We are rolling a dice three times; what is the probability of getting a total sum less than $17$? 
We are rolling a dice three times. What is the probability of getting
  a total sum less than $17$?

Let $A=$ $\{$ sum $<17$ $\}$ and $\bar{A}=$ $\{$ sum $\ge 17$ $\}$. If we find the probability $P(\bar{A})$, it's easy to find $P(A)=1-P(\bar{A})$. The number of total outcomes is $6^3=36\cdot6=216$. 
Now, I am thinking about the good outcomes (not sure if this is the right term in English $-$ the outcomes that satisfy our event). The max sum that is equal to $18=6\cdot 3$ can be reached when the three dice show $6$ $(6,6,6)$; this is $1$ good outcome. What about sum equal to $17$? Well, there are $3$ good outcomes: $(6,6,5);(6,5,6);(5,6,6)$. If what I say is true, $P(A)=\dfrac{212}{216}=\dfrac{53}{54}.$ 
I want to ask why when we have $2$ different dice (e.g. blue and green), we accept $(6,2)$ and $(2,6)$ as different outcomes, but $(2,2)$ as only one? 
 A: When you are counting the valid outcomes you are counting the different ones. So you count $(6,2)$ and $(2,6)$ as different outcomes when you can distinguish one dice from the other. Note that $6^2$ are the possible different outcomes that you can get after you throw two distinguishable dice. In $6^2$ you are counting implicitly $(2,6)$ and $(6,2)$ as different outcomes, and of course just one $(2,2)$ (there is other different $(2,2)$?)
However suppose that you throw two indistinguishable dice, then there is a unique outcome $\{2,6\}$ (because you cannot distinguish what dice have what number). But then the total number of possible outcomes that you can count aren't $6^2=36$, because you cannot distinguish one dice from the other. The total number of possible outcomes in this case would be $\binom{6}{2}+6=21$, where $\binom62$ is the number of possible different throws where the two dice shows different numbers, and $6$ is the possible number of throws where the two dice shows the same number.
A: If we have a blue die and a green die, each outcome can be written as an ordered pair $(b, g)$, where $b$ is the number that appears on the blue die and $g$ is the number that appears on the green die.  Since there are six possible outcomes for each die, there are $6 \cdot 6 = 36$ such ordered pairs.  
Notice that $(2, 1) \neq (1, 2)$ since $(a, b) = (c, d) \iff a = c$ and $b = d$. On the other hand, $(2, 2) = (2, 2)$ since $a = c$ and $b = d$.  
Clearly, the events $(2, 1)$ and $(1, 2)$ are distinguishable (to people who can distinguish blue from green) since $(2, 1)$ means we have a $2$ on the blue die and a $1$ on the green die, while $(1, 2)$ means we have a $1$ on the blue die and a $2$ on the green die.  On the other hand, the only way $(2, 2)$ can occur is if we have a $2$ on both dice. Thus, there are two ways to get a $1$ and a $2$ when we roll two distinguishable dice, but only one way to get two $2$'s.
