Two players, $$P_1$$ and $$P_2$$, 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 $$P_1$$ and $$P_2$$, respectively. If $$x > y$$, then $$P_1$$ scores 5 points and $$P_2$$ scores 0 point. If $$x = y$$, then each player scores 2 points. If $$x < y$$, then $$P_1$$ scores 0 point and $$P_2$$ scores 5 points. Let $$X_i$$ and $$Y_i$$ be the total scores of $$P_1$$ and $$P_2$$, respectively, after playing the $$i^{th}$$ round.

List-I	List-II
(I) Probability of $$(X_2 \geq Y_2)$$ is	(P) $$\frac{3}{8}$$
(II) Probability of $$(X_2 > Y_2)$$ is	(Q) $$\frac{11}{16}$$
(III) Probability of $$(X_3 = Y_3)$$ is	(R) $$\frac{5}{16}$$
(IV) Probability of $$(X_3 > Y_3)$$ is	(S) $$\frac{355}{864}$$
	(T) $$\frac{77}{432}$$

The correct option is:

For each round of the game let $$x$$ and $$y$$ be the numbers shown by $$P_1$$ and $$P_2$$, respectively.
If $$x\gt y$$ the score pair is $$(5,0)$$; if $$x=y$$ it is $$(2,2)$$; if $$x\lt y$$ it is $$(0,5)$$.

Because the dice are fair and independent, among the $$36$$ ordered pairs $$(x,y)$$:
• $$x\gt y$$ occurs in $$15$$ pairs  $$\Rightarrow$$ probability $$\dfrac{15}{36}=\dfrac{5}{12}$$.
• $$x=y$$ occurs in $$6$$ pairs    $$\Rightarrow$$ probability $$\dfrac{6}{36}=\dfrac{1}{6}$$.
• $$x\lt y$$ also occurs with probability $$\dfrac{5}{12}$$ (symmetry).

Define the “score difference” for one round as $$D=({\text{score of }}P_1)-({\text{score of }}P_2).$$ Hence $$D$$ takes three values

$$D=+5\;(\text{prob. }5/12),\quad D=0\;(\text{prob. }1/6),\quad D=-5\;(\text{prob. }5/12).$$

Totals after $$n$$ rounds satisfy $$X_n-Y_n=D_1+D_2+\dots +D_n,$$ with $$D_1,D_2,\dots$$ i.i.d. as above.

Case 1: Two rounds (distribution of $$S_2=D_1+D_2$$)

All ordered pairs $$(D_1,D_2)$$ and their probabilities:

$$\begin{array}{c|c} S_2 & \text{Probability} \\ \hline +10 & (5/12)^2 = 25/144\\ +5 & 2\,(5/12)(1/6)=20/144\\ 0 & 2\,(5/12)^2 + (1/6)^2 = 50/144+4/144 = 54/144\\ -5 & 2\,(1/6)(5/12)=20/144\\ -10 & (5/12)^2 = 25/144 \end{array}$$

Probability $$X_2\ge Y_2$$ (i.e. $$S_2\ge0$$): $$\dfrac{25+20+54}{144}=\dfrac{99}{144}=\dfrac{11}{16}.$$

Probability $$X_2\gt Y_2$$ (i.e. $$S_2\gt0$$): $$\dfrac{25+20}{144}=\dfrac{45}{144}=\dfrac{5}{16}.$$

Case 2: Three rounds (distribution of $$S_3=D_1+D_2+D_3$$)

Write $$(W,T,L)$$ for the counts of wins, ties, losses (so $$W+T+L=3$$ and $$S_3=5(W-L)$$).

For $$S_3=0$$ we need $$W=L$$. Two possibilities:

1. $$(0,3,0)$$ - three ties.   Probability $$\left(\dfrac{1}{6}\right)^3=\dfrac{1}{216}.$$

2. $$(1,1,1)$$ - one win, one tie, one loss.
  Number of permutations $$=3!/(1!1!1!)=6.$$
  Probability of each permutation $$\;(5/12)(1/6)(5/12)=\dfrac{25}{864}.$$
  Total probability $$6\times\dfrac{25}{864}=\dfrac{150}{864}=\dfrac{25}{144}.$$

Hence $$P(S_3=0)=\dfrac{1}{216}+\dfrac{25}{144}= \dfrac{4}{864}+\dfrac{150}{864}= \dfrac{154}{864}= \dfrac{77}{432}.$$

Because the game is symmetric, $$P(S_3\gt0)=P(S_3\lt0).$$ Therefore

$$P(S_3\gt0)=\dfrac{1-P(S_3=0)}{2}= \dfrac{1-\dfrac{77}{432}}{2}= \dfrac{355}{432}\times\dfrac{1}{2}= \dfrac{355}{864}.$$

We now list the required probabilities:

(I) $$P(X_2\ge Y_2)=\dfrac{11}{16} \; (Q)$$
(II) $$P(X_2\gt Y_2)=\dfrac{5}{16} \; (R)$$
(III) $$P(X_3=Y_3)=\dfrac{77}{432} \; (T)$$
(IV) $$P(X_3\gt Y_3)=\dfrac{355}{864} \; (S)$$

Thus the correct matching is:
(I) → (Q), (II) → (R), (III) → (T), (IV) → (S).

Hence the correct option is:
Option A which is: (I) → (Q); (II) → (R); (III) → (T); (IV) → (S).