average number of tossing coins till 'Head-Tail-Head' series of observation comes up I know that we use 'geometric distribution' to solve problems such as 'Expected number of tosses till first head comes up' or 'expected number of rolling dice till first 'number 6' comes up'.
However, this question 'average number of tossing coins till 'Head-Tail-Head' series of observation comes up' is a little bit different because the observation that we want is not just one event, but a SERIES of events.
It seems like we will still use the geometric distribution to solve this problem. But is the answer for this question 8 because the expected number is '1/p', and 'Head-Tail-Head' probability is (1/2)(1/2)(1/2) = 1/8 ?
 A: No, unfortunately, you cannot do that! We'll think recursively. First, you have to toss it twice, note this down. Then, depending on your last two tosses, we'll formulate an expectation formula, i.e. $E[X|T_1T_2]$ will mean expected number of tosses ahead when your last two tosses are $T_1$ and $T_2$ respectively.
You've four possibilities: ${HH,HT,TH,TT}$. When your last two tosses are $HT$, if we toss Heads, we're done. So, with $0.5$ probability, we leave the game. But, if we toss Tails (with $0.5$ probability), we're left with $TT$ in the last two tosses, and our expectation becomes $E[X|TT]$. We could write our total expectation as follows, then: 
$$E[X|HT] = \frac{1}{2}.1+\frac{1}{2}(1+E[X|TT])$$
Similarly, we have:
$$
E[X|TT] = \frac{1}{2}(1+E[X|TH]) + \frac{1}{2}(1+E[X|TT]) \\
E[X|TH] = E[X|HH] = \frac{1}{2}(1+E[X|HH]) + \frac{1}{2}(1+E[X|HT])
$$
When solved, you'll get $E[X|HT] = 6, \ \ \ \ E[X|TH] = E[X|HH] = 8, \ \ \ \ E[X|TT] = 10$
Each of these cases occur with $\frac{1}{4}$ probability, adding the first two compulsory tosses would lead to $2 + \frac{6+8+8+10}{4} = 10$ expected tosses.
