jmc

For Manu to win on his first turn, the sequence of flips would have to be TTH, which has probability ${\left(\frac{1}{2}\right)}^3$. For Manu to win on his second turn, the sequence of flips would have to be TTTTTH, which has probability ${\left(\frac{1}{2}\right)}^6$. Continuing, we find that the probability that Manu wins on his $n$th turn is ${\left(\frac{1}{2}\right)}^{3n}$. The probability that Manu wins is the sum of these probabilities, which is $$\frac{1}{2^3}+\frac{1}{2^6}+\frac{1}{2^9}+\cdots =\frac{\frac{1}{2^3}}{1-\frac{1}{2^3}}=\boxed{\frac{1}{7}},$$ where we have used the formula $a/(1-r)$ for the sum of an infinite geometric series whose first term is $a$ and whose common ratio is $r$.