I was given a problem in class:
Three unusual dice, A,B and C are different from other dice, A has the numbers 3,3,4,4,8,8 B has the numbers 1,1,5,5,9,9 and C has the numbers 2,2,6,6,7,7
a) If dice A and B are rolled, find the probability that B beats A (the number on dice B is greater than the number on A)
I don't quite understand how to go about this question but I figured that it was just the number of numbers greater in dice B/total? i.e 5,5 beats 3,3, and 4,4 however 5,5 does not beat 8 if it is rolled. But 9,9 beats all of A - so initially I thought it was 4/6 since 5,5,9,9 beat A but the 8 can still beat the 5's so would the probability only be 2/6?
There are nine equally probable outcomes when rolling dice A and B.
There are five of these where $B > A$.
The probability that B beats A is thus $\dfrac 5 9$.
$$\begin{bmatrix}A\backslash B & \mid & 1 & 5 & 9 \\ \hline 3 & \mid & A\!>\!B & \color{blue}{B\!>\!A} & \color{blue}{B\!>\!A} \\ 4 & \mid & A\!>\!B & \color{blue}{B\!>\!A} & \color{blue}{B\!>\!A} \\8 & \mid & A\!>\!B & A\!>\!B & \color{blue}{B\!>\!A}\end{bmatrix}$$
Or $$\begin{array}{rcl}\mathcal{P}(B\!>\!A) &=& \mathcal{P}(A\!=\!3) \mathcal{P}(B\!>\!3) + \mathcal{P}(A\!=\!4) \mathcal{P}(B\!>\!4) + \mathcal{P}(A\!=\!8) P(B\!>\!8) \\ ~ & = & \frac 1 3 \frac 2 3+ \frac 1 3 \frac 2 3 + \frac 1 3 \frac 1 3 \\ ~ & = & \frac 5 9 \end{array}$$