Let $(X_{n})_{n\in\mathbb{N}}$ be a Markov Chain such that $X_{n}$ is the maximum score obtained after $n$ throws of a fair dice. Therefore, the state set is $M=\left\{1,2,3,4,5,6\right\}$.
I know that if we consider the random variable $Y_{k}$ which are the result of $k$th throw of our fair dice, then $(Y_{k})_{k\in\mathbb{N}}$ is iid with uniform distribution over $\left\{1,2,3,4,5,6\right\}$, so, we have $X_{n}=\max\left\{Y_{1},\ldots,Y_{n}\right\}$. Furthermore, the cumulative distribution of $X_{n}$ is $$F_{X_{n}}(k)=\left\{\begin{array}{ll} 0 & \mathrm{if}\: k<1 \\ \frac{i^{n}}{{6}^{n}} & \mathrm{if}\: i\leq k <i+1 \:\: \mathrm{for}\: i=1,2,3,4,5.\\ 1 & \mathrm{if}\:\: k\geq 6 \end{array}\right. $$ For other hand, the transition matrix is $$\Pi=\left[\begin{array}{rrrrrr} \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ 0 & \frac{2}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ 0 & 0 & \frac{3}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ 0 & 0 & 0 & \frac{4}{6} & \frac{1}{6} & \frac{1}{6} \\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array}\right].$$ My first objective is to find the invariant distribution with respect to our Markov chain, we recall
Definition: $\pi$ is an invariant measure if $\pi \Pi=\pi$. If in addition $\sum_{i\in M}\pi(i)=1$, then we say it is an invariant distribution
Find $\pi$ was easy, this comes from the uniform distribution, ie, $\pi(i)=\frac{1}{6}$ for all $ i \in M $.
My question: I need to show directly that $$\left\|\Pi^{n}(x,\cdot)-\pi\right\|_{TV} \overset{n\rightarrow\infty }{\longrightarrow} 0\tag{*}$$ for all $x\in M$, where $TV$ means total variation distance.
My attempt: By this link (page 48) we have $$\left\|\Pi^{n}(x,\cdot)-\pi\right\|_{TV} = \frac{1}{2}\sum_{i=1}^{6}\left|\Pi^{n}(x,i)-\pi(i)\right|.$$ For other hand, we have that $$\Pi^{n}=\left[\begin{array}{rrrrrr} \left(\frac{1}{6}\right)^{n} & \left(\frac{2}{6}\right)^{n}-\left(\frac{2}{6}\right)^{n} & \left(\frac{3}{6}\right)^{n}-\left(\frac{2}{6}\right)^{n} & \left(\frac{4}{6}\right)^{n}-\left(\frac{3}{6}\right)^{n} & \left(\frac{5}{6}\right)^{n}-\left(\frac{4}{6}\right)^{n} & 1-\left(\frac{5}{6}\right)^{n} \\ 0 & \left(\frac{2}{6}\right)^{n} & \left(\frac{3}{6}\right)^{n}-\left(\frac{2}{6}\right)^{n} & \left(\frac{4}{6}\right)^{n}-\left(\frac{3}{6}\right)^{n} & \left(\frac{5}{6}\right)^{n}-\left(\frac{4}{6}\right)^{n} & 1-\left(\frac{5}{6}\right)^{n} \\ 0 & 0 & \left(\frac{3}{6}\right)^{n} & \left(\frac{4}{6}\right)^{n}-\left(\frac{3}{6}\right)^{n} & \left(\frac{5}{6}\right)^{n}-\left(\frac{4}{6}\right)^{n} & 1-\left(\frac{5}{6}\right)^{n} \\ 0 & 0 & 0 & \left(\frac{4}{6}\right)^{n} & \left(\frac{5}{6}\right)^{n}-\left(\frac{4}{6}\right)^{n} & 1-\left(\frac{5}{6}\right)^{n}\\ 0 & 0 & 0 & 0 & \left(\frac{5}{6}\right)^{n} & 1-\left(\frac{5}{6}\right)^{n}\\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array}\right].$$ In this sense, for example, for $x=6$ we have $$\left\|\Pi^{n}(6,\cdot)-\pi\right\|_{TV} = \frac{1}{2}\left(\frac{5}{6}\right).$$ But this last one contradicts (*), which can not happen since it contradicts the Ergodic Theorem for Markov Chains (see Page 6 Theorem 1.3).
What is the mistake I am making? Or am I misinterpreting some of the concepts?
Your chain is absorbing at $i=6$ therefore $\pi(i)\neq 1/6$. Invariant distribution is $(0,0,0,0,0,1)$. Manual solution of $\pi \Pi=\pi$ give this answer immediately.