Chapter 1.4, Problem 10E

Let $D_1,D_2,D_3$ be three four-sided dice whose sides have been labeled as follows: $D_1:0333D_2:2225D_3:1146$

The three dice are rolled at random. Let A, B, and C be the events that the outcome on die $D_1$ is larger than the outcome on $D_2$, the outcome on $D_2$ is larger than the outcome on $D_3$, and the outcome on D3 is larger than the outcome on $D_1$, respectively. Show that (a) $P\left(A\right)=\frac{9}{16}$,

(b) $P\left(B\right)=\frac{9}{16}$, and (c) $P\left(C\right)=\frac{10}{16}$.

Do you find it interesting that each of the probabilities that $D_1$ “beats” $D_2$, $D_2$ ‘beats” $D_3$, and $D_3$ “beats” is greater than $\frac{1}{2}$? Thus, it is difficult to determine the best” die.