Assume there are $n$ people and a successful derangement has been drawn. If $1$ drew $2$, what is the probability that $2$ drew $1$?

Question

Assume there are $n$ people and a successful derangement has been drawn. If $1$ drew $2$, what is the probability that $2$ drew $1$?

Answer 1

Assume there are $n$ people, and a successful derangement has been drawn. If $1$ drew $2$, what is the probability that $2$ drew $1$?

Note that $*$ denotes the person to be deranged.

Considering the venn diagram,

\begin{align*} P(21**\cdots|*1**\cdots)&=\frac{N(21**\cdots)}{N(*1**\cdots)}\\ \end{align*}

METHOD $1$

where, by the inclusion-exclusion principle, \begin{align*} N(21**\cdots)&=\sum_{k=0}^{n-2}(-1)^k{n-2\choose k}(n-2-k)!=(n-2)!\sum_{k=0}^{n-2}\frac{(-1)^k}{k!}=!(n-2)\\ N(*1**\cdots)&=\sum_{k=0}^{n-2}(-1)^k{n-2\choose k}(n-1-k)!=(n-1)[!(n-2)]+(n-2)[!(n-3)]=\frac{!n}{n-1}\\ \end{align*} after rigorous simplification including breaking the summation into two.

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METHOD $2$

In a simpler method, since $m$ persons can be deranged in $!m=m!\left(\frac1{2!}-\frac1{3!}+\cdots+(-1)^m\right)$ ways, \begin{align*} N(21**\cdots)&=!(n-2)\\ N(*1**\cdots)&=\frac{!n}{n-1}\\ \end{align*} Explanation for $N(*1**\cdots)=\frac{!n}{n-1}$:

In deranging all $n$ persons, there are $n-1$ possible places $1$ could have been moved to, for the derangement method doesn't count the place of $1$. Hence, there are $\frac{!n}{n-1}$ ways of deranging $n$ persons such that $1$ moves to the place of $2$ and not to that of $3,4,\ldots n$.

Answer 2

Hint:

How many possibilities are there in that situation for persons $3,4,\dots,n$?

Be aware that each of them must be placed on a spot that differs from their original spot and that they cannot be placed on the spots 1 and 2. So again we recognize the possibilities as derangements.