In Theorem 4.9 of the book Markov Chains and Mixing Times by Levin & Peres, there is a convergence theorem for ergodic Markov chains which states that there exist constants $C > 0$ and $\alpha \in (0,1)$ such that $$\sup_{x \in \mathcal{X}}\|P^t(x) - \pi\|_{TV} \leq C \alpha^t$$
Here $\mathcal{X}$ is the state space of the Markov chain, and $\pi$ is its stationary distribution. Is there a stronger mixing condition than this, where we can say that the chain converges to its stationary distribution in finite time with probability 1? In other words, under what conditions can we state that there exists a time step $\tau > 0$ such that
$$\sup_{x \in \mathcal{X}}\|P^t(x) - \pi\|_{TV} \leq C I\{t \leq \tau\}$$
where $I(A)$ is the indicator function on a set $A$?
I don't have a solution to your problem, but here is an interesting example. Consider the following evolution on the state space $S = \{0,1\}^n \times \{ 1\ldots n \}$: $$ (b_1b_2\ldots b_n, i) \to (b_1 \ldots b_{i-1} B b_{i+1} \ldots b_n, i+1) $$ where $B \in \{0,1\}$ is a chosen uniformly and the addition is assumed to be modulo $n$. Hence the chain randomly flips the $i^{th}$ bit, then $i+1^{th}$ bit and so on (repeating from the start after reaching $n$). This chain exactly mixes in time $n$. But this chain is periodic. It will be interesting to consider aperiodic chains.