jmc

The probability of rolling 1 or 6 is $\frac{2}{6}$, and the probability of flipping heads is $\frac{1}{2}$. Therefore, the probability that the game will end on the first turn is $\frac{2}{6}\cdot \frac{1}{2}=\frac{1}{6}$. The probability that the game will not end on the first turn is $1-\frac{1}{6}=\frac{5}{6}$. Given that the game is still going after 1 turn, the probability that the game will not end on the second turn is also $\frac{5}{6}$. Therefore, the probability that the game does not end by the end of the second turn is ${\left(\frac{5}{6}\right)}^2$. Similarly, the probability that the game is still going after 3 turns is ${\left(\frac{5}{6}\right)}^3=\frac{125}{216}$. So the probability that the game is over by the end of the third turn is $1-\frac{125}{216}=\boxed{\frac{91}{216}}$.