Answer

To answer how much you should pay to play this, we need expected value of the game. So what are all the possible outcomes and payoffs? We can get H straight away with probabiliy $\frac{1}{2}$, in which case we receive the payoff of $2^1$. We can finish the game with 2 flips, if we get TH. The chance of this is $\frac{1}{2^2}$ and the payoff is $s^2$. The game could finish after three flips only if we get TTH, as we must have only Ts before the first H. This is one of the $2^3$ possible outcomes of a 3-coin flip, so the chance of this is $\frac{1}{2^3}$ and the payoff is $2^3$.

Put these together to notice the pattern:$$E(U) = \frac{1}{2}2 + \frac{1}{2^2}2^2 + \frac{1}{2^3}2^3 + ... = \sum_{\infty} 1$$So we should pay an infitie amount to play this game? Of course not, but we haven't made a mistake. This game is a well-known case of the St. Petersburg paradox, demonstrating the limits of using expected value. Daniel Bernoulli proposed a solution to the paradox by introducing the utility function, which discounts marginal utility as our payoff grows. you can read more here.