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You are right to be skeptical as there are some conditions missing. For example, $\mathbb{P}(T_{s_1}<T_{s_2}\mid X_1=s_1)=1$, but there is no guarantee that $b_{s_1}=1$. The equation $b_i=\sum_{j\in S}b_j\, p_{ij}$ will be true if the initial state $i$ is so "far away" from $s_1$ that $X_1=s_1$ is impossible.
Once you have got 3 consecutive heads, your goal is attained, and the task is over. 1.The probabilities in question obtained after N trials are the cumulative probabilities. 2. No, you can't say that. $P(N \le 8)$ is the probability that at most $8$ flips are needed. you can understand from the probability values for trials 3 through 8 $3\quad 0.125 000 ... 1 Say your generator matrix is: $$G = \begin{bmatrix} -\lambda_0 & \lambda_{01} & \lambda_{02} & \cdots \\ \lambda_{10} & -\lambda_{1} & \lambda_{12} & \cdots \\ \lambda_{20} & \lambda_{21} & -\lambda_{2} & \cdots \\ \cdots \\ \end{bmatrix}$$ where, as usual, for each ... 1 maybe this helps. page 153 ff. https://www.math.ucdavis.edu/~gravner/MAT135B/materials/ch13.pdf You will also find the definitions for the transition rates there by summing over probabilities P[X_n=j|X_0=i] and the expected mean time of recurrence. So if some probabilities P[X_n=j|x0=i] of returning back to i in n steps are <1, by definition you ... 1 The generator matrix is: $$G= \begin{bmatrix} -2 & 2 & 0 & 0 \\ 3 & -5 & 2 & 0 \\ 0 & 3 & -5 & 2 \\ 0 & 0 & 3 & -3 \\ \end{bmatrix}$$ The balance equations in vector form are:$\mathbf{\pi G = 0}, so, working down each of the four columns in turn: \begin{align} ... 1 To see a counterexample to your argument, let for example the transition probabilities:6 \to 5= 0.01$and$6 \to \{3,4\}=0.99$,$5 \to 6= 0.01$and$5 \to \{1,2\}=0.99\$. Now, a simple calculation shows that $$P_6[T_{1,2}<+\infty]\le 0.01 < 0.99 \le P_5(T_{1,2}<+\infty)$$ For them to be equal, you need certain symmetry conditions on the ...