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.
 A: 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)
$$
\sum_{k=1}^7\overbrace{\frac18}^{k\text{ on d}8}\cdot\overbrace{\frac{k-1}6}^{\lt k\text{ on d}6}
+\overbrace{\frac18}^{8\text{ on d}8}\cdot\overbrace{1\vphantom{\frac16}}^{\lt8\text{ on d}6}
=\frac{9}{16}
$$
A: Let $d_8$ be the result of the 8-sided die, and $d_6$ the result of the the 6-sided die.
$$P(d_8<d_6) = \sum_{i=1}^6 P((d_8 < d_6) \cap(d_6=i)) = \sum_{i=1}^6 P(d_8 < d_6|d_6=i)\cdot P(d_6=i)$$
$$=\sum_{i=1}^6 \frac{i-1}{8} \cdot \frac{1}{6} = \frac{1}{48}\sum_{i=1}^6 (i-1) = \frac{\frac{6\cdot 7}{2}-6}{48} = \frac{15}{48} = \frac{5}{16}$$
You can generalise for any $n,m$ sided dice. Assume $n>m$ we have:
$$P(d_n<d_m) = \sum_{i=1}^m P((d_n < d_m) \cap(d_m=i)) = \sum_{i=1}^m P(d_n < d_m|d_m=i)\cdot P(d_m=i)$$
$$=\sum_{i=1}^m \frac{i-1}{k} \cdot \frac{1}{m} = \frac{1}{nm}\sum_{i=1}^m (i-1) = \frac{\frac{m\cdot (m+1)}{2}-m}{nm} = \frac{\frac{m+1}{2}-1}{n} = \frac{\frac{m-1}{2}}{n} = \frac{m-1}{2n}$$
A: This is rather easy. You don't even need a grid. We just have to count the number of tuples $(x,y) ; x \subset A$ and $y\subset B$ [where A is the est of all possible numbers of the six sided dice(1-6) and B is the  set of all possible numbers of the eight sided dice(1-8)] such that $x>y$(by the problem). There are 15 such tuples: 5 for x = 6, 4 for x = 5 so on till 1 for x = 2. Thus the probability is P= $\frac{15}{48}$.
PS: I don't know how your edit makes sense with the probability being $\frac{120}{48}$,which i believe is greater than 1.
A: $$\begin{align}\mathsf P(X_6>X_8) ~=~& \mathsf P(X_8<7)~\mathsf P(X_6>X_8\mid X_8<7)+\mathsf P(X_8>6)~\mathsf P(X_6>X_8\mid X_8>6) \\ ~=~& \frac 68\cdot\frac {15}{36}+\frac 28\cdot 0  \\ ~=~& \frac {5}{16} \end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

From the OP:
  
  
*
  
*" Say one person rolls an $8$ sided die and the other rolls a six ".
  
*" What is the probability that the six sided die
  $\underline{is\ higher\ than}$ the $8$ ? ".
  

$$\bbx{\ds{\large%
\mbox{Hereafter,}\ \bracks{\cdots}\ \mbox{is the}\ Iverson\ Bracket}}
$$

The answer to $2.$ is given by:
\begin{align}
{1 \over 8}\sum_{d_{8} = 1}^{8}{1 \over 6}\sum_{d_{6} = 1}^{6}
\bracks{d_{6} > d_{8}} & =
{1 \over 48}\sum_{d_{8} = 1}^{8}\braces{\bracks{d_{8} \leq 5}
\sum_{d_{6} = d_{8} + 1}^{6}} =
{1 \over 48}\sum_{d_{8} = 1}^{8}\bracks{d_{8} \leq 5}\pars{6 - d_{8}}
\\[5mm] & =
{1 \over 48}\sum_{d_{8} = 1}^{5}\pars{6 - d_{8}} =
{1 \over 48}\bracks{5 \times 6 - {5\pars{5 + 1} \over 2}} =
\bbx{\ds{5 \over 16}} = 0.3125
\end{align}
A: If there is a die with $m$ sides and a die with $n$ sides, assuming that $m<n$, 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.
