Puzzle 15 of 26 · Hard

The lost boarding pass

100 passengers board a plane with 100 seats, one at a time. The first has lost their boarding pass and sits in a random seat. Everyone after sits in their own seat if it’s free, and in a random free seat if it isn’t. What’s the chance the last passenger gets their own seat?

Seat 1

\displaystyle 1

Seats 2–99

\displaystyle \dots

Seat 100

\displaystyle 100

The seat the first passenger lost, and the last passenger’s seat.

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Answer: 1/2 = 50%

Step 1 · Only two seats matter
$\displaystyle \text{the first passenger's seat and the last passenger's seat}$
Whenever someone sits at random, they either take seat 1 (everyone after is fine), seat 100 (the last passenger loses), or push the problem on.
Step 2 · At every random pick
$\displaystyle \text{seat } 1 \text{ and seat } 100 \text{ are equally likely}$
Both seats are free until one of them is taken, and a random pick treats them the same. The other seats only delay the choice.
Step 3 · Chance seat 1 is taken first
$\displaystyle 1 \text{ of the } 2 \text{ seats that matter } = \frac{1}{2}$
The game ends the moment either seat is taken, and each is equally likely to go first.

Answer $\displaystyle \frac{1}{2} = 50\%$

Step 3 of 3

The method · Unit 3, Conditional probabilitySplitting into casesFind a chance by adding up every case

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