$\frac{5}{12}$
Assume a die has numbers 1 through 6 and a probability of $1/6$ of landing on each number. Suppose we roll a red die and a green die (that are independent). What is the probability that the number on the red die is larger than the number on the green die?
$\frac{5}{12}$
When we roll two dice, a single outcome can be expressed as an ordered-pair,e.g., ($\textcolor{red}{5},\textcolor{green}{4}$). We can enumerate allpossible outcomes—the sample space $\Omega$—is the following table:
| ($\textcolor{red}{1},\textcolor{green}{1}$) | ($\textcolor{red}{1},\textcolor{green}{2}$) | ($\textcolor{red}{1},\textcolor{green}{3}$) | ($\textcolor{red}{1},\textcolor{green}{4}$) | ($\textcolor{red}{1},\textcolor{green}{5}$) | ($\textcolor{red}{1},\textcolor{green}{6}$) |
| ($\textcolor{red}{2},\textcolor{green}{1}$) | ($\textcolor{red}{2},\textcolor{green}{2}$) | ($\textcolor{red}{2},\textcolor{green}{3}$) | ($\textcolor{red}{2},\textcolor{green}{4}$) | ($\textcolor{red}{2},\textcolor{green}{5}$) | ($\textcolor{red}{2},\textcolor{green}{6}$) |
| ($\textcolor{red}{3},\textcolor{green}{1}$) | ($\textcolor{red}{3},\textcolor{green}{2}$) | ($\textcolor{red}{3},\textcolor{green}{3}$) | ($\textcolor{red}{3},\textcolor{green}{4}$) | ($\textcolor{red}{3},\textcolor{green}{5}$) | ($\textcolor{red}{3},\textcolor{green}{6}$) |
| ($\textcolor{red}{4},\textcolor{green}{1}$) | ($\textcolor{red}{4},\textcolor{green}{2}$) | ($\textcolor{red}{4},\textcolor{green}{3}$) | ($\textcolor{red}{4},\textcolor{green}{4}$) | ($\textcolor{red}{4},\textcolor{green}{5}$) | ($\textcolor{red}{4},\textcolor{green}{6}$) |
| ($\textcolor{red}{5},\textcolor{green}{1}$) | ($\textcolor{red}{5},\textcolor{green}{2}$) | ($\textcolor{red}{5},\textcolor{green}{3}$) | ($\textcolor{red}{5},\textcolor{green}{4}$) | ($\textcolor{red}{5},\textcolor{green}{5}$) | ($\textcolor{red}{5},\textcolor{green}{6}$) |
| ($\textcolor{red}{6},\textcolor{green}{1}$) | ($\textcolor{red}{6},\textcolor{green}{2}$) | ($\textcolor{red}{6},\textcolor{green}{3}$) | ($\textcolor{red}{6},\textcolor{green}{4}$) | ($\textcolor{red}{6},\textcolor{green}{5}$) | ($\textcolor{red}{6},\textcolor{green}{6}$) |
So, $|\Omega|=6\times6=36$.
Since the two dice are independent, the probability of each ordered pair is $\frac{1}{6}\frac{1}{6}=\frac{1}{36}$. the probabilityof any event $A$ can then be calculated as follows:
$$P(A)=\frac{|A|}{36}.$$
Let us denote by $L$ the event containing all outcomes such that the number on the red die is larger than the number on the green die. We immediately note that such outcomes are the entries in the table below the diagonal. And,$|L|=15$. Therefore,
$$P(L)=\frac{15}{36}=\frac{5}{12}.$$