For a particular peculiar pair of dice, the probabilities of rolling $1$, $2$, $3$, $4$, $5$, and $6$, on each die are in the ratio $1:2:3:4:5:6$. What is the probability of rolling a total of $7$ on the two dice? 
$\mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7}$
Let $x$ be the [probability](https://artofproblemsolving.com/wiki/index.php/Probability) of rolling a $1$. The probabilities of rolling a 
$2$, $3$, $4$, $5$, and $6$ are $2x$, $3x$, $4x$, $5x$, and $6x$, respectively.
The sum of the probabilities of rolling each number must equal 1, so
$x+2x+3x+4x+5x+6x=1$
$21x=1$
$x=\frac{1}{21}$
So the probabilities of rolling a $1$, $2$, $3$, $4$, $5$, and $6$ are respectively $\frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}$, and $\frac{6}{21}$. 
The possible combinations of two rolls that total $7$ are: $(1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ; (6,1)$
The probability P of rolling a total of $7$ on the two dice is equal to the sum of the probabilities of rolling each combination. 
$P = \frac{1}{21}\cdot\frac{6}{21}+\frac{2}{21}\cdot\frac{5}{21}+\frac{3}{21}\cdot\frac{4}{21}+\frac{4}{21}\cdot\frac{3}{21}+\frac{5}{21}\cdot\frac{2}{21}+\frac{6}{21}\cdot\frac{1}{21}=\frac{8}{63} \Rightarrow C$