Probability of Equal Points Scored by Teams in Two Matches

To find the probability $P(X = Y)$, we need to consider the possible outcomes for team $T_1$ in each of the two independent games and the points accumulated by both teams, $T_1$ and $T_2$.

For a single match, the possible outcomes from the perspective of $T_1$, their respective probabilities, and the points awarded to $(T_1, T_2)$ are given as follows:

• Win ($W$): Probability $P(W) = \frac{1}{2}$, Points awarded: $(3, 0)$
• Draw ($D$): Probability $P(D) = \frac{1}{6}$, Points awarded: $(1, 1)$
• Loss ($L$): Probability $P(L) = \frac{1}{3}$, Points awarded: $(0, 3)$

Let the outcome of the two matches be represented as an ordered pair $(M_1, M_2)$, where $M_1, M_2 \in \{W, D, L\}$. The total points accumulated after two matches are:

• $X =$ Points awarded to $T_1$
• $Y =$ Points awarded to $T_2$

We need to determine the outcomes where the total points scored by both teams are equal, i.e., $X = Y$.

Let's evaluate all possible combined outcomes for the two matches:

1. $(W, L)$:

   • Total points for $T_1$, $X = 3 + 0 = 3$
   • Total points for $T_2$, $Y = 0 + 3 = 3$
   • Here, $X = Y = 3$.
   • Probability: $P(W, L) = P(W) \times P(L) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

2. $(D, D)$:

   • Total points for $T_1$, $X = 1 + 1 = 2$
   • Total points for $T_2$, $Y = 1 + 1 = 2$
   • Here, $X = Y = 2$.
   • Probability: $P(D, D) = P(D) \times P(D) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

3. $(L, W)$:

   • Total points for $T_1$, $X = 0 + 3 = 3$
   • Total points for $T_2$, $Y = 3 + 0 = 3$
   • Here, $X = Y = 3$.
   • Probability: $P(L, W) = P(L) \times P(W) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$

For all other combinations, $X \neq Y$:

• $(W, W) \implies X = 6, Y = 0$
• $(W, D) \implies X = 4, Y = 1$
• $(D, W) \implies X = 4, Y = 1$
• $(D, L) \implies X = 1, Y = 4$
• $(L, D) \implies X = 1, Y = 4$
• $(L, L) \implies X = 0, Y = 6$

Thus, the total probability $P(X = Y)$ is the sum of the probabilities of these mutually exclusive event outcomes:

$P(X = Y) = P(W, L) + P(D, D) + P(L, W)$

Substitute the calculated values into the equation:

$P(X = Y) = \frac{1}{6} + \frac{1}{36} + \frac{1}{6}$

$P(X = Y) = \frac{6}{36} + \frac{1}{36} + \frac{6}{36} = \frac{13}{36}$

Therefore, the correct option is C.