The roll of the dice Find the probability of rolling a sum of three or less on two dice.
(Yes, I know this is an "easy" question but I second quess myslef and like second opinions.) 
 A: Given two dice: both rolled...Take the result/outcome to be an ordered pair consisting of the number appearing on each die: (number on die 1, number on die 2).
$X$: How many total possible outcomes are there? 


*

*$X = 6\times 6 = 6^2 = 36$


Of these, how many sum to $3$, or to $2$? Let's call the number of outcomes (ordered pairs) whose values sum to $3:\; Y_3,\,$ and the number of outcomes (ordered pairs) whose values sum to $2:\;Y_2$.


*

*For example rolling a $1$ on the first die and 2 on the second die will work: $(1, 2):\, 1 + 2 = 3$. 

*

*What is another ordered pair in which the sum of the entries will be $3$: That would be $\,(2, 1)$: rolling a 2 on the first die, and a 1 on the second die. So $Y_3 = 2$. 


*Then there's only one outcome/ordered pair that for which the values will sum to $2$: $(1, 1)$. So $Y_2 = 1$


Probability of obtaining a sum of $3$ or less when rolling two dice: $$\dfrac{Y_3 + Y_2}{X} = \dfrac{2 + 1}{36} = \dfrac{3}{36} = \dfrac{1}{12}$$
A: Imagine that the dice are red and green, Christmas dice.
Let us record the outcome of the tossing as an ordered pair $(a,b)$, where $a$ is the number on the red die, and $b$ is the number on the green die.
There are $6^2$ possible records, all equally likely.
We get a sum of $3$ or less with the following outcomes: $(1,1)$, $(1,2)$, and $(2,10$. So there are $3$ "favourable" outcomes, and therefore our probability is $\dfrac{3}{36}$. 
A: $$
   \begin{array}{c|cccccc} X+Y & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \\ 
                             1 & \color\green{2} & \color\green{3} & 4 & 5 & 6 & 7 \\
                             2 & \color\green{3} &  4 & 5 & 6 & 7 & 8 \\ 
                             3 & 4 & 5 & 6 & 7 & 8 & 9 \\
                             \ddots & \vdots & & & &  & \vdots
\end{array}
$$
Since each entry in the above table is equally likely:
$$ \Pr\left(X+Y \leq 3 \right) = \frac{3}{6 \times 6} = \frac{1}{12}$$
