Zou and Chou are practicing their $100$-meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou:


$LWWWW$
$WLWWW$
$WWLWW$
$WWWLW$
$WWWWL$

We proceed with casework:
Case (1): Sequences #1-4, in which Zou does not lose the last race.
The probability that Zou loses a race is $\frac13,$ and the probability that Zou wins the next race is $\frac13.$ For each of the three other races, the probability that Zou wins is $\frac23.$
There are four sequences in this case. The probability of one such sequence is $\left(\frac13\right)^2\left(\frac23\right)^3.$
Case (2): Sequence #5, in which Zou loses the last race.
The probability that Zou loses a race is $\frac13.$ For each of the four other races, the probability that Zou wins is $\frac23.$
There is one sequence in this case. The probability is $\left(\frac13\right)^1\left(\frac23\right)^4.$
Answer
The requested probability is \[4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},\] from which the answer is $16+81=\boxed{097}.$
~MRENTHUSIASM