The first thing I notice when I start to solve this problem is that the distance traveled by the man and dog combined by the time of their first reunion is 1250, the distance from the house to the end of the road and back. Let's say the dog travels "f" times as fast as the human. Let d₁ be the distance the dog travels in his first sortie, and h₁ be the distance traveled by the human.
d₁+h₁=1250
d₁/h₁=f

Solving these two equations for d₁ and h₁, we get
d₁=(1250)(f)/(f+1)
h₁=(1250)(1)/(f+1)

Now that the dog and the human are together again, the puzzle becomes a whole new game. The distance to the end of the road and back is now 1250-2h₁, and that's the distance traveled by the dog plus the human during the dog's second sortie.  Let d₂ be the distance traveled by the dog in his second excursion, and h₂ be the distance traveled by the human.
d₂+h₂ = 1250-2h₁
d₂+h₂ = 1250-(1250)(2)/(f+1)
d₂+h₂ = (1250)(f-1)/(f+1)
and, of course,
d₂/h₂=f

Solving these, we get
d₂=(1250) (f-1)/(f+1) (f)/(f+1)
h₂=(1250) (f-1)/(f+1) (1)/(f+1)

Now the dog and human are together again after two dog-trips to the end of the road. Once again, it's a whole new game, where the distance to the end of the road and back is now 1250-2h₁-2h₂. Using the same logic once again, we get
d₃+h₃ = (1250) (f-1)²/(f+1)² 
d₃/h₃ = f

By now, I'm sure you have spotted the pattern. Using a recursive proof, it's not hard to show that during the n^th sortie of the dog, the distance covered by the dog plus the human, d_n+h_n is given by this formula:
d_n+h_n = (1250) (f-1)^(n-1)/(f+1)^(n-1) 

The distance to the end of the road and back after the dog has made four sorties is given in the problem as twice 27 yards, or 162 feet. This is the same as the distance that would have been covered by dog and man combined during the fifth sortie,
d₅+h₅ = (1250) (f-1)⁴/(f+1)⁴ = 162
Solving this for (f-1)⁴/(f+1)⁴, we get the following:
162/1250 = (f-1)⁴/(f+1)⁴ 
81/625 = (f-1)⁴/(f+1)⁴ 
(81/625)^(1/4) = (f-1)/(f+1)
3/5 = (f-1)/(f+1)
f = 4

Now we know the dog runs four times as fast as the man. The statement of the problem tells us the man goes 4 miles an hour.  So the dog goes 16 miles an hour.