Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Let $P_n$ be the probability of getting consecutive $1,2,3$ rolls in $n$ rolls and not rolling $1,2,3$ prior to the nth roll. 
Let $x = P_3+P_5+...=1-(P_4+P_6+..)$. Following Solution 2, one can see that
\[P_{n+1}=P_{n}-\frac{P_{n-2}}{6^3}\]
for all positive integers $n \ge 5$. Summing for $n=5,7,...$ gives
\[(1-x)-\frac{1}{6^3}=x-\frac{1}{6^3}-\frac{x}{6^3}\]
\[\implies x = \frac{m}{n} = \frac{216}{431} \implies m+n=216+431= \boxed{647}\]