It is possible to count the different configurations of interest to calculate the probability directly, by formalizing some of the ideas already presented.
In an allowed configuration, denote a displacement of one or more passengers with the diagram $i\rightarrow j$ whenever passenger $i$ displaces passenger $j$ from their assigned seat ($i < j$).
Suppose C is an allowed configuration with a displacement D of passengers, say, $...i\rightarrow j...$ Clearly i has a predecessor (which can be added to the diagram) or i is passenger 1, since the problem states that only passenger 1 is free to displace passengers without being displaced themselves. Since each predecessor must represent a passenger who boarded earlier, of which there are a finite number, using this argument at most i times, shows that D must begin with passenger 1. By a similar argument, D must have a last passenger which chooses passenger 1’s seat to end the displacement.
If E is a displacement in C, by the same argument it must start with passenger 1, followed by the choices already indicated in D, so that E is the same as D. Clearly, two allowed configurations are the same if and only if their displacements are the same. Additionally, note that a displacement of the form, $\textstyle\ 1\rightarrow i \rightarrow … \rightarrow j$ where {i,…,j} is a subset of {2,…,100} in increasing order, always specifies a valid configuration. Hence,
There is a bijection between the set of allowable configurations and diagrams of the form $1\rightarrow i \rightarrow … \rightarrow j$ where {i,…,j} is a subset of {2,…,100} in increasing order.
Now count configurations by counting the diagrams. Write 1 as the diagram for the null displacement in the configuration where all passengers sit in their assigned seats. Note that this is same as counting all subsets of a two particular sets, so the probability that the final passenger is sitting in their assigned seat is:
$\textstyle \frac{\textrm{Number of all diagrams of the form } 1\rightarrow i \rightarrow … \rightarrow j\textrm{ where {i,…,j} is a subset of {2,…,99} in increasing order}}{\textrm{Number of all diagrams of the form } 1\rightarrow k \rightarrow … \rightarrow l\textrm{ where {k,…,l} is a subset of {2,…,100} in increasing order}} = \frac{2^{98}}{2^{99}} = \frac{1}{2}\\$