A passenger aircraft has 100 seats and 100 passengers are about to board it.
The first passenger is drunk and has lost his boarding pass. He has no idea which is his seat, so he sits in a random seat.
All the other passengers have got boarding passes and follow the same process. If their seat is unoccupied they seat in their own seat. If their seat is occupied they sit in a random seat.
What is the probability that the 100th and last passenger to board will sit in their own seat?
Consider an aircraft with two seats and two passengers. The drunk passenger has a 50% chance of sitting in his seat, and so the last passenger has a 50% chance of sitting in their seat.
With three seats and three passengers, there is a 1/3 chance that the drunk sits in his seat, and the last passenger will sit in their own seat. Similarly, there is a 1/3 chance that the drunk sits in the last passenger's seat, and the last passenger will not sit in their own seat. And there is a 1/3 chance that the drunk sits in the second passenger's seat, in which case there is now a 50% chance that the last passenger sits in their own seat.
Combining these, there is still a 50% that the last passenger sits in their own seat.
And this continues all the way to 100 seats and 100 passengers, so the probability is 50%.