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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?

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Perhaps I'm thinking about this the wrong way, but isn't this just the probability that three tails occur on the first three tosses? Or maybe you mean a win occurs if TTTH is rolled. – Patrick Mar 27 '12 at 2:42
I'm assuming that combinations such as TTTTTTH or THTTTH are valid for winning. But I'm not sure if I'm making a correct interpretation of the problem. – bacchus Mar 27 '12 at 2:54
To me the answer to the problem as posed is clearly $\left(\frac{1}{2}\right)^3$. "Up to" means might as well quit as soon as you have your $100$ dollars. With THTTTH you didn't get three tails before encountering a head. – André Nicolas Mar 27 '12 at 3:14
@Andre That was my first solution. But after seeing the final solution value presented by the book's author I decided to consider other possible combinations. But you are right when you state that THTT... does not respect the problem statement. – bacchus Mar 27 '12 at 3:24
@AndréNicolas: According to your interpretation, the only instance that yields winning seems TTTH (if I correctly understood). Then why the probability is $(1/2)^{3}$ instead of $(1/2)^4$? – Sangchul Lee Mar 27 '12 at 3:25

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.

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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 $5/32$

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