Decisional Diffie-Hellman assumption vs decisional bilinear Diffie-Hellman assumption For the Decisional Diffie-Hellman (DDH) assumption we know that:

Given $g^a$ and $g^b$ for uniformly and independently chosen $a,b \in Z_p$ the value of $g^{ab}$ looks like a random value in group $\mathbb{G}$.

For the Decisional Bilinear Diffie-Hellman (DBDH) assumption we know that:

In a group $\mathbb{G}_0$ of prime order $p$, let $a,b,c \in \mathbb{Z}_p$ be chosen at random and $g$ be a generator of $\mathbb{G}_0$. The adversary when given $(g,g^a,g^b,g^c)$ must be able to distinguish a valid tuple $e(g,g)^{abc} \in \mathbb{G}_T$ from a random element $R \in \mathbb{G}_T$.

I cannot clearly understand the difference between both. Can someone state the difference of each and when to consider each one. I know that DBDH is used when considering pairings but I am still confused about the difference of both.
I don't know if it can helps. But actually, you can win the DDH game if you are using pairings. 

Suppose you are given $X = g^x$ and $Y = g^y$, you can apply $e(X,Y) = e(g,g)^{xy}$. 

Given $Z$, your goal is to know if $(X,Y,Z)$ is a DDH triplet. 

You can test, using useful properties of pairings, if $e(g,Z) = e(X,Y)$ and so you can distinguish from a random element. 

So I think that's why we consider 3 numbers to distinguish. 

You can also think about Key-exchange Diffie-Hellman and Joux "Tripartie DH" as described in this PDF.