Even though all 52 cards of a deck are distinguishable, we can simplify matters by
considering all 4 aces as indistinguishable from each other, and the 48 non-aces also as
indistinguishable from each other.

I will first justify this assumption of indistinguishability by what is meant by
"expectation".

The expectation in this case is the average number of adjacent aces per shuffle.

Suppose we performed the experiment many times of shuffling a deck of cards and after
each shuffle, we counted and kept a running average of the number of adjacent pairs of
aces per shuffle. The expectation of the number of adjacent pairs of aces is the number
that should be close to the running average of adjacent pairs of aces per shuffle.

My justification is by thinking of the way we would perform each experiment of going
through the cards to count the number of adjacent pairs of aces. As we would go through
the deck of cards looking for adjacent pairs of aces, we would pay no attention to the
suit of any of the cards, and we would also pay no attention to the rank of any
of the cards except merely to determine whether it is an ace or not.

So I will go by my assumption. It is the same as if we had a deck of 4 white cards and
48 black cards and we wanted to find the expected number of adjacent pairs of white
cards per shuffle. But I go back to aces and non-aces.

We begin by laying out the 48 "indistinguishable" non-aces in a circle, leaving a space
between each non-ace.  We will now place the 4 "indistinguishable" aces in some of those
48 spaces between cards.  Remember, we can put more than 1 ace in a single space between
two non-aces.
 
We can insert the 4 aces in some of 48 available spaces between non-aces in the
following cases, which correspond to the 5 ways to write 4 as a sum of positive
integers or as a single positive integer:

1+1+1+1 = 2+1+1 = 2+2 = 3+1 = 4

Case 1.  4 separated aces (no adjacent pairs) [That's 0 adjacent pairs]
Case 2.  1 adjacent pair of aces and 2 aces separated. [That's 1 adjacent pair]
Case 3.  2 adjacent pairs of aces.
Case 4.  3 aces together and 1 ace separated. [That's also 2 pairs of adjacent aces]
Case 5.  All 4 aces together. [That's 3 pairs of adjacent aces]

Case 1. There are 48C4=194580 ways to place the 4 aces, no two together.
Case 2. There are 48 ways to place the adjacent pair and 47C2=1081 ways to place the
        separated two.  That's (48)(1081)=51888 ways.
Case 3. There are 48C2=1128 ways to place the two pairs of adjacent aces.
Case 4. There are 48 ways to place the 3 aces together and then 47 ways to place the
        separated ace.  Thats (48)(47)=2256 ways.
Case 5. There are 48 ways to place all 4 aces together.

The total number of ways to place the aces is

194580 + 51888 + 1128 + 2256 + 48 = 249900

The probability that there are 0 (no) adjacent pairs = 194580/249900 = 3243/4165
The probability that there is exactly 1 adjacent pair = 51888/249900 = 4324/20825
The probability that there are exactly 2 adjacent pairs = (1128+2256)/249900 = 282/20825
The probability that there are exactly 3 adjacent pairs = 48/249900 = 4/20825

The expected number of pairs of aces together is
0(3243/4165) + 1(4324/20825) + 2(282/20825) + 3(4/20825) = 4/17  <--answer

A recap of what this 4/17 now means:

Out of every 17 shuffles you expect to average counting about 4 pairs of adjacent aces. 

Edwin