Your version has two errors.
$(1)\;$You didn't choose which die is the solo one. To fix it, you need a factor
of $3$.
$(2)\;$Of your two factors of $5$, the second one is incorrect. It should be $1$, not $5$, since there is no choice of color for the third die.
With those corrections, you get the product
$$(3)(6)(5)(1) = 90$$
which matches the official answer.
For $0\le k\le 2$, let $p_k$ be the probability that the first occurrence of the roll sequence $1,2,3$ occurs after an odd number of (future) rolls, given that the prior $k$ rolls (from oldest to most recent) match the first $k$ terms of $1,2,3$.
Our goal is to find $p_0$.
Then we have the following system
$$
\begin{cases}
p_0=\left({\large{\frac{1}{6}}}\right)(1-p_1)+\left({\large{\frac{5}{6}}}\right)(1-p_0)\\[4pt]
p_1=
\left({\large{\frac{1}{6}}}\right)(1-p_2)
+\left({\large{\frac{1}{6}}}\right)(1-p_1)
+\left({\large{\frac{2}{3}}}\right)(1-p_0)
\\[4pt]
p_2={\large{\frac{1}{6}}}
+\left({\large{\frac{1}{6}}}\right)(1-p_1)
+\left({\large{\frac{2}{3}}}\right)(1-p_0)
\end{cases}
$$
of $3$ linear equations, in $3$ unknowns.
Solving the system, we get $p_0={\large{\frac{216}{431}}}$.
Explanation:
Initially, there are no prior live terms to keep track of.
At any point before the first occurrence of the roll sequence $1,2,3$, the number of prior live terms is either $0$, $1$, or $2$.
- If the prior roll is $1$, then $k=1$ (we have one prior live term).$\\[4pt]$
- If the last two rolls (oldest to most recent) are $1,2$, then $k=2$ (we have two prior live terms).$\\[4pt]$
- Otherwise, we have $k=0$ (the number of prior live terms is zero).
The number of prior live terms (i.e., the value of $k$) can be regarded as the state of the game.
Note that $p_k$ is the probability of success, given state $k$, in an odd number of future rolls. Hence, after the next roll, we want the probability of failure, not success. That explains the expressions of the form $(1-p_j)$ on the right-hand side of each equation.
Now let's consider the equations, one at a time . . .
When $k=0$, there are two possibilies:
- If the next roll is $1$ (probability ${\large{\frac{1}{6}}}$), the state changes to $k=1$.$\\[4pt]$
- Otherwise (probability ${\large{\frac{5}{6}}}$), the state remains at $k=0$.
Therefore we have $p_0=\left({\large{\frac{1}{6}}}\right)(1-p_1)+\left({\large{\frac{5}{6}}}\right)(1-p_0)$.
When $k=1$, there are three possibilies:
- If the next roll is $2$ (probability ${\large{\frac{1}{6}}}$), the state changes to $k=2$.$\\[4pt]$
- If the next roll is $1$ (probability ${\large{\frac{1}{6}}}$), the state remains at $k=1$.$\\[4pt]$
- Otherwise (probability ${\large{\frac{2}{3}}}$), the state changes to $k=0$.
Therefore we have
$
p_1=
\left({\large{\frac{1}{6}}}\right)(1-p_2)
+\left({\large{\frac{1}{6}}}\right)(1-p_1)
+\left({\large{\frac{2}{3}}}\right)(1-p_0)
$.
When $k=2$, there are three possibilies:
- If the next roll is $3$ (probability ${\large{\frac{1}{6}}}$), the game is over, and we have success.
- If the next roll is $1$ (probability ${\large{\frac{1}{6}}}$), the state changes to $k=1$.$\\[4pt]$
- Otherwise (probability ${\large{\frac{2}{3}}}$), the state changes to $k=0$.
Therefore we have
$
p_2={\large{\frac{1}{6}}}
+\left({\large{\frac{1}{6}}}\right)(1-p_1)
+\left({\large{\frac{2}{3}}}\right)(1-p_0)
$.
Best Answer
Since you are wanting a shortcut method,
count the number of ways $4$ balls can be placed in $6$ bins marked $1-6$, using stars and bars
Note that each of the $\binom{4+6-1}{6-1}$ results thus obtained can yield only one non-decreasing sequence.
A result of $\;\;\fbox{2}\fbox{0}\fbox{0}\fbox{1}\fbox{0}\fbox{1}\;$, e.g. means obtaining $1-1-4-6$ in sequence.
Thus $Pr = \dfrac{\binom95}{6^4}$