First, Alice chooses minimal $n_a$ divisible by 3 such that her bits at positions $n_a, n_a + 1, n_a + 2$ are not all the same, and Bob similarly chooses $n_b$.
Looking at triplet $A_{n_a}, A_{n_a + 1}, A_{n_a + 2}$. Alice chooses $m_a$ according to following rule: {001: 0, 010: 2, 011: 0, 100: 1, 101: 1, 110: 2}. Bob chooses $m_b$ in the same way.

Now Alice says $n_a + m_a$, and Bob says $n_b + m_b$. It's easy to check that probability of $n_a = n_b$ is $\frac{3}{5}$, probability of winning in this case is $\frac{5}{12}$. If $n_a \neq n_b$, then probability of winning is default $\frac{1}{4}$. So winning probability for this strategy is $\frac{3}{5} \cdot \frac{5}{12} + \frac{2}{5} \cdot \frac{1}{4} = 0.35$.