Probability that the absolute difference of two dice is equal or less than 2. The scenario: I roll two dice, what's the probability that the absolute difference between the two dice is equal or less than 2?

I know the easiest way is to create a truth table and count the possibilities

Dice 1 | Dice 2
1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;      |  1
2 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;      |  2
...
6 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;      |  6

But, the caveat is that I need to find the answer without having to count the possibilities.

There has to be a quicker/formulaic way of doing this, without having to write out the sample spaces/count the possibilities.
$$P(|X_1-X_2|\leq2)=\frac16\sum_{i=1}^6P(|X_1-X_2|\leq2\mid X_2=i)=$$
$$=\frac16\sum_{i=1}^6P(|X_1-i|\leq2)$$
because of independence.

$$P(|X_1-i|\leq2)=P(|X_1-i|\leq2\ \cap X_i\geq i)+P(|X_1-i|\leq2\ \cap X_1&lt;i)=$$
$$=P(X_1\leq2+i\ \cap X_i\geq i)+P(i-2\leq X_1 \cap X_1&lt;i)=$$
$$=\begin{cases}P(X_1=6)+P(X_1=4)+P(X_1=5)&amp;\text{ if }&amp; i=6\\
P(X_1=6)+P(X_1=5)+P(X_1=3)+P(X_1=4)&amp;\text{ if }&amp; i=5\\
P(X_1=4)+P(X_1=5)+P(X_1=6)+P(X_1=2)+P(X_1=3)&amp;\text{ if }&amp; i=4\\
P(X_1=3)+P(X_1=4)+P(X_1=5)+P(X_1=1)+P(X_1=2)&amp;\text{ if }&amp; i=3\\
P(X_1=2)+P(X_1=3)+P(X_1=4)+P(X_1=1)&amp;\text{ if }&amp; i=2\\
P(X_1=1)+P(X_1=2)+P(X_1=3)&amp;\text{ if }&amp; i=1\\
\end{cases}=$$
$$=\begin{cases}\frac12&amp;\text{ if }&amp; i=6\\
\frac23&amp;\text{ if }&amp; i=5\\
\frac56&amp;\text{ if }&amp; i=4\\
\frac56&amp;\text{ if }&amp; i=3\\
\frac23&amp;\text{ if }&amp; i=2\\
\frac12&amp;\text{ if }&amp; i=1\\
\end{cases}.$$

So

$$P(|X_1-X_2|\leq2)=\frac23.$$
Upon many observations, i got this distribution function: $  \Pr(X=i)
          =  2\tfrac{6-i}{36}$ for $i \in \{1,2,3,4,5\}$, $\Pr(X=0)=\tfrac{1}{6}$.
Now to find $\Pr(X\leq2)$:
$$\Pr(X\leq 2)=\Pr(X=0) + \Pr(X=1) + \Pr(X=2)= \tfrac{1}{6} + 2\tfrac{6-1}{36} + 2\tfrac{6-2}{36}=\tfrac{2}{3}$$