If I have a Matrix like the one below, what is the probability $p_t$ that at a certain time $t$, we are still not able to arrive at state $z$
$$ \begin{array}{c|lcr} \text{States} & x & y & \text{$z$} \\ \hline \text{$x$} & (1-\beta) & \beta & 0 \\ \text{$y$} & (1-\beta) & 0 & \beta \\ \text{$z$} & 0 & 0 & 1 \end{array} $$
I tried using the law of total probability with the formula,
$\text{Pr}[p_t]=\text{Pr}[p_t|x]\text{Pr}[x]+\text{Pr}[p_t|y]\text{Pr}[y]+\text{Pr}[p_t|z]\text{Pr}[z]$
but I am not sure how to proceed if there is an $\epsilon$ chance of making a mistake. The way I wrote it is,
$\text{Pr}[p_t]=(1-\epsilon)\beta^{t-1}[(1-\beta)^{t-2}(1-\epsilon)+(1-\beta)^{t-2}(1-\epsilon)+0]$
The first term above is for the probability of staying in state $x$, then for the state $y$ and lastly for the state $z$.
$t=0$ starts at state $x$ and I want to get the probability when the state is in state $y$
Please tell me what to modify to make my question clear.
Thank you very much.
Here's a hint.
Denote $p^x_t$ the probability to not reach $z$ in the next $t$ turns, given that we are now on $x$. Similarly, define $p_t^y$ the probability to not reach $z$ in $t$ turns, given we are now on $y$.
Then you can write the following recursive relation $$ p_t^x = (1-\beta)p_{t-1}^x + \beta p_{t-1}^y $$ $$ p_t^y = (1-\beta)p_{t-1}^x $$
Or combine the two written above to be:
$$ p_t^x = (1-\beta)p_{t-1}^x + \beta (1 - \beta) p_{t-2}^x $$
This is a recursive formula that should be solved for general $t$. Initial conditions would be $p_1^x = 1, p_2^x = 1 - \beta^2$