The probability that $1$ is a fixed point of a random permutation of $\{1,2,\ldots,n\}$ (with uniform distribution) is $1/n.$ This is easy to prove since there are $(n-1)!$ permutations that have $1$ as a fixed point and $(n-1)!\,/\,n!=1/n.$ A more direct proof is simply to observe that the image of $1$ under a random permutation is equally likely to be any of the $n$ elements of $\{1,2,\ldots,n\}.$

The probability that $2$ immediately follows $1$ in a random permutation is also $1/n,$ and there is a proof similar to the first proof above in that it involves the computation $(n-1)!\,/\,n!=1/n.$ In a permutation $\pi_1\pi_2\ldots\pi_n$ such that $\pi_i\pi_{i+1}=12$ there are $n-1$ possible values of $i$ and there are $(n-2)!$ ways to permute $\{3,4,\ldots,n\}$ among the remaining $n-2$ positions. Hence there are $(n-1)!$ permutations in which $2$ immediately follows $1.$ Perhaps simpler is to observe that there are $(n-1)!$ permutations of $\{12,3,4,\ldots,n\},$ where $12$ is regarded as a single object.

My question: is there a direct way of seeing this, similar to the direct way of seeing that the probability that $1$ is a fixed point is $1/n?$