before the last passenger is seated, then the last passenger is displaced and must take seat $1$. Let $E_k$ be the event that the $k$th passenger to be seated is the first to take one of seats $1$ and $N$. Then
$$
\Pr(\text{$N$ gets their assigned seat})=\sum_{k=1}^{N-1}\Pr(\text{$k$ sits in $1$}\mid E_k)\Pr(E_k)=\frac{1}{2}\sum_{k=1}^{N-1}\Pr(E_k)=\frac{1}{2}.
$$
Diagrammatically, this calculation is shown below. The probability that $N$ gets their assigned seat is the sum of the probabilities that each of passengers $1$ through $N-1$ takes seat $1$.

**Solution using a bijection between cycles:** (This completes a final, needed step in hunter's answer.) The seating process results in a permutation of the passengers, and this permutation always consists of a single cycle containing passenger $1$ followed by displaced passengers, an ascending order. So for $N=100$ the cycle $(1,46,53,75,88)$ represents the situation where passenger $1$ displaces passenger $46$, $46$ displaces $53$, $53$ displaces $75$, $75$ displaces $88$, and $88$ sits in passenger $1$'s seat. Passengers $89$ and up, and passenger $100$ in particular, will sit in their assigned seats. The cycle $(1,46,53,75,88,100)$ represents a similar situation, except that passenger $88$ displaces passenger $100$. Passengers $89$ through $99$ will sit in their assigned seats, and passenger $100$ will have only one choice of seat: passenger $1$'s seat.

The probability of occurrence of these two cycles is the same:
$$
\frac{1}{100}\frac{1}{55}\frac{1}{48}\frac{1}{26}\frac{1}{13}.
$$
In general, when displaced passenger $k$ chooses a seat, they choose uniformly at random from $101-k$ possible seats. Once passenger $88$ has chosen either seat $1$ or seat $100$ (each of which happens with probability $\frac{1}{13}$), the seating of the remaining passengers standing in line is determined. Now every cycle that does not contain $100$ is paired in this way with an equally probable cycle that does contain $100$. Hence passenger $100$ sits in their assigned seat with probability $\frac{1}{2}$.