Step 1: Analyze the probabilities of outcomes in a single round.

Let $x$ be the reading on the die rolled by player $P_1$ and $y$ be the reading on the die rolled by player $P_2$. Both are fair six-sided dice. The total number of possible outcomes in a single round is $6 \times 6 = 36$.

The scoring rules lead to three possible events in a round:

1. $P_1$ wins (W): This happens if $x > y$. The number of favorable outcomes is calculated by summing the possibilities for each value of $y$:

   • If $y=1$, $x \in \{2,3,4,5,6\}$ (5 outcomes)
   • If $y=2$, $x \in \{3,4,5,6\}$ (4 outcomes)
   • If $y=3$, $x \in \{4,5,6\}$ (3 outcomes)
   • If $y=4$, $x \in \{5,6\}$ (2 outcomes)
   • If $y=5$, $x \in \{6\}$ (1 outcome) Total outcomes for $x>y$ is $5+4+3+2+1=15$. The probability is $p_1 = P(W) = \frac{15}{36} = \frac{5}{12}$. Scores: $P_1$ gets 5 points, $P_2$ gets 0 points.

2. Draw (D): This happens if $x = y$. The favorable outcomes are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. There are 6 such outcomes. The probability is $p_d = P(D) = \frac{6}{36} = \frac{1}{6} = \frac{2}{12}$. Scores: Both players get 2 points.

3. $P_2$ wins (L, for $P_1$ loses): This happens if $x < y$. By symmetry with the $x>y$ case, there are 15 favorable outcomes. The probability is $p_2 = P(L) = \frac{15}{36} = \frac{5}{12}$. Scores: $P_1$ gets 0 points, $P_2$ gets 5 points.

Check: $p_1 + p_d + p_2 = \frac{5}{12} + \frac{2}{12} + \frac{5}{12} = \frac{12}{12} = 1$.

Step 2: Calculate probabilities for 2 rounds.

Let the outcomes of the two rounds be represented by a sequence of two letters (e.g., WD means $P_1$ wins round 1, and round 2 is a draw). $X_2$ and $Y_2$ are the total scores after 2 rounds.

• WW: $X_2=5+5=10, Y_2=0+0=0 \implies X_2 > Y_2$. Prob: $p_1^2 = (\frac{5}{12})^2 = \frac{25}{144}$.
• WD, DW: $X_2=5+2=7, Y_2=0+2=2 \implies X_2 > Y_2$. Prob: $2 p_1 p_d = 2(\frac{5}{12})(\frac{2}{12}) = \frac{20}{144}$.
• WL, LW: $X_2=5+0=5, Y_2=0+5=5 \implies X_2 = Y_2$. Prob: $2 p_1 p_2 = 2(\frac{5}{12})(\frac{5}{12}) = \frac{50}{144}$.
• DD: $X_2=2+2=4, Y_2=2+2=4 \implies X_2 = Y_2$. Prob: $p_d^2 = (\frac{2}{12})^2 = \frac{4}{144}$.
• DL, LD: $X_2=2+0=2, Y_2=2+5=7 \implies X_2 < Y_2$. Prob: $2 p_d p_2 = 2(\frac{2}{12})(\frac{5}{12}) = \frac{20}{144}$.
• LL: $X_2=0+0=0, Y_2=5+5=10 \implies X_2 < Y_2$. Prob: $p_2^2 = (\frac{5}{12})^2 = \frac{25}{144}$.

(II) Probability of $(X_2 > Y_2)$: This corresponds to the outcomes WW, WD, DW. $P(X_2 > Y_2) = P(WW) + P(WD) + P(DW) = \frac{25}{144} + \frac{20}{144} = \frac{45}{144}$. Simplifying the fraction: $\frac{45}{144} = \frac{9 \times 5}{9 \times 16} = \frac{5}{16}$. This matches List-II option (R). So, (II) $\rightarrow$ (R).

(I) Probability of $(X_2 \geq Y_2)$: This is $P(X_2 > Y_2) + P(X_2 = Y_2)$. $P(X_2 = Y_2)$ corresponds to WL, LW, DD. $P(X_2 = Y_2) = P(WL) + P(LW) + P(DD) = \frac{50}{144} + \frac{4}{144} = \frac{54}{144}$. $P(X_2 \geq Y_2) = \frac{45}{144} + \frac{54}{144} = \frac{99}{144}$. Simplifying the fraction: $\frac{99}{144} = \frac{9 \times 11}{9 \times 16} = \frac{11}{16}$. This matches List-II option (Q). So, (I) $\rightarrow$ (Q).

Step 3: Calculate probabilities for 3 rounds.

Let $N_W, N_D, N_L$ be the number of wins for $P_1$, draws, and losses for $P_1$ in 3 rounds, respectively. We have $N_W+N_D+N_L=3$. The total scores are $X_3 = 5N_W + 2N_D$ and $Y_3 = 5N_L + 2N_D$.

(III) Probability of $(X_3 = Y_3)$: The condition $X_3 = Y_3$ implies $5N_W + 2N_D = 5N_L + 2N_D$, which simplifies to $N_W = N_L$. Possible combinations for $(N_W, N_D, N_L)$ satisfying $N_W+N_D+N_L=3$ and $N_W = N_L$:

1. $(N_W, N_D, N_L) = (0, 3, 0)$: All three rounds are draws (DDD). The number of such sequences is 1. Probability is $p_d^3 = (\frac{2}{12})^3 = \frac{8}{1728}$.
2. $(N_W, N_D, N_L) = (1, 1, 1)$: One win, one draw, one loss. The number of sequences (permutations of WDL) is $3! = 6$. Probability for one sequence is $p_1 p_d p_2 = \frac{5}{12} \frac{2}{12} \frac{5}{12} = \frac{50}{1728}$. Total probability for this case is $6 \times \frac{50}{1728} = \frac{300}{1728}$. Total probability: $P(X_3 = Y_3) = \frac{8}{1728} + \frac{300}{1728} = \frac{308}{1728}$. Simplifying: $\frac{308}{1728} = \frac{154}{864} = \frac{77}{432}$. This matches List-II option (T). So, (III) $\rightarrow$ (T).

(IV) Probability of $(X_3 > Y_3)$: The condition $X_3 > Y_3$ implies $5N_W + 2N_D > 5N_L + 2N_D$, which simplifies to $N_W > N_L$. We sum the probabilities for all combinations $(N_W, N_D, N_L)$ that satisfy this condition:

1. $(3, 0, 0)$: (WWW) $\binom{3}{3}p_1^3 = 1 \cdot (\frac{5}{12})^3 = \frac{125}{1728}$.
2. $(2, 1, 0)$: (WWD) $\binom{3}{2,1,0}p_1^2 p_d = 3 \cdot (\frac{5}{12})^2(\frac{2}{12}) = \frac{150}{1728}$.
3. $(2, 0, 1)$: (WWL) $\binom{3}{2,0,1}p_1^2 p_2 = 3 \cdot (\frac{5}{12})^2(\frac{5}{12}) = \frac{375}{1728}$.
4. $(1, 2, 0)$: (WDD) $\binom{3}{1,2,0}p_1 p_d^2 = 3 \cdot (\frac{5}{12})(\frac{2}{12})^2 = \frac{60}{1728}$. Total probability: $P(X_3 > Y_3) = \frac{125 + 150 + 375 + 60}{1728} = \frac{710}{1728}$. Simplifying: $\frac{710}{1728} = \frac{355}{864}$. This matches List-II option (S). So, (IV) $\rightarrow$ (S).

Step 4: Conclusion

The matching is as follows:

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

This corresponds to option A.