The non-uniform probability of sums from the throw of multiple dice I'm reading The Drunkards Walk by Leonard Mlodinow. In the book, the author writes:

From a throw of three dice, a sum of 9 and 10 can be constructed in an equal combinations. However, the outcome (6, 3, 1) is more probable than the outcome (3, 3, 3) because there is just one way you can throw 3 threes, yet there are 6 ways you can throw a 6, 3 and a 1.

This is how I understand probability theory. However, the statement led me to ask the question: why is there is just one way to throw 3 threes? Despite the fact that the outcome is the same, there are a number of different ways to throw three 3s? That is, each of the three dice are distinct from the others, so why do we exclude the other combinations of rolling 3 threes?
 A: Six ways to get $1,3,6$:
$$
\begin{array}{c|c|c|c|c|c}
\overbrace{
\begin{array}{rc}
\text{first die:} & 1 \\
\text{second die:} & 3 \\
\text{third die:} & 6
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 1 \\
\text{second die:} & 6 \\
\text{third die:} & 3
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 3 \\
\text{second die:} & 1 \\
\text{third die:} & 6
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 3 \\
\text{second die:} & 6 \\
\text{third die:} & 1
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 6 \\
\text{second die:} & 1 \\
\text{third die:} & 3
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 6 \\
\text{second die:} & 3 \\
\text{third die:} & 1
\end{array}}
\end{array}
$$
One way to get $3,3,3$:
$$
\overbrace{\begin{array}{rc}
\text{first die:} & 3 \\
\text{second die:} & 3 \\
\text{third die:} & 3
\end{array}}
$$
What would be the other ways to get $3,3,3$ besides that one?
A: Yes, you can colour the dice, but that actually makes it more, not less clear that $6+3+1$ counts six times and $3+3+3$ counts once: Now you have the six partitions $6R+3Y+1B$, $3R+6Y+1B$, etc., whereas you still have only the one partition $3R+3Y+3B$. There's nothing to rearrange; since the numbers are all $3$, nothing changes if you swap the colour labels among them.
A: The similar and more simple situation is to throw two dice. We have 36 possible outcomes:  
         1,1  1,2  1,3  1,4  1,5  1,6
         2,1  2,2  2,3  2,4  2,5  2,6  
         3,1  3,2  3,3  3,4  3,5  3,6
         4,1  4,2  4,3  4,4  4,5  4,6  
         5,1  5,2  5,3  5,4  5,5  5,6
         6,1  6,2  6,3  6,4  6,5  6,6

Here for example there are two outcomes $(2,1)$ and $(1,2)$ and only one outcome $(1,1)$.
There is no problem with coloured dice. For example suppose that you have red and yellow dice. You can consider the above outcomes if the order of throws is not important for you. 
Now, if you consider outcomes $(R1,Y1)$ and $(Y1,R1)$ as distinct (which means that the order of throws is important for you), then you have to consider outcomes $(R1,Y2)$, $(Y1,R2)$, $(R2,Y1)$ and $(Y2,R1)$ as distinct too. However this consideration leads to the same probabilities of events, because the set of all possible outcomes have 72 elements.       
