Unable to derive reason/formula for permutation problem What is the probability of $n$ preceding $1$ and $n$ preceding $2$ when we randomly
select a permutation of ${1, 2, . . . , n}$ where $n ≥ 4$?
I wrote out examples of n! when n equals some number like $3$ and $4$, and I see that the answer should be $\frac{1}{3}$, but I'm unable to understand why this is the answer through a more formulaic way. 
 A: Consider the six permutations of $1,2,5(=n)$
$1-2-5, 1-5-2, 2-1-5, 2-5-1, 5-1-2, 5-2-1$
Clearly, in any such permutation of $3, \dfrac13$ satisfy the condition that $n$ precedes both $1$ and $2$
The intervening numbers ($3, 4$ here) can be put anywhere and permuted in each sequence, they will only serve as padding, they won't change the fraction that have $n$ before $1$ and $2$,
hence $Pr = \dfrac13$ 
A: The result holds for all $n \geq 3$. 
What matters here is that of the $3! = 6$ permutations of $1, 2, n$, two of them have $n$ appearing before both $1$ and $2$.  The placement of the other $n - 3$ numbers in the sequence has no effect on whether $n$ appears before both $1$ and $2$.  Hence, the probability that in a permutation of $1, 2, \ldots, n$ that $n$ appears before both $1$ and $2$ is $p = 1/3$, as you concluded.  
More formally, there are $n!$ permutations of $1, 2, \ldots, n$.  Let's count the number of permutations in which $n$ appears before both $1$ and $2$.  We choose three of the $n$ positions for the numbers $1$, $2$, and $n$.  Since $n$ appears before both $1$ and $2$, it must go in the leftmost of these three slots.  We then have $2!$ ways of arranging $1$ and $2$ in the remaining selected spots.  The remaining $n - 3$ numbers can be arranged in the remaining $n - 3$ open slots in $(n - 3)!$ ways.  Thus, the probability that $n$ appears before both $1$ and $2$ is 
$$p = \frac{\binom{n}{3} \cdot 2! \cdot (n - 3)!}{n!} = \frac{\frac{n(n - 1)(n - 2)}{3 \cdot 2} \cdot 2 \cdot (n - 3)!}{n!} = \frac{\frac{n!}{3}}{n!} = \frac{1}{3}$$
To connect the two arguments given above, note that we could also count the number of possible permutations of $1, 2, \ldots, n$ in the following manner.  We choose $3$ of the $n$ slots for $1$, $2$, and $n$.  We can arrange them in these slots in $3!$ ways, then arrange the remaining $n - 3$ numbers in the $n - 3$ open slots in $(n - 3)!$ ways.  Hence, the probability that $n$ appears before both $1$ and $2$ is 
$$p = \frac{\binom{n}{3} \cdot 2! \cdot (n - 3)!}{\binom{n}{3} \cdot 3! \cdot (n - 3)!} = \frac{2}{6} = \frac{1}{3}$$
