A twelve or two sevens

Idea. The game has two states: the last total was a 7, or it was not (which includes the start). Write the probability that A wins from each state and condition on the next roll.

Step 1: the equations. Let $a$ be the probability that A wins from the start (or after any total other than 7), and $b$ the probability after a 7. A roll gives 12 with probability $\frac{1}{36}$, 7 with probability $\frac{6}{36}$, and something else with probability $\frac{29}{36}$. $$\begin{array}{rl}a& =\frac{1}{36}+\frac{6}{36}b+\frac{29}{36}a,\\ b& =\frac{1}{36}+\frac{6}{36}\cdot 0+\frac{29}{36}a.\end{array}$$ In the second line, a 7 right after a 7 means B wins.

Step 2: solve. The first equation gives $7a=1+6b$, the second $36b=1+29a$. Substituting, $$42a=6+1+29a,a=\frac{7}{13},$$ and $b=\frac{6}{13}$.

So A wins with probability $\frac{7}{13}\approx 0.538$. A twelve is six times rarer than a seven, but B needs two sevens in a row, which is rarer still: a pair of consecutive sevens has probability $\frac{1}{36}$, the same as a twelve, and B's sevens often get interrupted.