What's the Probability when throwing $2$ dice, the sum equals $12$ before $7$? When throwing $2$ dice and taking their sum what's the probability the first $12$ occurs before the first $7$?
My approach:
Let $A$ denote the sum equalling $12$ and $B$ equalling $7$
\begin{align}
P(A|A\text{ or }B) &= \frac{P(A \text{ and } (A \text{ or } B))}{P(A \text{ or } B)}\\
&= \frac{P(A)}{P(A\text{ or }B)}\\
&= \frac{1/36}{1/36 + 1/6}\\
&= 0.14
\end{align}
Have I gone wrong somewhere?

Let me demonstrate a different approach leading to the same answer.
Let the sum of two throws be denoted by $X$.
We're doing multiple experiments in a sequence: $X_1, X_2, X_3, \cdots$
Each $X_i$ can take integer values from $2$ to $12$ given by the following distribution
$$\begin{gather*}
  P(x) = \begin{cases}
  \dfrac{1}{36} & \text{if $x \in \{2,12\}$}\\
  \dfrac{2}{36}=\dfrac{1}{18} & \text{if $x \in \{3,11\}$}\\
  \dfrac{3}{36}=\dfrac{1}{12} & \text{if $x \in \{4,10\}$}\\
  \dfrac{4}{36}=\dfrac{1}{9} & \text{if $x \in \{5,9\}$}\\
  \dfrac{5}{36} & \text{if $x \in \{6,8\}$}\\
  \dfrac{6}{36} =\dfrac{1}{6} & \text{if $x = 7$}\\
  \end{cases}
\end{gather*}$$

$\color{white}{.}$
$$  \begin{align} 
&\;\;\;\;\ P (\text{sum equals 12 before 7}) \\
&= P(X_1=12) + P(X_1\neq 12, 7 ; X_2=12) + P(X_1, X_2\neq 12, 7 ; X_3=12)  + \ldots \\
&= (1/36) + (1-(1/36+1/6))(1/36) + (1-(1/36+1/6))^2(1/36)  + \ldots \\
&= \dfrac{1/36}{1-(1-(1/36+1/6))} \\
&= \dfrac{1/36}{1/36+1/6} \\
\end{align}
$$
