## The Airplane Seat problem

I thought I’d drop a little problem to think about for a while. If you’ve seen it before, please don’t ruin it for those who haven’t by posting a solution in the comments.

Let’s imagine an airplane with 100 seats. Instead of the cattle-call boarding procedures we know, passengers board the plane one at a time. On this flight, the first passenger has lost his ticket, so he just picks a seat at random to sit in. When the second passenger boards, if her own seat is available she sits in it. If not, *she* picks a random empty seat. This repeats until the plane fills. What is the probability that the 100th passenger gets to sit in his own seat?

Assume that all the random choices are made with equal probability of picking any currently-empty seat.

A nice problem that I hadn’t seen before. I understand that the new Airbus 380 will carry up to 853 passengers (if they’re all willing to sit in tiny, economy class seats). Do you know, the probability seems to be the same for that aircraft as well. I’d just toss a coin for it.

Chris Hobbs

Comment by Chris Hobbs | February 5, 2007 |

[...] The Airplane Seat solution And now, the solution to the Airplane Seat problem. [...]

Pingback by The Airplane Seat solution « The Unapologetic Mathematician | October 26, 2007 |

[...] is the original presentation and solution of the problem in “Car talk”. The problem and its solution was described also in John Amstrong’s blog “the Unapologetic [...]

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