There is an old question here on the Elchanan Mossel's Dice Paradox:
You throw a fair six-sided die until you get 6. What is the expected number of throws (including the throw giving 6) conditioned on the event that all throws gave even numbers?
There are plenty of good answers to that question. I have a related but different question, namely, why does the Markov chain procedure give a different answer?
If I wanted to set up this solution as a Markov chain, I would have three states: the initial state $S$ before we throw the dice, also equivalent to rolling an odd number, the "even-but-not-6" state which I'll label $2|4$ and the absorbing state $6$.
Therefore, for states $\{S, 2|4, 6 \}$, the transition matrix $P$ would look something like:
$$P = \{p_{ij}\} = \begin{pmatrix}3/6 & 2/6 & 1/6 \\\ 3/6 & 2/6 & 1/6 \\\ 0 & 0 & 1\end{pmatrix}$$
The expected time to reach the absorption state starting from state $i$, which I denote $\mu_i$, is the unique solution to the equations $\mu_6 = 0$ and $\mu_i = 1 + \sum_{j=1}^3 p_{ij} \mu_j$. The absorption state here is the final state 6.
Turning the handle gives $\mu_S = 6$
Can anyone provide any guidance on where I am going wrong?
Thanks