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The last passenger

Chance and uncertaintyLevel 4 · Advanced · ●●●●○

An airplane has $n$ seats ($n\ge 2$) and $n$ passengers with assigned seats.
Passenger 1 loses his card and sits in a random seat.
From passenger 2 onwards:

if your seat is free, you sit in it;
if it's busy, choose a free seat at random.

What is the probability that the last passenger sits in his own seat?

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Due to symmetry between the two, each occurs with probability 1/2.
When a passenger is forced to choose randomly, it can happen: Choose seat 1: from there the chaos ends and the last one ends well.
Therefore: $\mathbb{P}(\text{last passenger sits in their own seat})=\tfrac12$.

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Answer:

$$ \mathbb{P}(\text{last passenger sits in their own seat})=\tfrac12 \quad (n\ge2). $$

Why this is true: the random process effectively ends when one of two seats is chosen first: seat 1 or seat $n$.

If seat 1 is chosen first, the last passenger gets seat $n$.
If seat $n$ is chosen first, the last passenger cannot get their own seat.
Any other choice just defers the same structure to a later step.

By symmetry, seat 1 and seat $n$ are equally likely to be the first critical seat chosen, so the probability is $1/2$.

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