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Taking Seats on a Plane

There are 100 seats on a plane and 100 passengers, each with his ticket. However, the first person to enter the plane discovers he has lost his ticket, so he picks a seat at random. Afterwards, every new passenger sits in his place if it is free, and otherwise picks a vacant seat at random.

You are the last to enter the plane. What is the probability you'll sit in your seat?

I managed to solve this using induction (i.e. marking by $A(n)$ the probability where $n$ is the number of passengers and then finding a recursive formula for $A(n)$ which is quite simple). However, I want to know if there are more "instantly obvious" or one-liner solutions.

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This is problem 69 in Bollobas "The Art of Mathematics" - so there is an elegant solution. Bollobas gives no hint, but maybe someone else can give one. –  Mark Bennet Jun 1 '12 at 16:13
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So, you can just google it and expect to find an answer. I googled the first sentence and got several links with a solution. It's 50%. –  Graphth Jun 1 '12 at 16:21
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As I said, you can just google it and get one. I don't mean you can google it to get the answer. I mean you can google it to get a solution that is very simple. Google the first sentence, look at the first two links. Although, Andre's solution is better than those! –  Graphth Jun 1 '12 at 16:27
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Voting to close as duplicate of Byron's link. This is why titles more descriptive than "Nice riddle"(!) are important. –  Rahul Jun 1 '12 at 16:58
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marked as duplicate by Rahul, Byron Schmuland, leonbloy, Hans Lundmark, Sasha Jun 1 '12 at 17:41

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1 Answer

up vote 10 down vote accepted

Let the first passenger to board have ticket for seat $p$, and suppose your ticket is for seat $q$. Either (i) $p$ is filled before $q$ is or (ii) $q$ is filled before $p$ is. In case (i), you will get seat $q$, and in case (ii) you won't.

These two cases are equally likely. For it is equally likely that the first passenger will choose $p$ or $q$. And if she chooses neither, then by symmetry $p$ and $q$ remain equally likely to be filled first, since they are the correct seat for none of the remaining passengers. So the required probability is $\frac{1}{2}$.

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