I am trying to find the distriubtion of X when $M_X(t)=\frac 16 e^t+ \frac{2}6e^{2t}+\frac{3}6e^{3t}$ With some simple computations, I found that $Var(x)=5/9$, and $E(x)= 7/3$
However, since the given MGF does not match any common forms I found in a text, I was not able to match it to a random variable. I even tried to use the definition, $\frac 16 e^t+ \frac{2}6e^{2t}+\frac{3}6e^{3t}=E(e^{tX})$
Along with this MGF, I could not match $e^t/ (2-e^t)$ with a random variable X.
Having the answer to these would be nice, but I am more interested in the $process$ one would use to match a moment generating function to a random variable, aside from just looking at common forms of moment generating functions.
Read off the coefficients of the moment generating function. Note that $$ M_{X}(t)=E(e^{tX})=\sum_{x} e^{tx}P(X=x). $$ Hence $$ P(X=1)=1/6;\quad P(X=2)=2/6;\quad P(X=3)=3/6. $$