Two players, P₁ and P₂, play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let x and y denote the readings on the die rolled by P₁ and P₂, respectively. If x > y, then P₁ scores 5 points and P₂ scores 0 point. If x = y, then each player scores 2 points. If x < y, then P₁ scores 0 point and P₂ scores 5 points. Let X_i and Y_i be the total scores of P₁ and P₂, respectively, after playing the i^th round.

List-I	List-II
(I) Probability of X2 ≥ Y2 is	(P) $\frac{3}{8}$
(II) Probability of X2 > Y2 is	(Q) $\frac{11}{16}$
(III) Probability of X3 = Y3 is	(R) $\frac{5}{16}$
(IV) Probability of X3 > Y3 is	(S) $\frac{355}{864}$
	(T) $\frac{77}{432}$

The correct option is :

The correct option is (I) → (Q); (II) → (R); (III) → (T); (IV) → (S).

Understanding the Game Dynamics:
In each round, two players P₁ and P₂ roll a standard fair 6-sided die once. Let the outcome of P₁'s roll be x and P₂'s roll be y.
Since the die has 6 distinct faces and is fair, each of the 36 possible ordered pairs (x, y) is equally likely with a probability of $\frac{1}{36}$.

The rules for scoring in any single round are:

1. Case 1: x = y (Tie)
There are 6 such outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
Probability of tie: P(x = y) = $\frac{6}{36}=\frac{1}{6}$.
Points awarded: P₁ gets 2 points, P₂ gets 2 points.

2. Case 2: x > y (P₁ wins the round)
Number of outcomes where x > y is $\frac{36-6}{2}=15$.
Probability of P₁ winning: P(x > y) = $\frac{15}{36}=\frac{5}{12}$.
Points awarded: P₁ gets 5 points, P₂ gets 0 points.

3. Case 3: x < y (P₂ wins the round)
Number of outcomes where x < y is 15.
Probability of P₂ winning: P(x < y) = $\frac{15}{36}=\frac{5}{12}$.
Points awarded: P₁ gets 0 points, P₂ gets 5 points.

Part (I): Probability of X₂ ≥ Y₂
Let X_i and Y_i be the total scores of P₁ and P₂ after round i.
By symmetry between players P₁ and P₂:

$P(X_2>Y_2)=P(X_2<Y_2)$

Also, the total probability space covers:

$P(X_2>Y_2)+P(X_2=Y_2)+P(X_2<Y_2)=1$

Thus, $P(X_2\ge Y_2)=P(X_2>Y_2)+P(X_2=Y_2)$.
Using symmetry, $P(X_2\ge Y_2)=\frac{1+P(X_2=Y_2)}{2}$.

X₂ = Y₂ occurs only if:
- Both rounds are ties: Probability = $\left(\frac{1}{6}\right)\times \left(\frac{1}{6}\right)=\frac{1}{36}$
- One round is won by P₁ and one round is won by P₂ (in any order): Probability = $2\times \left(\frac{5}{12}\right)\times \left(\frac{5}{12}\right)=\frac{25}{72}$

Therefore,

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

Now, calculate P(X₂ ≥ Y₂):

$P(X_2\ge Y_2)=\frac{1+\frac{3}{8}}{2}=\frac{11}{16}$

Hence, (I) → (Q).

Part (II): Probability of X₂ > Y₂
Using the values derived above:

$P(X_2>Y_2)=\frac{1-P(X_2=Y_2)}{2}=\frac{1-\frac{3}{8}}{2}=\frac{5}{16}$

Hence, (II) → (R).

Part (III): Probability of X₃ = Y₃
After 3 rounds, P₁ and P₂ will have equal total scores (X₃ = Y₃) if and only if:
- All 3 rounds end in ties: Probability = ${\left(\frac{1}{6}\right)}^3=\frac{1}{216}$
- Exactly 1 round is a tie, 1 round won by P₁, and 1 round won by P₂. The number of permutations for these 3 outcomes is 3! = 6:
Probability = $6\times \left(\frac{1}{6}\right)\times \left(\frac{5}{12}\right)\times \left(\frac{5}{12}\right)=\frac{25}{144}$

Summing these probabilities:

$P(X_3=Y_3)=\frac{1}{216}+\frac{25}{144}=\frac{2+75}{432}=\frac{77}{432}$

Hence, (III) → (T).

Part (IV): Probability of X₃ > Y₃
By symmetry between P₁ and P₂:

$P(X_3>Y_3)=\frac{1-P(X_3=Y_3)}{2}$

Substituting $P(X_3=Y_3)=\frac{77}{432}$:

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

Hence, (IV) → (S).

Combining all the matches:
(I) → (Q)
(II) → (R)
(III) → (T)
(IV) → (S)