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Possible Duplicate:
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.

Community
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Gadi A
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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 01 '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%. – GeoffDS Jun 01 '12 at 16:21
  • As I said, I know the solution and have a basic method of finding it. I want something elegant and beautiful. – Gadi A Jun 01 '12 at 16:25
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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! – GeoffDS Jun 01 '12 at 16:27
  • The reason I posted this question here is exactly the hope that someone here knows an "optimal" solution, as far as elegance is concerned. – Gadi A Jun 01 '12 at 16:32
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    See also http://math.stackexchange.com/questions/5595/taking-seats-on-a-plane –  Jun 01 '12 at 16:50
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    Voting to close as duplicate of Byron's link. This is why titles more descriptive than "Nice riddle"(!) are important. –  Jun 01 '12 at 16:58

1 Answers1

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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 $\frac{1}{2}$.

André Nicolas
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