Assume we want to model the game where we roll a die until the sequence $65656$ shows up using Markov chains.
Assume the state space $\mathbb{X} = \{0, 1,2, 3\}$, where:
- $0$ : No $65$'s show up
- $1$ : One $65$ shows up
- $2$ : Two $65$'s show up
- $3$ : One $6$ shows up
If we define $p: \mathbb{X} \times \mathbb{X} \rightarrow [0,1]$ as $$ p(x,y) = \mathbb{P}[X_{n+1} = y | X_n = x] $$
then for $a \in \{0,1\}$:
$$ p(a, b) = \begin{cases} \frac{35}{36}, \quad a = b \\ \frac{1}{36}, \quad b = a + 1 \\ 0, \quad \text{elsewise} \end{cases} $$ as the probability for a $65$ to show up after two successive rolls is $\frac{1}{36}$.
For $a=2$:
$$ p(a, b) = \begin{cases} \frac{5}{6}, \quad a = b \\ \frac{1}{6}, \quad b = a + 1 \\ 0, \quad \text{elsewise} \end{cases} $$ as the probability for a $6$ to show up after one roll is $\frac{1}{6}$.
Considering the above, the transition matrix is: $$ P = \left[\begin{array}{cccc} \frac{35}{36} & \frac{1}{36} & 0 & 0 \\ 0 & \frac{35}{36} & \frac{1}{36} & 0 \\ 0 & 0 & \frac{5}{6} & \frac{1}{6} \\ 0 & 0 & 0 & 1 \end{array}\right] $$
Does this chain describe the game correctly?
You can’t model the game like this. If the first two rolls don’t yield $65$, they could still yield $X6$, and then you wouldn’t be back in the initial state, but in a different state in which a single $5$ would get you to the state where $65$ has occurred. You need a state for each prefix of the desired sequence, and thus six states; the corresponding transition matrix is
$$ P=\left[\matrix{ \frac56&\frac16&0&0&0&0\\ \frac46&\frac16&\frac16&0&0&0\\ \frac56&0&0&\frac16&0&0\\ \frac46&\frac16&0&0&\frac16&0\\ \frac56&0&0&0&0&\frac16\\ 0&0&0&0&0&1 }\right]\;. $$
Since the later states form a simple chain, you can combine some of them if you like, but you can’t combine the first two like you did, since they can be reached separately from other states.