Airplane Seating Problem

100 passengers are boarding an airplane with 100 seats. Everyone has a ticket with his seat number. These 100 passengers boards the airplane in order. However, the first passenger lost his ticket so he just take a random seat. For any subsequent passenger, he either sits on his own seat or, if the seat is taken, he takes a random empty seat. What's the probability that the last passenger would sit on his own seat?

Answer: 1/2

Solution:
When 100th passenger arrives, only seat 1 or 100 is vacant (rest all must be non-empty). To see this, let seat 50 is vacant. Since passenger 50 came before passenger 100, he would sit in his seat only. Hence seat 50 cannot be vacant. So all permutations of the seating arrangement would result in last person sitting in either seat 1 or seat 100. Both the options are equi-likely.

Tip: Its wiser to solve problems like these for small values and then try to generalize. For e.g. let there be three passengers p1, p2, p3. p1 can sit in seat 1 with prob=1/3. In this case p3 sites in seat 3. p1 can sit in seat 2 with probability = 1/3, p2 can sit in either seat 1 or 3 (with prob=1/2). so p3 gets to sit in his seat with probability = 1/3 * 1/2 = 1/6. p1 can sit in seat 3 with probability 1/3. In this case p3 can sit in seat 3 with probability = 0. So the total probability for p3 to sit in his seat = 1/3 + 1/6 + 0 = 1/2