A fair $6$-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number? ($2011$ AMC $8$ Problem, Question \#$18$) 
Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
There are $6\times6=36$ ways to roll a die twice, and $6$ of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first time is greater than the second should be the same as the number of rolls where the second time is greater than the first. In other words, there are $\dfrac{30}{2}=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is $\dfrac{15+6}{36}=\dfrac{21}{36}=\frac{7}{12}$. 