I did this. Please see why I am getting the answer in a strange way!!
Let Probability of getting tail = p, Probability of getting head = (1-p)
Bob wins if the outcome is (TH) or(TT+HH)TH or (TT+HH)(TT+HH)TH and so on.
So, P(Bob winning) = p(1-p) + [p^{2} + (1-p)^{2}]p(1-p) + [p^{2 }+ (1-p)^{2}]^{2}p(1-p) + ...........
Let [p^{2 }+ (1-p)^{2}] be x.
So, Probaility = p(1-p)[1 + x + x^{2} . .....] = p(1-p)($\frac{1}{1-x}$) = $\frac{p-p^{2}}{1-[p^{2}+1+p^{2}-2p]}$ = $\frac{p-p^{2}}{2p-2p^{2}}$ = 0.5
Will this question give 0.5 for every cases?? I have not even substituted the values.
@prajwal @air1 please see