2011
Problem - 1167
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?
There are totally $6\times 6=36$ possible scenarios among which $6$ will have the same numbers in two rolls. The remaining cases are symmetric. Therefore the answer is $$\frac{(36-6)/2+6}{36}=\frac{7}{12}$$