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Tim and Rich each roll a die. Whoever gets the higher number wins; if they both roll the same number, neither wins. Could you please check my answers to following questions and give hints about $d$ and $e$?

a. What is the probability that Tim wins?

b. If Rich rolls a $3$, what is the probability that he wins?

c. If Rich rolls a $3$, what is the probability that Tim wins?

d. If Rich wins, what is the probability that Tim rolled a $2$?

e. If Rich wins, what is the probability that Tim rolled a $3$?

Answers

a. I think it is $15/36$.

b. It is $2/6$. Because Tim could only choose $1$ or $2$.

c. It is $3/6$. Because Tim could choose $4$, $5$ or $6$.

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Notice that the problem is symmetric: both players win with equal probability. The chance of a tie is simply $\frac16$, so for (a) the chance that Tim wins is $\frac12 \cdot Pr(\neg tie) = \frac12 \cdot (1 - \frac16) = \frac5{12}$.

For (b) Rich wins if Tim rolls below a 3, so $\frac13$ is correct. Likewise, (c) is $\frac12$.

For (d) and (e) we use Bayes' theorem: $P(A|B) = \frac{P(A\cap B)}{P(B)}$:

$$Pr(Tim=2 | Rich wins) = \frac{Pr(Tim=2 \cap Rich wins)}{Pr(Rich wins)} = \frac{\frac16\cdot\frac23}{\frac5{12}} = \frac4{15}$$

(e) is basically the same problem.

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Your answers to (a)-(c) are correct as far as I can see!

The way I'd approach this problem is by considering the outcome space $(x,y)$ where $x$ is the score on Tim's dice and $y$ is the score on Rich's dice. Since we know that each individual outcome $(x,y)$ has probability $\frac{1}{36}$, we get something like this...

Those highlighted in red are those where $y>x$ and so Rich wins / Tim loses. Those highlighted in green are those in which $x<y$ and so Tim wins.

Then I'd just count up the relevant outcomes (so for example for e. I'd look at those in which Rich won (of which there are $(6\times 5) / 2=15$ as you pointed out), and count the relevant ones. Part d is similar.

$\begin{array}{cccccc} (1,1)&\color{red}{(1,2)}&\color{red}{(1,3)}&\color{red}{(1,4)}&\color{red}{(1,5)}&\color{red}{(1,6)}\\ \color{green}{(2,1)}&(2,2)&\color{red}{(2,3)}&\color{red}{(2,4)}&\color{red}{(2,5)}&\color{red}{(2,6)}\\ \color{green}{(3,1)}&\color{green}{(3,2)}&(3,3)&\color{red}{(3,4)}&\color{red}{(3,5)}&\color{red}{(3,6)}\\ \color{green}{(4,1)}&\color{green}{(4,2)}&\color{green}{(4,3)}&(4,4)&\color{red}{(4,5)}&\color{red}{(4,6)}\\ \color{green}{(5,1)}&\color{green}{(5,2)}&\color{green}{(5,3)}&\color{green}{(5,4)}&(5,5)&\color{red}{(5,6)}\\ \color{green}{(6,1)}&\color{green}{(6,2)}&\color{green}{(6,3)}&\color{green}{(6,4)}&\color{green}{(6,5)}&(6,6)\\ \end{array}$

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