5. The faces of a 12 -sided die are numbered $1,2,3,4,5,6,7,8,9,10,11$, and 12 such that the sum of the numbers on opposite faces is 13 . The die is meticulously carved so that it is biased: the probability of obtaining a particular face $F$ is greater than $1 / 12$, the probability of obtaining the face opposite $F$ is less than $1 / 12$ while the probability of obtaining any one of the other ten faces is $1 / 12$. When two such dice are rolled, the probability of obtaining a sum of 13 is $29 / 384$. What is the probability of obtaining face $F$ ?
Answer: $7 / 48$
Solution: The probabilities that the die lands on its corresponding faces are $\frac{1}{12}, \ldots, \frac{1}{12}, \frac{1}{12}+x$, and $\frac{1}{12}-x$, where the last two probabilities are for the face $F$ and its opposite face, respectively, while the rest are for the other faces (since these probabilities must sum up to 1 ). Now, the sum of the results of the two dice can only be 13 if the results shown on both dice are such that they are opposites of one another. Hence,
$$
10\left(\frac{1}{12}\right)\left(\frac{1}{12}\right)+2\left(\frac{1}{12}+x\right)\left(\frac{1}{12}-x\right)=\frac{29}{384} \text {. }
$$

This leads to the quadratic equation
$$
\frac{1}{144}-x^{2}=\frac{7}{2304} .
$$

Solving for $x$, we get $x=\frac{1}{16}$ as the only plausible solution. Therefore, the probability of obtaining face $F$ must be $\frac{1}{12}+\frac{1}{16}=\frac{7}{48}$.