imc

The outcome of the first dice can be $1$, $2$, $3$, or $4$. For each of these $4$ cases, we can find the possible outcomes for the second dice that makes the sum of the top two numbers $5$, $7$, or $9$, and then calculate the respective probabilities. Case 1 - the first dice is 1: the outcome of the second dice can be $4$, $6$, or $8$. There is a $\frac{1}{6}$ probability of rolling a $1$ with the first dice and a $\frac{1}{2}$ probability of rolling a $4$, $6$, or $8$ with the second dice. $\frac{1}{6}\ast \frac{1}{2}=\frac{1}{12}.$ Case 2 - the first dice is 2: the outcome of the second dice can be $3$ or $5$. There is a $\frac{1}{3}$ probability of rolling a $2$ with the first dice and a $\frac{1}{3}$ probability of rolling a $3$ or $5$ with the second dice. $\frac{1}{3}\ast \frac{1}{3}=\frac{1}{9}.$ Case 3 - the first dice is 3: the outcome of the second dice can be $4$ or $6$. There is a $\frac{1}{3}$ probability of rolling a $3$ with the first dice and a $\frac{1}{3}$ probability of rolling a $4$ or $6$ with the second dice. $\frac{1}{3}\ast \frac{1}{3}=\frac{1}{9}.$ Case 4 - the first dice is 4: the outcome of the second dice can be $1$, $3$, or $5$. There is a $\frac{1}{6}$ probability of rolling a $4$ with the first dice and a $\frac{1}{2}$ probability of rolling a $1$, $3$, or $5$ with the second dice. $\frac{1}{6}\ast \frac{1}{2}=\frac{1}{12}.$ $\frac{1}{12}+\frac{1}{9}+\frac{1}{9}+\frac{1}{12}=\frac{14}{36}=\frac{7}{18}$, so the answer is $\boxed{7/18}$.