What is the probability of getting TTHH before HHH in repeated fair coin toss?

$ \begingroup $ I would have guessed that this interrogate is basically a duplicate, but I was n’t able to find a interrogate with an answer that applies to this particular problem .
We can reformulate this problem using a Markov range : explicitly we ‘re looking to compute the probabilities $ \Bbb P ( HHH ), \Bbb P ( TTHH ) $ that the Markov chain settles into absorbing states corresponding respectively to $ HHH $ and $ TTHH $ occurring first .
The ( eight ) states are the possible stages of progress toward one of the two goal sequences ( $ HHH $, $ TTHH $ ).

The ephemeral states are :

The absorb states are :

In each state transient state there are two possible following states, corresponding to flipping $ H $ and $ T $ next, respectively, both of which thus have probability $ \frac { 1 } { 2 } $. explicitly, the conversion matrix of the Markov chain, with deference to the above ordain of states, is $ $ P=\left ( \begin { align } { cccccc|cc } \cdot & \frac { 1 } { 2 } & \frac { 1 } { 2 } & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \frac { 1 } { 2 } & \frac { 1 } { 2 } & \cdot & \cdot & \cdot & \cdot \\ \cdot & \frac { 1 } { 2 } & \cdot & \cdot & \frac { 1 } { 2 } & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \frac { 1 } { 2 } & \cdot & \frac { 1 } { 2 } & \cdot & \cdot \\ \cdot & \frac { 1 } { 2 } & \cdot & \cdot & \cdot & \cdot & \frac { 1 } { 2 } & \cdot \\ \cdot & \frac { 1 } { 2 } & \cdot & \cdot & \cdot & \cdot & \cdot & \frac { 1 } { 2 } \\ \hline & & & & & & 1 & \cdot \\ & & & & & & \cdot & 1 \\ \end { array } \right ). $ $ Since the absorb states are at the end of our list, the above matrix is in a favored form : The upper-left block, call it $ Q $, specifies the transitions between transient states, and the upper-right jam, call it $ R $, specifies those from ephemeral states to absorbing states .
By construction, if we start in the $ one $ thorium ephemeral state the probability of settling in the $ j $ thursday absorbing state is the $ ( i, j ) $ -entry $ B_ { ij } $ of $ B : = N R $, where $ N : = ( I – q ) ^ { -1 } $ is the alleged cardinal matrix of the Markov chain. Since we start in the initial department of state, $ \emptyset $, it ‘s enough to compute $ B_ { 11 } $, i.e., the probability that the sequence $ HHH $ occurs first, and $ B_ { 12 } $, i.e., the probability that $ TTHH $ occurs first ; since there are entirely $ 2 $ absorbing states, $ B_ { 12 } = 1 – B_ { 11 } $, and so it ‘s adequate to compute $ $ \Bbb P ( HHH ) = B_ { 11 } = \sum_ { k=1 } ^6 N_ { 1k } R_ { k1 }. $ $ The lone nonzero introduction in the first base column of $ R $ is $ R_ { 51 } = \frac { 1 } { 2 } $, then $ \Bbb P ( HHH ) = N_ { 15 } R_ { 51 } = \frac { 1 } { 2 } N_ { 15 } $, and using Cramer ‘s convention we can compute $ N_ { 15 } = \frac { 5 } { 6 } $ without computing the other entries of $ N $. Substituting yields the claim probabilities : $ $ \Bbb P ( HHH ) = \frac { 5 } { 12 } \qquad \textrm { and } \qquad \Bbb P ( TTHH ) = \frac { 7 } { 12 }. $ $

reference : https://leowiki.com
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