Dice roll combinations in Monopoly with doubles, using total combinations and sums of dice

Question

So in Monopoly, when doubles are rolled with a pair of dice, one can roll again to reach a new sum. For example, 5 can be reached with (1,4), (4,1), (2,3), (3,2) or with doubles - (1,1) and (1,2) or (1,1) and (2,1). The probability of getting a 5 is thus $\frac {1}{36}+\frac {1}{36}+\frac {1}{36}+\frac {1}{36}+(\frac {1}{36})^2+(\frac {1}{36})^2 = 0.11265 $. One method I tried to use to model this was by taking the total number of ways to get 5 divided by the total number of possible sum combinations in monopoly. Including doubles, I found that there are $((36\times 6)+30)\times 6)+30 = 1506$ possible sums (I can try to explain this later if anyone would like), and of this there are 6 ways to get 5. But $\frac {6}{1506} = 0.003984 $ and not $0.11265$. Is my fundamental logic in approaching this wrong? I would appreciate any help, thanks a lot!