Probability of Selecting Permutations of Set Elements Given a set $\{1, 2, \ldots, n\}$ where $n\ge4$, find the probability that $n$ precedes 1 and $n-1$ precedes 2 when a permutation of this set is selected at random.  
When $n$ immediately precedes 1 and $n-1$ immediately precedes 2, the probability is $(n-2)!/n!$, but I'm not sure how to reason about the cases where $n$ and $n-1$ do not immediately precede 1 and 2.  A suggestion as to how I should think about this case would be appreciated.
 A: The events $1$ precedes $n$ and $n$ precedes $1$ are equally likely, so each has probability $\frac{1}{2}$. The same is true of the events  $2$ precedes $n-1$ and $n-1$ precedes $2$. 
The events $1$ precedes $n$ and $2$ precedes $n-1$ are independent. So the required probability is $\frac{1}{2}\cdot\frac{1}{2}$.
Remark: You can also do a counting argument, using the fact that all permutations of the numbers $1$ to $n$ are equally likely. 
A: Another way to think about this situation: The $n=4$ case and $n > 4$ case are actually the same, because one can think about generating a permutation of $n$ elements and tossing out every number other than $1,2,n-1,n$, and the result is a permutation of $(1,2,n-1,n)$ with each permutation equally likely by symmetry. So $n=4$ is the only case to consider. For $n=4$ a simple counting argument will work.
One can also show that all permutations for $n=4$ can be generated by first choosing the positions where $1,4$ will go: $\binom{4}{2}$ ways, and then arranging $(1,4)$ and $(2,3)$ in $2! \times 2!$ ways. Either way, no matter which two slots $(1,4)$ are in, $1$ out of the $4$ configurations satisfies what you are looking for.
