A Markov chain on{0,1,2,3} has $$P=\begin{pmatrix}1/2&1/2&0&0\\1/2&1/4&1/4&0\\0&0&1/4&3/4\\0&0&0&1\end{pmatrix}.$$ Starting at1, let p be the probability of visiting0 before3. Find6p.
From2, state0 is unreachable, so its success probability is0. From1, the first step goes directly to0 with probability1/2, returns to1 with1/4, or goes to2 with1/4. Hence p=1/2+p/4, giving p=2/3 and6p=4.
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