$\textbf{Answer}$

$50\%$.

$\textbf{Analysis}$

Let's start with simpler cases:

• If there are only two passengers, then the answer is obviously $50\%$. Whether the last (i.e. the $2^{nd}$ passenger) can sit on the $2^{nd}$ seat depending on whether the first one chooses the first or the second seat. That is a half-half chance.
• If there are three passengers, then:
  1. If the first passenger chooses the $1^{st}$ seat, then everyone will sit on their assigned seat.
  2. If the first passenger chooses the $2^{nd}$ seat, then depending on the $2^{nd}$ passenger's choice, the last one has a half-half chance to sit on the $3^{rd}$ seat
  3. If the first passenger chooses the $3^{rd}$ seat, then the last one will have to sit on a wrong seat (i.e. the $1^{st}$ seat)

Consider this $3$-passenger scenario described above, there is still a half-half chance. This is because the $1^{st}$ and the $3^{rd}$ cases are symmetric (probability of $100\%$ v.s. $0\%$) and the probability for the $2^{nd}$ case is $50\%$.

This analysis makes us wonder whether the answer is always $50\%$. Indeed, it is. This result can be obtained using symmetric and recursive arguments.

Let's consider the first passenger's choice. The two cases when he sits on the first seat and the last seat are symmetric. In the former case, the last passenger will sit on his assigned seat. In the latter case, the last passenger has to sit on the $1^{st}$ seat. This means the overall probability in these two scenarios is $50\%$. (This is similar to the first and last situations in the $3$-passenger case.)

If the first passenger chooses a seat in the middle, e.g. the $k^{th}$ seat where $k\ne 1, 100$, then passengers $2$, $3$, $\cdots$, $(k-1)$ will sit on their assigned seats. Ignoring all these passengers, we find that the case becomes this: there are $(k+1)$ passengers where the $1^{st}$ passenger took the $2^{nd}$ seat and it is the second passenger's turn to pick up a seat. Next, we can also artificially swap the first and the second passengers (i.e. swap their tickets). Now, the case becomes that the second passenger always take his assigned seat. Ignoring this passenger again, we find the situation is exactly the same as a problem of $k$ passengers  where $k < n$. Repeating this passenger number reduction process, an $n$-passenger problem will eventually become equivalent to a $2$-passenger problem. Hence, we conclude that the final answer is $50\%$.

In fact, if we review the second case above in the $3$-passenger scenario, we will find it exactly the same as a $2$-passenger problem (i.e.the first passenger can be ignored). This is consistent with the above described number reduction process.

$\textbf{Note}$

This example again demonstrates the power of starting analysis using simpler cases. The number reduction process described here is the same as the recursive method. The essence of these techniques is to continuously simplify a complex problem to a simpler equivalent till the answer can be readily obtained. ($n\rightarrow k\rightarrow\cdots\rightarrow 2$ ).