Two “loaded” dice each have the property that a $2$ or a $4$ is three times as likely to appear as a $1,3,5$, or $6$ on each roll. What is the probability that a $7$ will be the sum when the two dice are rolled?
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
If the dice are not loaded, each number $1,2,3,\cdots\!,6$ is equally likely to come up. The probability of rolling each number on a single die is $1/6$. Note that $1/6+1/6+1/6+1/6+1/6+1/6=1$.
What if the dice are loaded in favor of $2$ and $4$?
If the probability of rolling a $1,3,5$, or $6$ equals $1/x$, the probability of rolling a $2$ or $4$ equals $3/x$.
$1/x+1/x+1/x+1/x+3/x+3/x=1$
$10/x=1$
$x=1/10$
The probability of rolling a $1,3,5$, or $6$ on a single die equals $1/10$ and the probability of rolling a $2$ or $4$ on a single die equals $3/10$.
The sum of $7$ occurs when two dice are rolled as follows
$1,6$
$2,5$
$3,4$
$4,3$
$5,2$
$6,1$
The probability of a sum of $7$
$\dfrac{1}{10}\cdot\dfrac{1}{10}+\dfrac{3}{10}\cdot\dfrac{1}{10}+\dfrac{1}{10}\cdot\dfrac{3}{10}+\dfrac{3}{10}\cdot\dfrac{1}{10}+\dfrac{1}{10}\cdot\dfrac{3}{10}+\dfrac{1}{10}\cdot\dfrac{1}{10}=\dfrac{14}{100}=\dfrac{7}{50}$

Answer: $7/50$