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 $n^{\text{th}}$nth 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)?

I encourage you to grab a pen and a piece of paper, and give this problem a go. When you are ready, keep reading for the solution.

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Solution

I will reveal the answer first. The 100th person can only take seat #1 or #100. That gives a probability of 1/2.

Let’s see why.

If passenger 1 chooses seat 1, this means everyone else also sits correctly, this means passenger 100 gets seat 100.

If passenger 1 chooses seat 100, then passenger 100 will never get their seat.

What about any other seat from seat 2 to seat 99? That’s where it gets interesting.

If passenger 1 took seat k, where 2 ≤ k ≤ 99, then person k will have to choose a random seat, but every other person from 2 to 99 can still get their desired seat. Of course, when it’s k’s turn to pick a seat, they might pick j, where k < j ≤ 99. This means person j will now have to pick a random seat when it’s their turn.

Eventually someone will be forced to choose between seat 1 and seat 100, with no bias.

In any of the situations, the 100th person will end up in their own seat with a probability of 0.5.

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Two Bonus Questions

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

What’s the probability that the k-th person will sit on their seat?