The answer you gave in your edit is not the answer to your original question. It's the (almost but not quite correct) answer to the question "What is the expected value of the number of sequences 4213433 occurring if I throw a four-sided die $4.5\times 10^6$ times?" This is in fact a much easier question because of the linearity of expectation. You don't have to worry about correlations, you can just add up the expectation values for each of the slots at which the sequence might occur. However, there aren't $4.5\times10^6$ of these, only $4.5\times10^6-6$, so the expected number of occurrences is actually $(4.5\times10^6-6)/4^7$, but that's the same up to the decimal places you specified.
Your original question is a bit harder, since to find the probability of at least one such sequence occurring, you need to take into account that the events of the sequence occurring in overlapping slots aren't independent. For instance, the probability of getting the sequence 44 at least once if you roll the die $3$ times is $7/4^3$, whereas the probability of getting the sequence $43$ at least once is $8/4^3$, even though the expected number of occurrences in both cases is $2/4^2$.
You have an unbiased six-sided die and an unbiased three-sided die, so are drawing pairs uniformly distributed on $\{1,2,3,4,5,6\}\times\{0,1,2\}$.
$$\mathsf P(X+Y=1) ~=~ \mathsf P((X,Y)\in\{(1,0)\qquad\;\;\,\qquad\;\;\,\})~=~ 1/18
\\\mathsf P(X+Y=2) ~=~ \mathsf P((X,Y)\in\{(2,0),(1,1)\qquad\;\;\,\})~=~ 2/18
\\\mathsf P(X+Y=3) ~=~ \mathsf P((X,Y)\in\{(3,0),(2,1),(1,2)\})~=~ 3/18
\\\mathsf P(X+Y=4) ~=~ \mathsf P((X,Y)\in\{(4,0),(3,1),(2,2)\})~=~ 3/18
\\\mathsf P(X+Y=5) ~=~ \mathsf P((X,Y)\in\{(5,0),(4,1),(3,2)\})~=~ 3/18
\\\mathsf P(X+Y=6) ~=~ \mathsf P((X,Y)\in\{(6,0),(5,1),(4,2)\})~=~ 3/18
\\\mathsf P(X+Y=7) ~=~ \mathsf P((X,Y)\in\{\qquad\;\;\,(6,1),(5,2)\})~=~ 2/18
\\\mathsf P(X+Y=8) ~=~ \mathsf P((X,Y)\in\{\qquad\;\;\,\qquad\;\;\,(6,2)\})~=~ 1/18
\\$$
Best Answer
This sentence is false.
Imagine that instead of a 6-sided die and a 4-sided die, you had a 1000000-sided die, numbered 1-1000000, and a 4-sided die, numbered 1-4. You select one of the two dice uniformly at random, and roll it, and the result is a 2. Do you really claim that you have no information about which die was selected? Intuitively, it is very probable that the 4-sided die was selected; if the 1000000-sided die had been selected, the result could have been anything between 1 and 1000000, and it would look like quite an unprobable coincidence that the result was a 2.
A great tool to formalise this intuition and turn it into precise numbers, and make it work with a 6-sided die instead of a 1000000-sided die, is Bayes' theorem.
Let $D_6$ be the event "we picked the 6-sided die first" and $D_4$ the event "we picked the 4-sided die first". Let $R_2$ be the event "we rolled a 2 on the first roll". Then, in the absence of information, the probability of picking the 6-sided die first is $P(D_6) = \frac{1}{2}$; but the probability of having picked the 6-sided die, knowing that we rolled a 2, is noted $P(D_6 \,|\, R_2)$ and can be calculated using Bayes' theorem:
\begin{align*} P(D_6 \,|\, R_2) & = \frac{P(R_2 \,|\, D_6)P(D_6)}{P(R_2)} \\ P(D_6 \,|\, R_2) & = \frac{P(R_2 \,|\, D_6)P(D_6)}{P(R_2 \,|\, D_6)P(D_6) + P(R_2 \,|\, D_4)P(D_4)} \\ \end{align*}
Filling in these values:
We get: \begin{align*} P(D_6 \,|\, R_2) & = \frac{\frac{1}{6}\times\frac{1}{2}}{\frac{1}{6}\times\frac{1}{2} + \frac{1}{4}\times\frac{1}{2}} \\ P(D_6 \,|\, R_2) & = \frac{2}{5} \\ \end{align*} Thus, after observing that we rolled a 2, the probability that we picked the 6-sided die is $\frac{2}{5}$, and the probability that we picked the 4-sided die is $\frac{3}{5}$.
(If we had used a 1000000-sided die and a 4-sided die, using the same reasoning we would have found that after observing a 2, the probability of having picked the 1000000-sided die is only $\frac{1}{250001}$, or $0.0004\%$.)
Now you can calculate the expectation of the second roll after having observed a 2 on the first roll. Calling $X$ the random variable equal to the result of the second roll:
\begin{align*} \mathbb{E}(X \,|\, R_2) & = \mathbb{E}(X \,|\, D_6)P(D_6 \,|\, R_2) + \mathbb{E}(X \,|\, D_4)P(D_4 \,|\, R_2) \\ \mathbb{E}(X \,|\, R_2) & = 2.5 \times \frac{2}{5} + 3.5 \times \frac{3}{5} \\ \mathbb{E}(X \,|\, R_2) & = 3.1 \\ \end{align*}
Thus the expectation of the second roll, having observed a 2 on the first roll, is 3.1. This is slightly higher than the expectation of the second roll we would have calculated if we hadn't observed the first roll: $$\mathbb{E}(X) = \mathbb{E}(X \,|\, D_6)P(D_6) + \mathbb{E}(X \,|\, D_4)P(D_4) = 2.5 \times \frac{1}{2} + 3.5 \times \frac{1}{2} = 3$$