When Bendeguz boarded the 78-seat passenger train car with his valid ticket, he was shocked to realize that every seat was already occupied. This happened because Domotor had boarded without a ticket. The other 77 passengers, including Elek, had all purchased tickets, but they did not necessarily sit in the seats assigned to them. Bendeguz confronts the person occupying his seat. The person who stands up now also wants to sit in their own seat, and so on. This continues until Domotor is eventually exposed. What is the probability that Elek will be able to watch the events unfold from a seated position?
Let's track the events until one of Dömötör and Elek has to stand up. We will show that in exactly half of the 78! cases, Elek (unfavorably) has to change his seat. To this end, we pair the 78! seating orders such that in each pair, exactly one order is favorable for Elek. The pairing is as follows: the pair of any given order is the one obtained by swapping Dömötör and Elek in it. This is indeed a pairing, since every order differs from its pair, and the members of the pairs find each other reciprocally, as the double swap of Dömötör and Elek restores the original order. It is clear that an order is favorable for Elek if and only if its pair is unfavorable. Therefore, the probability that Elek can watch the events leading to Dömötör's unveiling while seated is $\frac{1}{2}$.

Based on the work of Béky Bence (Fazekas M. Főv. Gyak. Gimn., 11th grade), Dömötör Csilla (Győr, Révai M. Gimn., 11th grade), Horvàth Illés (Fazekas M. Főv. Gyak. Gimn., 11th grade), and Várkonyi Dániel (Fazekas M. Főv. Gyak. Gimn., 12th grade).

Remarks. 1. The majority of those who solved the problem realized that from the perspective of the problem, it is enough to examine only Dömötör, Elek, and the "other passengers," and the positions of Dömötör and Elek are symmetric, interchangeable.

2. The final result can be easily guessed if we calculate the desired probability for 3 passengers instead of 79, and then for 4, using our fingers. After that, we can prove by complete induction that the result is $\frac{1}{2}$ regardless of the number of passengers.
3. Many tried to enumerate the cases: errors arose here from those who incorrectly added the probabilities of non-exclusive events.