YES

We show the termination of R/S, where R is

  average(s(x),y) -> average(x,s(y))
  average(x,s(s(s(y)))) -> s(average(s(x),y))
  average(0,0) -> 0
  average(0,s(0)) -> 0
  average(0,s(s(0))) -> s(0)

and S is:

  rand(x) -> x
  rand(x) -> rand(s(x))

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  average#(s(x),y) -> average#(x,s(y))
  average#(x,s(s(s(y)))) -> average#(s(x),y)

and Q consists of the rules:

  average(s(x),y) -> average(x,s(y))
  average(x,s(s(s(y)))) -> s(average(s(x),y))
  average(0,0) -> 0
  average(0,s(0)) -> 0
  average(0,s(s(0))) -> s(0)
  rand(x) -> x
  rand(x) -> rand(s(x))

The weakly monotone algebra (N^2, >_lex) with

  average#_A((x1,y1),(x2,y2)) = (x1 + x2, y1 + y2)
  s_A((x1,y1)) = (x1, y1 + 1)
  average_A((x1,y1),(x2,y2)) = (x1 + x2 + 1, y1 + y2)
  0_A = (1, 0)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  average#(x,s(s(s(y)))) -> average#(s(x),y)

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^2, >_lex) with

  average#_A((x1,y1),(x2,y2)) = (x1, y1)
  s_A((x1,y1)) = (x1, y1 + 1)
  average_A((x1,y1),(x2,y2)) = (x1 + x2 + 1, y1 + y2)
  0_A = (1, 1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  average#(s(x),y) -> average#(x,s(y))

We remove them from the set of dependency pairs.

No dependency pair remains.