1. Possible stopping sums

When two fair dice are rolled, the sum can be from $2$ to $12$.

We stop the process when the sum is either:

• a prime number, or
• a perfect square.

In the range $2$ to $12$:

• Prime sums: $2,3,5,7,11$
• Perfect square sums: $4,9$

So the process continues until one of the sums in $\{2,3,4,5,7,9,11\}$ appears.

2. Interpretation of the condition

We are given that the process stops because a perfect square appears before any prime appears.

Thus, among the stopping sums, the first one observed is either $4$ or $9$.

We need: $p=P(\text{odd perfect square} \mid \text{perfect square occurs before prime})$

Since the only odd perfect square here is $9$, this becomes $p=P(9 \mid \text{first among }\{2,3,4,5,7,9,11\}\text{ is a square})$

That is, $p=\frac{P(\text{first stopping sum is }9)}{P(\text{first stopping sum is }4\text{ or }9)}$

3. Key idea: geometric waiting with relative probabilities

On each roll:

• If the sum is one of $\{2,3,4,5,7,9,11\}$, the process stops.
• Otherwise, it repeats.

Therefore, conditional on a stopping sum occurring, the probability that it is a particular stopping sum is proportional to its one-roll probability.

So, $p=\frac{P(\text{sum }=9)}{P(\text{sum }=4)+P(\text{sum }=9)}$

because we are conditioning on the event that the stopping sum is a perfect square.

4. Compute one-roll probabilities

Number of outcomes for sum $4$: $(1,3),(2,2),(3,1)$ So, $P(4)=\frac{3}{36}$

Number of outcomes for sum $9$: $(3,6),(4,5),(5,4),(6,3)$ So, $P(9)=\frac{4}{36}$

Hence, $p=\frac{4/36}{3/36+4/36}=\frac{4}{7}$

5. Find $14p$

$14p=14\cdot \frac{4}{7}=8$

6. Comparison with stored answer

Our derived answer is $8$, which matches the stored correct answer.