We can solve this problem by dividing it into cases. If Markov rolls a 1 or 2 on the first turn, he will flip a coin on the second turn. He must flip a heads to flip a coin on his third turn. There is a $ \frac{2}{6}\cdot \frac{1}{2}=\frac{1}{6} $ chance of this case happening. If Markov does not roll a 1 or 2 on the first turn, he will roll the die on the second turn. He must roll a 1 or 2 on the second turn to flip a coin on the third turn. There is a $ \frac{4}{6}\cdot \frac{2}{6}=\frac{2}{9} $ chance of this case happening. The total probability that Markov will flip a coin on the third turn is then $ \frac{1}{6}+\frac{2}{9}=\boxed{\frac{7}{18}} $.