Coin sequence paradox from Martin Gardner's book "An event less frequent in the long run is likely to happen before a more frequent event!"
How can I show that THTH is more likely to turn up before HTHH with a probability of
9/14, even though waiting time of THTH is 20 and HTHH, 18!
I would be very thankful if you could show me the way of
calculating the probability of turning up earlier, 
and the waiting time. Thank you!

Here is a somewhat tedious, but sure, approach. Consider the pairs of prefixes of the two target words 
$$
A=THTH\qquad\text{and}\qquad B=HTHH,
$$ 
and the process of the longest prefixes reached at each time. 
This yields a Markov chain, with every transition probability $1/2$, on the states $(O,O)$,  $(T,O)$, $(O,H)$,  $(TH,H)$,  $(T,HT)$,  $(THT,HT)$,   $(TH,HTH)$, and the two states $A$, to denote the fact that the target word $A$ is reached, and $B$ to denote the fact that the target word $B$ is reached. 
To compute the probabilities of hitting $A$ and $B$, the transition graph of this chain can be somewhat simplified, leading to
$$
\begin{array}{cccccc}
O,O & \rightarrow & TH,H & \stackrel{\longleftarrow}{\rightarrow} & THT,HT & \rightarrow & A
\\
& \searrow & \downarrow\uparrow& &\uparrow & & 
\\
& & T,HT & \rightarrow & TH,HTH & \rightarrow & B
\end{array}
$$
The probability $p_{00}$ to reach $A$ before $B$ starting from $(O,O)$ is such that $2p_{00}=p_{21}+p_{12}$, $2p_{21}=p_{32}+p_{12}$, $2p_{12}=p_{23}+p_{21}$, $2p_{32}=1+p_{21}$ and $2p_{23}=0+p_{32}$, with notations one hopes to be self-explanatory. This yields $p_{00}=9/14$. Obviously, $p_{00}\gt1/2$ because of the transition from $(TH,HTH)$ to $(THT,HT)$.
To compute mean waiting times, the method is similar. One uses for each word the Markov chain describing the length of the longest prefix realized. 
For the word $A$, the nonzero transition probabilities are from $0$ to $0$ and $1$, from $1$ to $1$ and $2$, from $2$ to $3$ and $0$, and from $3$ to $1$ and $4$ (word $A$ completed). In terms of waiting times, this yields the system of equations $t_0=1+\frac12(t_0+t_1)$, $t_1=1+\frac12(t_1+t_2)$, $t_2=1+\frac12(t_3+t_0)$, and $t_3=1+\frac12(t_1+0)$, hence $(t_0,t_1,t_2,t_3)=(20,18,16,10)$ and the mean waiting time is $t_0=20$.
For the word $B$, the nonzero transition probabilities from $0$ to $0$ and $1$, from $1$ to $1$ and $2$, from $2$ to $3$ and $0$, and from $3$ to $2$ and $4$ (word $B$ completed). The only difference is the transition from $3$ to $2$ (instead of $1$) hence the waiting time $t_0$ should be smaller here. To wit, the affine system is now $t_0=1+\frac12(t_0+t_1)$, $t_1=1+\frac12(t_1+t_2)$, $t_2=1+\frac12(t_3+t_0)$, $t_3=1+\frac12(t_2+0)$, hence $(t_0,t_1,t_2,t_3)=(18,16,14,8)$ and the mean waiting time is $t_0=18$.
The cause of the difference between these two waiting times is easily located: in the mean, the chain uses once the transition from $3$ to $1$ respectively to $2$. To reach $2$ from $1$ in case $A$, the transitions are $1/2$ from $1$ to $1$ and $1/2$ from $1$ to $2$. Thus, one will lose in the mean one unit of time in the loop $1\to1$ and one unit of time to perform the step $1\to2$. After that one is back to the situation of chain $B$ and apart from that the two paths are identical. Hence chain $A$ needs, in the mean, two more steps to reach destination.
