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 both nontrivial and advanced solution.
Consider an ideal gambling where a dealer Alice tosses a fair coin repeatedly. After she made her $(n-1)$-th toss (or just at the beginning of the game if $n = 1$), the $n$-th player Bob joins the game. He bets $2^0\$$ that the $n$-th coin is $T$. If he loses, he leaves the game. Otherwise he wins $2^1\$$, which he bets that the $(n+1)$-th coin is $H$. If he loses, he leaves the game. Otherwise he wins $2^2\$$, which he bets that the $(n+2)$-th coin is $T$. This goes on until he wins $2^4\$$ for $THTH$ and leaves the game, or the game stops.
Thus let $X^{(n)}_k$ be the r.v. of the winning the $n$-th player made when $k$-th coin toss is made. If we let
$$X_n = X^{(1)}_n + \cdots + X^{(n)}_n,$$
Then it is easy to see that $X_n - n$ is a martingale null at 0. Thus if $S$ denotes the stopping time of the first occurrence of $THTH$, and if we assume that $\mathbb{E}[S] < \infty$, then
$$0 = \mathbb{E}[X_{S} - S],$$
thus we have
$$\mathbb{E}[S] = \mathbb{E}[X_{S}] = 2^4 + 2^2 = 20.$$
Let $Y_n$ be the total winning corresponding to $HTHH$, and $T$ be the stopping time of the first occurrence of $HTHH$. Then we have
$$\mathbb{E}[T] = \mathbb{E}[Y_{T}] = 2^4 + 2^1 = 18.$$
Finally, let $U = S \wedge T$ be the minimum of $S$ and $T$. Then it is also a stopping time. Now let $p = \mathbb{P}(S < T)$ be the probability that $THTH$ precedes $HTHH$. Then
$$ \mathbb{E}[U] = \mathbb{E}[X_{U}] = \mathbb{E}[X_{S}\mathbf{1}_{\{S < T\}}] + \mathbb{E}[X_{T}\mathbf{1}_{\{S > T\}}] = 20p + 0(1-p),$$
and likewise
$$ \mathbb{E}[U] = \mathbb{E}[Y_{U}] = \mathbb{E}[Y_{S}\mathbf{1}_{\{S < T\}}] + \mathbb{E}[Y_{T}\mathbf{1}_{\{S > T\}}] = (2^3 + 2^1)p + 18(1-p).$$
Therefore we have $p = 9/14 \approx 64.2857 \%$.
