Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.0 AIME, harder Prove it Philippines

Problem:

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 $F$ ?

Solution

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,
$10\left(\frac{1}{12}\right)\left(\frac{1}{12}\right)+2\left(\frac{1}{12}+x\right)\left(\frac{1}{12}-x\right)=\frac{29}{384}$
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 $\frac{1}{12}+\frac{1}{16}=\frac{7}{48}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.