<s>[INST] Dice outcomes probability 
Two dice are rolled. What is the probability that the sum of the
  numbers on the dice is at least 10

Let $Z$ denote the set of successful outcomes:
$Z=\{(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\}\\
\text{Sample space: }
S=6^2$
So the answer gives the probability as, $P=\frac{6}{36}$.
My question is why are the pairs $(5,5),(6,6)$ only included once? The way I see it for two dice $D_1,D_2$ is it may appear as $D_1: 5,D_2: 5$ or $D_2: 5,D_1: 5$, so why don't we count for this as two outcomes?
 [/INST] Your sample space $S=\{1,2,3,4,5,6\}^2$ has exactly 36 unique outcomes, with each having probability $\frac{1}{36}$. $(4,6)$ and $(6,4)$ are separate outcomes, whereas $(5,5)$ and $(5,5)$ are the same.

Intuitively, to get an outcome involving one $4$ and one $6$, I could roll either a $4$ or a $6$ for the first die, which has a probability of $\frac{2}{6}$ of success. If I was successful with the first die, I have a $\frac{1}{6}$ chance of rolling the number I'm missing with the second die. The total probability of success here is $\frac{2}{6}\cdot\frac{1}{6}=\frac{2}{36}$.
On the other hand, if I want an outcome with two $5$'s, I need to roll a $5$ on the first outcome, with probability $\frac{1}{6}$ of success. Given I succeed, I need to roll a $5$ again with the second die, with probability $\frac{1}{6}$. The total probability of success here is $\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$.
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