Coin toss probability question In the Chapter 12 of the book "Operations Research: An Introduction" written by Hamdy A. Taha, the following question is posed:

You can toss a fair coin up to 7 times. You will win $100 if three tails appear before a head is encountered. What are your chances of winning?

The solution for the problem according to the author is $\frac{5}{32}$. Nevertheless, after thinking about the problem and trying a number of possible ways of solving it, I cannot find that specific solution and therefore solve it.
Can anyone one help me to figure out how to solve it?
 A: Finally found an interpretation of the wording, admittedly somewhat far-fetched, that yields an answer of $5/32$.  It is an interpretation close to but seemingly not identical to those of sos440 and Patrick.
Let's say that Alicia wins if the following happens: somewhere in the tossing, there is a sequence of the shape TTTH which is not immediately preceded by a T.
So the winning sequences are TTTH, HTTTH, XHTTTH, and YXHTTTH, where X and Y are any of T or H.  The probability of TTTH is $1/16$, the probability of HTTTH is $1/32$, as is the probability of XHTTTH and of YXHTTTH, for a total of $1/16+3/32$, that is, $5/32$.
The fact of getting $5/32$ does not show this was the intended interpretation. More likely is another more reasonable interpretation followed by a typo or mistake.
A: The question has lots of ambiguity, according to comments I understood that head's number should not exceed the tail's number.
So we have these states as winning states: (T:tail, H:head, X:don't care)
TTTXXXX -> $\frac{2^4}{2^7}$
TTHTXXX -> $\frac{2^3}{2^7}$
TTHHTXX -> $\frac{2^2}{2^7}$
THTTXXX -> $\frac{2^3}{2^7}$
THTHTXX -> $\frac{2^2}{2^7}$
And the sum of above states is: $$\frac{2^4}{2^7} + \frac{2^3}{2^7} + \frac{2^2}{2^7}+\frac{2^3}{2^7}+\frac{2^2}{2^7}=\frac{4+2+1+2+1}{32}=\frac{10}{32}$$
If the sequence must end with a head (or at least one head must appear, I'm saying this because of André Nicolas comments), the total state count should divide by 2, so the result will be $5/32$
