Probability an 6 sided die will be higher than a 8 sided die? Say one person rolls an 8 sided die and the other rolls a six, what is the probability that the six sided die is higher than the 8?

I know that the expected value of the eight is 4.5 and the six is 3.5 but am having trouble figuring out how to find the probability. 

EDIT: Answer is 15/48 but still curious if there's a way of doing this without creating a grid.
Probability That The $\boldsymbol{6}$-Sided Die Will Be Higher (from the body of the question)
$$
\sum_{k=1}^6\overbrace{\frac16}^{k\text{ on d}6}\cdot\overbrace{\frac{k-1}8}^{\lt k\text{ on d}8}
=\frac{5}{16}
$$



Probability That The $\boldsymbol{8}$-Sided Die Will Be Higher (from the original title to the question)
$$
\sum_{k=1}^7\overbrace{\frac18}^{k\text{ on d}8}\cdot\overbrace{\frac{k-1}6}^{\lt k\text{ on d}6}
+\overbrace{\frac18}^{8\text{ on d}8}\cdot\overbrace{1\vphantom{\frac16}}^{\lt8\text{ on d}6}
=\frac{9}{16}
$$
If there is a die with $m$ sides and a die with $n$ sides, assuming that $m&lt;n$, the probability that $m$ will be greater than $n$ assuming that those are the names given to the outcomes when both of them are rolled is

$$\frac{m-1}{2n}$$

Full disclaimer: I made this up with help from the grid. This is a formula very specific to your question.