This seems to be a good opportunity to use a trick I figured out for approximating the probability of a binary event (I'm sure I'm not the first to discover it). The trick goes as follow: start with a probability distribution P(x) = 1. For each time the even happens, multiply P(x) by x, and for each time it doesn't, multiply by 1-x. After doing this for your data, renorm it so that the area beneath the curve is 1: you can do that by dividing your function by its integral from 0 to 1. If you have a range of outcomes, you can multiply by the function representing that to get expected utility. In this case, the function representing the outcomes can be modeled by O(x) = 4x - 3. If we assume the coin is biased towards heads (if it's biased towards tails, just substitute H and T in the following reasoning), we find through some math and puzzling that the formula that determines expected value for the sequence of coins is (N-4*T-4)/N+2) where H is the number of heads and T is the number of tails, and N represents the number of tosses. The probability of reaching any state is given by 1/N. For the first two flips, your expected value is negative (-$1 and -$.033 for the first and second respectively) no matter what happens, so we can count these as sunk costs. So our expected value is as follows here:

It took me 19 flips to come up with a positive expected value, though my answer with the original question is "as many as the person offering this can be convinced to give me".