Tossing a fair coin, we want to compute the probability of getting HTH before HTT?
From Markov state, we can see that state HTT is equivalent to the state HTH, then the probability should be 1/2. However we also know that the expectation steps arriving HTH is $E[HTH] = 2^3 + 2 = 10,\ E[HTT] = 2^3=8.$ It seems HTT should be easily arrived compared with HTH.
