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!

It is enough to devise a set of linear equations ($p=q=1/2$):
\begin{align*}
T &= pT+qTH \\
TH &= pTHT + qH \\
THT &= pT+q1 \\
H &= p(pT+q(pTHT+q0))+qH \\
X &= pT + qH
\end{align*}
and after solving it
\begin{align*}
T &= 5/7 \\
TH &= 5/7 \\
THT &= 6/7 \\
H &= 4/7 \\
X &= 9/14
\end{align*}
We get $X = 9/14$ which is what were you looking for. Let (1) means getting THTH before HTHH. What those equations mean is that probability of (1) starting from T is the same as $1/2$ of probability of (1) starting from TT (which is equivalent to T) and $1/2$ of probability of (1) starting from TH (which is not equivalent). The rest follows similar suit. 
Edit: To be more intuitive (but less strict) let us observe that if HTHH happens at position other than 0 (and that happens with $1-1/16 = 15/16$ probability), then with probability at least $1/2$ THTH happens before -- due to probability of T before HTHH. Just by that you know that probability of (1) is greater than $15/16 \cdot 1/2 = 15/32$.
Adding probability of THTHT (the last T is required so the events won't overlap) at position 0 (1/32) you got 1/2 total and we haven't counted all
the instances yet, so surely probability of (1) is strictly greater than 1/2.
Hope that helps ;-)
