medium | probability |

A line of 100 airline passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. For convenience, let's say that the $n^{\text{th}}$ passenger in line has a ticket for seat number 'n'. Being drunk, the first person in line picks a random seat (equally likely for each seat). All of the other passengers are sober, and will go to their assigned seats unless it is already occupied; If it is occupied, they will then find a free seat to sit in, at random. What is the probability that the last (100th) person to board the plane will sit in their own seat (#100)?

Can the last passenger arrive at any other seat than 1 or 100?

1/2

Note that the last passenger can only take seat #1 or #100.

- If any passenger takes seat #1, the cycle stops, and all the subsequent passengers take their own seats (including the last).
- Otherwise, if the #100 seat is taken before #1, the cycle is paused, i.e., the subsequent passengers do take their own seats, but the last passenger would take seat #1.

Now, any passenger from 1st to 99th who is picking a random vacant seat will choose between #1, #100 or any other seat equally likely. Thus, by symmetry, #1 or #100, anyone will be taken first - with equal probability.

Hence the last person ends up in their own seat with a probability of 0.5

### Follow-up Question

What's the probability that the second-last person sits on their seat?

### Follow-up Answer

$2/3$. The answer follows from the same logic of symmetry between the choice of seat #1, seat #100 and seat #99.

The probability that the $k^{\text{th}}$ person from the last will find their seat vacant is $k/(k+1)$.

### Script

Not convinced? Simulate using this Colab Script