HHHHH or TT

A successful string is a sequence of H's and T's in which HHHHH appears before TT does. Each successful string must belong to one of the following three types:

(a) those that begin with T, followed by a successful string that begins with H;

(b) those that begin with H, HH, HHH, or HHHH, followed by a successfully string that begins with T.

(c) the string HHHHH.

Let $P_h$ denote the probability of obtaining a successful string that begins with H, and let $P_t$ denote the probability of obtaining a successful string that begins with T. The three types of winning strings allow us to build recursive relations in the backward direction. More precisely, a winning string of type

(a) is of the form of T H . . ., which can be mapped one-to-one to a winning string of the form H . . .. Thus, $P_t=\frac{1}{2}P_h$ On the other hand, a winning string beginning with k (0 < k < 5) consecutive H's can also mapped one-to-one to a winning string beginning with T. It follows that

$P_h=(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16})P_t+\frac{1}{32}$

Solving the last two equations simultaneously, we find that

$P_h=\frac{1}{17}$

and

$P_t =\frac{1}{34}$

Thus the probability of obtaining five heads before obtaining two tails is

$P_h + P_t = \frac{3}{34}$ .

So A+B = 37.