The problem can be solved by recursion. Let $p_n$ be the probability that consecutive rolls of $1,4$ occur within the first $n$ rolls. Using inclusion/exclusion, this is
- the probability of obtaining $1,4$ on rolls $n-1$ and $n$;
- plus the probability of obtaining $1,4$ within rolls $1,\ldots,n-1$;
- minus the probability that both of the above occur.
If both occur then roll $n-1$ must be a $1$, so if $1,4$ occurs within the first $n-1$ rolls it must actually have occurred within the first $n-2$. Thus
$$p_n=\frac{1}{36}+p_{n-1}-\frac{1}{36}p_{n-2}\ .$$
Since we also have the conditions $p_0=p_1=0$, we can solve by standard methods to obtain
$$p_n=1-\frac{3}{\sqrt8}\left(\frac{3+\sqrt8}{6}\right)^{n+1}+\frac{3}{\sqrt8}\left(\frac{3-\sqrt8}{6}\right)^{n+1}\ ,$$
on the unlikely assumption that I have got the calculations correct.
Now A wins if for some $n=1,3,5,\ldots$, rolls $1$ to $n$ do not include $1,4$ and rolls $n+1,n+2$ are $1,4$. The probability of this happening is
$$\sum_{n=1\atop n\;\rm odd}^\infty \frac{1}{36}(1-p_n)$$
which is the sum of two geometric series and after a bit of work evaluates to $\frac{36}{73}$.