$A$ and $B$ throw a die. Find the probability that the sum of their throws is $6$ The answer comes up as $\frac6{36}$, but I get $\frac5{36}$.  The solution lists as possibilites: 
$$\{(1,5), (2,4),(3,3),(3,3),(4,2),(5,1)\}$$
I've always seen this answer as $\frac5{36}$ using a table.
 A: This is something that used to confuse me as well. If you get the same result on each die, that's only one of the $36$ possibilities. Because I'm too lazy to write out all the combinations for two 6-sided dice, I'll just demonstrate this via a 3-sided die. We know that there are $3\times 3=9$ possibilities. Let's list them:
$$\begin{matrix} (1, 1) & (1, 2) & (1, 3) \\ (2, 1) & (2, 2) & (2, 3) \\ (3, 1) & (3, 2) & (3,3) \end{matrix}$$
Notice that $(3, 3)$, $(1, 1)$, and $(2, 2)$ only appear once. It's weird to think about (because we have $(1, 2)$ and $(2, 1)$ for instance), but that's because rolling a $3$ on the first and a $3$ on the second is the same event as, well, a $3$ on the first and a $3$ on the second. On the other hand, getting a $1$ on the first and $2$ on the second is different from getting a $2$ on the first and a $1$ on the second.
The answer is $\frac5{36}$
A: $$
\{(1,5), (2,4),\underbrace{(3,3),(3,3)}_{\begin{smallmatrix} {} \\[2pt] \text{This should} \\[4pt] \text{appear only} \\[4pt] \text{once.}\end{smallmatrix}},(4,2),(5,1)\}
$$
