You’re on a full flight with a peculiar boarding process:

  • There are 100 passengers, each with a ticketed seat numbered 1 to 100.
  • The first passenger is drunk and picks a seat at random.
  • Each subsequent passenger:
    • Takes their own seat if it’s available.
    • If it’s taken, they choose a random available seat.

Question: What is the probability that the last passenger (100) ends up in their own seat?

Step 1: Analyze the Chaos

At first glance, this looks complicated—the drunk passenger disrupts the entire seating process.

But there’s a surprising pattern hiding beneath the randomness.

Let’s define \(P_n\) as the probability that passenger \(n\) finds seat \(n\) unoccupied when they board.

We want \(P_{100}\).

Step 2: Insight from Simpler Cases

Case \(n = 2\):

  • Passenger 1 (drunk) picks randomly: seat 1 or seat 2
    • If they pick seat 1 → passenger 2 gets their seat.
    • If they pick seat 2 → passenger 2 gets a random one.

So:

  • \(P_2 = \frac{1}{2}\)

Case \(n = 3\):

Carefully working through possibilities, you find:

  • \(P_3 = \frac{1}{2}\)

And for \(n = 4, 5, \ldots, 100\), simulations and theory confirm:

\[ P_{100} = \frac{1}{2} \]

Step 3: The Symmetry Argument

To solve this elegantly, notice what happens when any displaced passenger is forced to pick a random seat:

  • If they pick Seat 1: The chain of displacement ends immediately. Every remaining passenger will find their assigned seat empty, including Passenger 100.
  • If they pick Seat 100: Passenger 100’s seat is gone, and they will definitely lose.
  • If they pick Seat \(k\) (some other seat): The problem is simply deferred to Passenger \(k\), who will later find their seat taken and have to pick a random seat themselves.

Here is the magic symmetry: Whenever a passenger makes a random choice, Seat 1 and Seat 100 are always both available. Since they are picking randomly from the remaining pool, they are exactly equally likely to pick Seat 1 as they are to pick Seat 100.

The cycle of displacement bounces around until someone finally picks either Seat 1 or Seat 100. Because those two specific seats are perfectly symmetric in every random draw, there is exactly a 50/50 chance of which one gets picked first.

Final Answer

The probability that the last passenger gets their assigned seat is:

\[ \boxed{\frac{1}{2}} \]

Reference