Expected peaks in a random ordering of 1 to 14

The numbers $1,2,\dots,14$ are arranged in a uniformly random order in a row. Call a position a peak if its number exceeds every number in the neighbouring position(s); the two end positions have just one neighbour each. For instance, the ordering $513246$ of $1$–$6$ has peaks at positions $1$, $3$ and $6$. What is the expected number of peaks?

$5$. Use indicators and linearity of expectation. An end position is a peak with probability $1/2$ (it must beat its single neighbour) and an interior position with probability $1/3$ (it must be the largest of three). So the mean is $2\cdot\tfrac12+12\cdot\tfrac13=5$.