JEE Advanced 2022 — Paper 1 — Question 16

Let $x$ and $y$ be the outcomes of the dice rolled by $P_1$ and $P_2$, respectively. The sample space size for one round is $6 \times 6 = 36$.

The probabilities for a single round are: $P(P_1 \text{ wins}) = P(x > y) = \frac{15}{36} = \frac{5}{12}$ $P(\text{Draw}) = P(x = y) = \frac{6}{36} = \frac{1}{6}$ $P(P_2 \text{ wins}) = P(x < y) = \frac{15}{36} = \frac{5}{12}$ Case 1: Two Rounds ($n=2$)

Total outcomes = $36^2 = 1296$. The scores are equal ($X_2 = Y_2$) in the following scenarios: Two draws: $\left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{36}$ One win each: $2 \times \left(\frac{5}{12} \times \frac{5}{12}\right) = \frac{50}{144} = \frac{25}{72}$ Summing these:

$$P(X_2 = Y_2) = \frac{1}{36} + \frac{25}{72} = \frac{2 + 25}{72} = \frac{27}{72} = \frac{3}{8}$$

Using symmetry, $P(X_2 > Y_2) = P(X_2 < Y_2)$:

$$P(X_2 > Y_2) = \frac{1 - P(X_2 = Y_2)}{2} = \frac{1 - 3/8}{2} = \frac{5}{16}$$ $$P(X_2 \ge Y_2) = P(X_2 > Y_2) + P(X_2 = Y_2) = \frac{5}{16} + \frac{6}{16} = \frac{11}{16}$$

Case 2: Three Rounds ($n=3$)

The scores are equal ($X_3 = Y_3$) if: Three draws: $\left(\frac{1}{6}\right)^3 = \frac{1}{216}$ One draw, one $P_1$ win, and one $P_2$ win: $\frac{3!}{1!1!1!} \times \left(\frac{1}{6} \times \frac{5}{12} \times \frac{5}{12}\right) = 6 \times \frac{25}{864} = \frac{150}{864}$ Summing these:

$$P(X_3 = Y_3) = \frac{4}{864} + \frac{150}{864} = \frac{154}{864} = \frac{77}{432}$$

Now calculating $P(X_3 > Y_3)$ via symmetry:

$$P(X_3 > Y_3) = \frac{1 - P(X_3 = Y_3)}{2} = \frac{1 - 77/432}{2} = \frac{355/432}{2} = \frac{355}{864}$$

Final Matching (I) $\rightarrow$ (Q) (II) $\rightarrow$ (R) (III) $\rightarrow$ (T) (IV) $\rightarrow$ (S)