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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.2012-03-27
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    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.2012-03-27
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    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.2012-03-27
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    @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.2012-03-27
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    @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 $(1/2)^4$?2012-03-27
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    If we count all the instances that 'ends up with TTTH and no occurrence of TTTH before halting', the probability is still $\frac{1}{4}$, not $\frac{5}{32}$. I checked this first by using transition matrix, and then by direct counting with Mathematica. And more easily, notice that TTTH at each instance can occur at most once, otherwise the number of trials must exceed 8. So it is equivalent to insert TTTH into a random sequence of T and H with length 3, yielding total $2^3 \times 4 = 2^5$ possibilities. Thus again $\frac{2^5}{2^7} = \frac{1}{4}$.2012-03-27
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    No, under my interpretation one has already won if one gets $TTT$. But if they mean that a head *must* be encountered, then wins are TTTH, TTTTH, TTTTTH, TTTTTTH, which gives an answer not equal to $5/32$.2012-03-27
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    @AndréNicolas: Now I understood where the ambiguity arises.2012-03-27
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    My guess is that it means the sequence TTTH is encountered somewhere, and the game is over when that happens. So for example, TTHTTTH would be a win.2012-03-27

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