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I have this question: Let's say we have $n$ students sitting in $n$ seats, Now we get the students out of the seats and randomly assign them to the $n$ seats, what is the expected students sitting in their original seats?

For example, with $2$ students, the expected student would be $1$. As we have student $1$ in slot $1$, student $2$ in slot $2$,

we have $1/2$ chance of $\{1,2\}$ and $1/2$ chance of $\{2,1\}$ With that, the expected matched students would be $1/2 \cdot 2 + 1/2 \cdot 0 = 1$.

Mark Fantini
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faz
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  • I'm not sure this is as simple as it seems - find number of ways in which two children of the $n$ can be in wrong place, then do for three, four and so on. Then find the average. Finding number of ways in which $k$ children of $k$ can be in wrong place is also not an easy problem, so $k$ of $n$ is even more tricky. I am waiting for an answer, too. – tpb261 May 27 '14 at 04:31
  • The process if called derangement and this expectation can be computed using this. Wish somebody simplifies the whole thing :D – tpb261 May 27 '14 at 04:49

1 Answers1

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Let $X_i=1$ if student $i$ is sitting in her original seat, and let $X_i=0$ otherwise. Then the number $Y$ of students sitting in their original seats is given by $Y=X_1+X_2+\cdots+X_n$.

By the linearity of expectation, we have $E(Y)=E(X_1)+E(X_2)+\cdots +E(X_n)$.

We have $E(X_i)=\frac{1}{n}$, so $E(Y)=1$.

André Nicolas
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