in the tail of an increasing adjacency is $(n-1)!/n!=1/n$. Likewise, the probability that a letter $k<n$ is in the tail of a decreasing adjacency is $1/n$. Therefore, the probability that $k$ is in the tail of an adjacency is $$\label{eq:tail}\begin{cases}\frac{1}{n},&\mbox{ if } k=1 \mbox{ or }  n\\\frac{2}{n},&\mbox{ otherwise}\end{cases}.\nonumber$$ Summing over all letters $k=1,\ldots,n$ completes the proof.

An embedding in a permutation $\pi$ is likely to be normal if there is only a small proportion of letters in the tails of the adjacencies of $\pi$. Therefore, Lemma \[lem:adj\] indicates that the proportion of embeddings of $\lambda$ in a random permutation $\pi$ that are normal increases as the length of $\pi$ increases. If a permutation has no adjacencies of size $\ell>1$ then all embeddings will be normal, the proportion of such permutations tends to $1/e^2$ as the length of the permutations increase, see [@OEISA002464].

By Remark \[rem:tests\] we suspect that the second part of Equation vanishes for a significant proportion of intervals. A consequence of Proposition 3.3 in [@Smith14] is that if $\sigma$ and $\pi$ have the same number of descents then the second part of Equation vanishes. Note that although the definition of normal embeddings in [@Smith14] does not consider decreasing 