Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, drawing and losing a game against T2 are 1/2, 1/6 and 1/3, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 points for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2 respectively, after two games. P(X=Y) is

Let W, D, L denote the events that T1 wins, draws, and loses a game against T2, respectively. We are given that P(W) = 1/2, P(D) = 1/6, P(L) = 1/3.
The possible outcomes for two games are:

1. T1 wins both games (WW): X = 6, Y = 0
2. T1 wins, then draws (WD): X = 4, Y = 1
3. T1 wins, then loses (WL): X = 3, Y = 3
4. T1 draws, then wins (DW): X = 4, Y = 1
5. T1 draws, then draws (DD): X = 2, Y = 2
6. T1 draws, then loses (DL): X = 1, Y = 3
7. T1 loses, then wins (LW): X = 3, Y = 3
8. T1 loses, then draws (LD): X = 1, Y = 3
9. T1 loses both games (LL): X = 0, Y = 6

We are interested in the cases where X = Y. These are:

• Case 3: T1 wins, then loses (WL): X = 3, Y = 3. Probability = P(W)P(L) = (1/2)(1/3) = 1/6
• Case 5: T1 draws, then draws (DD): X = 2, Y = 2. Probability = P(D)P(D) = (1/6)(1/6) = 1/36
• Case 7: T1 loses, then wins (LW): X = 3, Y = 3. Probability = P(L)P(W) = (1/3)(1/2) = 1/6

Therefore, P(X=Y) = P(WL) + P(DD) + P(LW) = 1/6 + 1/36 + 1/6 = 6/36 + 1/36 + 6/36 = 13/36