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 _____

The outcome observed on each throw of the two fair dice is the sum of the two face-values.

List the “special” sums that can stop the experiment:

• Perfect squares within 2-12: $$4,\,9$$
• Primes within 2-12: $$2,\,3,\,5,\,7,\,11$$

Write the probabilities of these seven sums (using the usual $$36$$ equally-likely ordered pairs):

$$\begin{aligned} P(4)&=\frac{3}{36}, &\;\;&\text{(1,3), (2,2), (3,1)} \\[2mm] P(9)&=\frac{4}{36}, &\;\;&\text{(3,6), (4,5), (5,4), (6,3)} \\[2mm] P(\text{prime})&=\frac{15}{36}, &\;\;&1+2+4+6+2=15 \text{ ways} \end{aligned}$$

Denote

$$p_{\text{sq}}=P(\text{square})=P(4)+P(9)=\frac{7}{36},\qquad p_{\text{pr}}=P(\text{prime})=\frac{15}{36}.$$

On every independent throw we have three mutually exclusive possibilities: square (probability $$p_{\text{sq}}$$), prime (probability $$p_{\text{pr}}$$) or “other” (probability $$1-p_{\text{sq}}-p_{\text{pr}}$$).

The experiment stops the very first time a square or a prime appears. Because each throw is independent, the probability that the first stopping sum is a square equals

$$P(\text{square before prime})= \frac{p_{\text{sq}}}{p_{\text{sq}}+p_{\text{pr}}} =\frac{\tfrac{7}{36}}{\tfrac{7}{36}+\tfrac{15}{36}} =\frac{7}{22}.$$

Conditional on this event, we still have to know which square (4 or 9) has appeared. Since only these two outcomes can stop the experiment in the “square” category, and their relative frequencies are unaffected by primes, we simply renormalise:

$$P(9\mid\text{square before prime})= \frac{P(9)}{P(4)+P(9)}= \frac{\tfrac{4}{36}}{\tfrac{3}{36}+\tfrac{4}{36}}= \frac{4}{7}.$$

Thus $$p=\frac{4}{7}.$$

Finally, $$14p = 14 \times \frac{4}{7}=8.$$

Answer: 8