I was reading the second answer for this question. I wondered why the probability of encountering THT before TTH is 1/3 ?


Following the first T, we have:

  • T* with probability $\frac12$ (any number of T's followed by H yields TTH)
  • HT with probability $\frac14$ (yielding THT)
  • HH with probability $\frac14$ (returning to the initial state)

\begin{align}P(TTH) &= 0.5 + 0.25*P(TTH)\\ 0.75*P(TTH) &= 0.5\\ P(TTH) &= 2/3\\[3ex] P(THT) &= 0.25 + 0.25*P(THT)\\ 0.75*P(THT) &= 0.25\\ P(THT) &= 1/3\end{align}


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