There's GOT to be a theorem here:

In a random set of N unique numbers, there are (N-1)/2 sequence breaks
(where x[k-1] > x[k]).

Experimental results:

N,average sequence breaks in 1000000 random data sets
2,0.499
3,1.000
4,1.499
5,1.999
6,2.501
7,3.000
8,3.500
9,4.000
10,4.501
11,5.000
12,5.498
13,5.999
14,6.500
15,6.999
16,7.499
17,8.000
18,8.502
19,8.998
20,9.500

End note 6:

Proof provided by Dr. Ray Hamel of my department:  There are on average
(n-1)/2 sequence breaks in a sequence of n randomly selected distinct items.
Each sequence S of n items [where S = (S(1), . . . , S(n))] has its reverse
sequence R such that R(j) = S(n-j+1).  Both sequences together have n-1
sequence breaks:  for each adjacent pair of (distinct) values, the two
values are out of order either in R or in S (but not in both), and there are
n-1 adjacent pairs.  For randomly selected sequences, both sets are equally
likely, so that there are (n-1)/2 sequence breaks on average.