Fair four sided die is rolled twice, what is the possibility of the sum to be 4 or less? A fair four sided die is rolled twice, what is the possibility of the sum of the 2 die rolls to be 4 or less?
could you please explain in detail
 A: Record the result of the tossing as an ordered pair $(a,b)$, where $a$ and $b$ are integers between $1$ and $4$. Here the first coordinate records the result of the first toss, and the second coordinate records the result of the second toss.
(a) Let the number of such ordered pairs be $N$.  What is the value of $N$? 
There is a fast way of finding $N$. But one can do it slowly, by listing and counting. Here is a start: $(1,1)$, $(1,2)$, $(1,3)$, $(1,4)$.
(b) All these ordered pairs are equally likely.  How many ordered pairs give you a sum of $4$ or less? List and count them carefully. Let the number be $S$.
(c) Your probability is $\dfrac{S}{N}$.
A: The accepted answer didn't provide an actual numeric answer so I wanted to attempt to answer it myself to ensure I got it right.
$$\mathbb{P}(Sum \le 4) = \frac{\text{Number of Winning Pairs}}{\text{Number of Pairs}}$$
I'm pretty sure $\textit{Number of Pairs} = {4 \choose 1} ^ 2 = 16$
What I'm not sure about is if there's a smarter way to determine the number of winning pairs than just listing them: (1, 1), (1, 2), (1, 3), (2, 2) and their inverses so 8 in all.
So, $\mathbb{P}(Sum \le 4) = 8 / 16 = .5$
Is that what everybody else came up with?
