We call one side of a fair coin $1$ and the other side $0$. Now we throw the coin repeatedly until we get one of three specific sequences of $1s$ and $0s$. We stop throwing the coin when we have obtained one of these sequences. The sequences are: $$m_1=1111$$ $$m_2=1001$$ $$m_3=0011$$
What is the respective probability for each sequence $m_1$, $m_2$ and $m_3$ to be the one that appears and ends the coin-throwing?
My thoughts: As per usual with Markov Chains, the three sequences do not have to have have the same probability, that makes sense to me. As I thus far have only seen such problems with two such "Stopping-sequences", however, I am not sure how to solve it with Markov-Chains. Maybe there is a creative solution with Random Walks to it? Any proposals would be appreciated!
EDIT: Any referral to papers on methods on how to solve such problems with $n$ "stopping-sequences" would also be appreciated.
This is a finite-state, absorbing Markov chain, with $3$ absorbing states. For the states we simply have the prefixes of the absorbing states. The fist absorbing state is $m_1$ and the corresponding prefixes are $$m_{11}=1\\ m_{12}=11\\ m_{13}=111 $$ For $m_2$, we already have the prefix $1$, so we get $$m_{21}=10\\ m_{22}=100\\$$ $m_3$ gives $3$ new prefixes.
$$m_{31}=0\\ m_{32}=00\\ m_{33}=001 $$ The wiki article explains how to get the probabilities of absorption in each absorbing state when staring in any given transient state. Since we start in state $m_{11}$ or state $m_{31}$ with equal probability, we just have to average the two probabilities.