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: General Pattern and Intuition

Key realization:

  • The first passenger creates the only randomness.
  • The process preserves a surprising symmetry:
    • Whenever the first passenger chooses seat 1 → everyone else sits normally.
    • If they choose seat 100 → passenger 100 loses their seat.
    • Any other choice just “reshuffles” the issue among the remaining passengers.

The only time passenger 100 definitely loses is if passenger 1 chooses seat 100.

There are two equally likely endpoints:

  • Seat 1 is taken first → passenger 100 wins.
  • Seat 100 is taken first → passenger 100 loses.

Final Answer

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

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

Reference