2
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.