E(number of trials until HH)=(1-p^2) / (q*p^2) = (1+p)/(p^2)
E(number of trials until TT)=(1-q^2) / (p*q^2) = (1+q)/(q^2)

=> rate of HH = p^2 / (1+p)
rate of TT =q^2 / (1+q) = (1-p)^2 / (2-p)

=> P(HHTT)= [p^2/(1+p)] / {[p^2/(1+p)] + [(1-p)^2 / (2-p)]}

希望高手给检查下，谢谢！

First no matter what, the last one happening would be Tail. P(head)=p and P(Tail)=Q
So, there are three possible events could be happen before the 2nd tail.   HHTT, HTHT, THHT.
So, 3*(1-q)^2*(q^2)

Also, This an example of Negative Binomial Distribution, x=4 and n=3 p = p , and 1-p=q.  by plugging these into the formula, you should get the same result.  You could google the formula of Negative Binomial.

The key point here is "successive".

Let $E$ be the event of interest. Then HTHTHTHH... is an outcome in $E$ while HTHTHTT... is not.

If the first toss gives H, then an outcome belongs to $E$ if and only if there exists some positive integer $k$ such that the first $2(k-1)$ tossing give HTHT...HT and the $2k-1$ and $2k$ tossing give HH. The probability for this particular outcome with fixed $k$ is $p^{k+1}q^{k-1}$. Now sum over $k$ from 1 to $\infty$ to get $p^2/(1-pq)$.

A similar argument gives the probability of $E$ intersect the event that the first tossing gives T:  $p^2q/(1-pq)$. (The probability for a particular outcome with fixed $k$ is $p^{k+1}q^{k}$.)

Therefore the probability of $E$ is $p^2(1+q)/(1-pq)$, or in terms of $p$ only:
\[
  (2-p)p^2/(1-p+p^2).
\]