Probability of Rolling Odd Square Before Prime Sum on Dice

To find the required probability, let us first analyze all possible outcomes when two fair six-sided dice are rolled.

The total number of outcomes when two dice are rolled is $6 \times 6 = 36$. Let $S$ denote the sum of the numbers on the two faces. The possible values of $S$ range from $2$ to $12$.

We categorize the possible sums into prime numbers, perfect squares, and other non-terminating numbers:

1. Prime numbers: $S \in \{2, 3, 5, 7, 11\}$

   • $S = 2$: $(1,1) \implies 1$ outcome
   • $S = 3$: $(1,2), (2,1) \implies 2$ outcomes
   • $S = 5$: $(1,4), (2,3), (3,2), (4,1) \implies 4$ outcomes
   • $S = 7$: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \implies 6$ outcomes
   • $S = 11$: $(5,6), (6,5) \implies 2$ outcomes

   Total number of outcomes resulting in a prime number: $N_{\text{prime}} = 1 + 2 + 4 + 6 + 2 = 15$

2. Perfect squares: $S \in \{4, 9\}$ (note that $1$ and $16$ cannot be obtained)

   • Even perfect square ($S = 4$): $(1,3), (2,2), (3,1) \implies 3$ outcomes
   • Odd perfect square ($S = 9$): $(3,6), (4,5), (5,4), (6,3) \implies 4$ outcomes

   Total number of outcomes resulting in a perfect square: $N_{\text{square}} = 3 + 4 = 7$

3. Other numbers (non-terminating sums): $S \in \{6, 8, 10, 12\}$

   • Total number of non-terminating outcomes = $36 - (15 + 7) = 14$ outcomes.

Rolling continues until the sum obtained is either a prime number or a perfect square. Rolls resulting in non-terminating sums do not affect the relative probabilities of terminating on a specific number.

We are given that the sum turns out to be a perfect square before it turns out to be a prime number. Under this given condition, the process must terminate on one of the perfect square sums, i.e., $S \in \{4, 9\}$.

Let $p$ be the conditional probability that this perfect square is an odd number ($S = 9$).

Using the definition of conditional probability: $p = P(S = 9 \mid S \in \{4, 9\}) = \frac{P(S = 9)}{P(S = 4) + P(S = 9)}$

Substituting the number of favorable outcomes for each sum: $p = \frac{\frac{4}{36}}{\frac{3}{36} + \frac{4}{36}} = \frac{4}{3 + 4} = \frac{4}{7}$

We are asked to compute the value of $14p$: $14p = 14 \times \frac{4}{7} = 8$