Solution:

Let $a$ be the probability that the total number of rolls is odd; let $b$ be that probability given that the first roll is a $1,$ and $c$ given that the first two rolls are $1$-$2$ (in each case counting all rolls). Condition on the next roll, noting that whenever the count restarts, the rolls already used flip the required parity. Starting fresh: a $1$ leads to state $b;$ anything else uses one roll, after which an even continuation is needed. After a $1:$ another $1$ means the first roll is wasted, needing an even continuation of the $b$-type; a $2$ leads to $c;$ anything else wastes both rolls. After $1$-$2:$ a $3$ finishes in $3$ rolls (odd); a $1$ restarts at the $b$-state with two wasted rolls; anything else wastes all three. Thus $$a = \frac{1}{6}b + \frac{5}{6}(1 - a), \qquad b = \frac{1}{6}(1 - b) + \frac{1}{6}c + \frac{4}{6}a, \qquad c = \frac{1}{6}b + \frac{1}{6} + \frac{4}{6}(1 - a).$$

The first equation gives $b = 11a - 5;$ substituting the third into the second yields $41b = 11 + 20a,$ so $41(11a - 5) = 11 + 20a,$ giving $431a = 216$ and $a = \frac{216}{431}.$ Since $431$ is prime, $m + n = 216 + 431 = 647.$