Football teams $T_1$ and $T_2$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T_1$ winning, drawing and losing a game against $T_2$ are $\frac{1}{2},\frac{1}{6},\frac{1}{3}$ respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games. Find $$P(X>Y)$$

My Approach

I took simple approach that $X>Y$

$T_1~\text{wins and } T_2~\text{loses}=\frac{1}{2} \times\frac{1}{3} $

$T_1~\text{wins and } T_2~\text{draws}\frac{1}{2} \times\frac{1}{6}$

$T_1~\text{draws and } T_2~\text{loose}\frac{1}{6} \times\frac{1}{3}$

Required probability$$\frac{1}{6} +\frac{1}{12}+\frac{1}{18}=\frac{11}{36}$$

I know it is wrong. But I am unable to move on. Please help!

  • 1
    $\begingroup$ You can start enumerating the scenarios for which $X>Y$ (eg, T1 wins both matches), then calculate the probability of each scenario, and then sum them up. $\endgroup$ – user52227 May 25 '18 at 11:11
  • $\begingroup$ If $T_1$ wins, then $T_2$ must lose. Think in terms of $T_1$. For $T_1$ to win the series, either $T_1$ must win both games, win the first and draw the second, or draw the first and win the second. $\endgroup$ – N. F. Taussig May 25 '18 at 12:42

Lets think about the full distribution for $X$ and $Y$. What you have done is combined the distributions of $T_1$ and $T_2$. We want the combination of $X$ and $Y$. When $X=6$, $Y=0$. When $X=4$, $Y=1$. When $X=1$, $Y=4$. When $X=3$, $Y=3$. When $X=0$, $Y=6$. If $W$ is win, $D$ is draw and $L$ is loss, we can: $WW, LL, WL, LW, WD, DW, DD$

We want the combinations where $X \gt Y$ strictly. That is $X=6$, $Y=0$ AND $X=4$, $Y=1$. This is represented by $WW$ or $WD$ or $DW$. Sum this:

$2$ Wins: $\frac 12 \times \frac 12$

$1$ Win $1$ Draw: $\frac 12 \times \frac 16$

So $$P(X>Y)=P(X=6,Y=0)+P(X=4,Y=1)$$

$$=P(WW)+2 \times P(WD)$$

$$=\frac 12 \times \frac 12 \space + \space 2 \times\frac 12 \times \frac 16= \frac 14+ \frac 1{6}= \frac 5{12}$$

Edit ### - Looks like my first guess was right.

So 5/12 I hope is your answer.

  • $\begingroup$ Be careful. For the $X = 4, Y = 1$ case, team 1 can win the first game and draw the second or vice versa. $\endgroup$ – N. F. Taussig May 25 '18 at 12:34
  • $\begingroup$ I think I wrote out all the possibilities right? So that would be Y=4 and X=1 which is not in our set X>Y. $\endgroup$ – Tony Hellmuth May 25 '18 at 12:36
  • 1
    $\begingroup$ Evidently, I did not make myself clear. For the $X = 4$, $Y = 1$ case, team 1 could win the first game and draw the second game or team 1 could draw the first game and win the second game. You only accounted for one of those two possibilities. $\endgroup$ – N. F. Taussig May 25 '18 at 12:40
  • $\begingroup$ I think you may be right. That means my first edit was good hahaha! Thank you. $\endgroup$ – Tony Hellmuth May 25 '18 at 12:43

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