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50 people each purchase a ticket for concert A and a ticket for concert B. The concerts both occur in an auditorium of 50 seats. If the seat numbers for the tickets are distributed randomly, what's the probability that no one has the same seat number for both concerts?

I tried using random variables: Each person has a $(\frac{1}{50})(\frac{49}{50})$ chance of getting differing seat numbers, so the probability of no one getting the same seat number for both concerts is $(50)(\frac{49}{2500}) = \frac{49}{50}$. However, I am concerned about how the seat numbers are dependent; person X occupying seat 1 in concert A means that no one else can take that seat... is this a valid concern?

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Yes, your concern is valid. This, if we want an exact answer, is a classical problem that is not easy, unless one has met it before. It is the problem of derangements. Wikipedia has a reasonably thorough discussion. Ultimately the formula is very pretty. One can get a good approximation by neglecting the dependencies. –  André Nicolas Oct 9 '12 at 4:15
    
Thanks! Indeed, that Wikipedia article is good. How come using random variables here won't work though? –  John Hoffman Oct 9 '12 at 5:50
    
You could use random variable tricks to find, for example, the expected number of seat matches (it is $1$). By the way, the answer of Alex based on the technically incorrect independence assumption is numerically very close to the truth. –  André Nicolas Oct 9 '12 at 6:02

1 Answer 1

up vote 1 down vote accepted

If two people could somehow occupy one and the same seat the answer would be $$ \bigg(1-\frac{1}{50}\bigg)^{50} \approx e^{-1} $$ but since this is not possible we have to account for this, hence (this also assumes that for the first concert not two people got a ticket to one and the same place: $$ \frac{49}{50} \cdot \frac{48}{49} \cdots\frac{1}{2}\cdot1=\frac{1}{50} $$

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The ultimate answer is indeed close to $e^{-1}$. This is the problem of derangements (permutations with no fixed point). –  André Nicolas Oct 9 '12 at 5:12

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