Two fair dice, each with faces numbered $1, 2, 3, 4, 5$ and $6$, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If $p$ is the probability that this perfect square is an odd number, then the value of $14p$ is $\_\_\_\_\_\_$.

$\text{Two fair dice; observe the sum}$

$\text{Repeat till the sum is prime or a perfect square}$

$\text{Given: the stop is a perfect square before a prime}$

$p = P(\text{odd square} \mid \text{square before prime})$

$\text{To find: } 14p$

$\text{Sums: } 2,3,4,5,6,7,8,9,10,11,12$

$\text{Primes: } 2, 3, 5, 7, 11$

$\text{Perfect squares: } 4,\ 9$

$\text{Prime sums: } 2{:}1,\ 3{:}2,\ 5{:}4,\ 7{:}6,\ 11{:}2$

$\text{Total} = 1+2+4+6+2 = 15$

$P(\text{prime}) = \dfrac{15}{36}$

$\text{Square sums: } 4{:}3,\ 9{:}4 \Rightarrow 3+4 = 7$

$P(\text{square}) = \dfrac{7}{36}$

$\text{Others (repeat): } 6{:}5,\ 8{:}5,\ 10{:}3,\ 12{:}1$

$P(\text{repeat}) = \dfrac{14}{36}$

$\text{Re-rolls are irrelevant} \Rightarrow \text{race is } \text{prime vs square}$

$P(\text{square before prime}) = \dfrac{7/36}{7/36 + 15/36} = \dfrac{7}{22}$

$\text{Given a square: } 4 \to 3 \text{ ways},\quad 9 \to 4 \text{ ways}$

$\text{Odd square} = 9 \text{ only}$

$p = \dfrac{4}{3+4} = \dfrac{4}{7}$

$14p = 14 \times \dfrac{4}{7} = 8$