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".
This seems to be a good opportunity to use a trick I figured out for approximating the probability of a binary event
That's just Bayesian updating. P(H|E)=P(H)*P(E|H)/P(E)
P(x)=P(H), x=P(E|H), and the integral of the previous step is P(E), since that's the whole point of your probability function.
I came up with this puzzle after reading Vaniver's excellent post on the Value of Information. I enjoyed working it out over Thanksgiving and thought I'd share it with the rest of you.
Your friend holds up a curiously warped coin. "Let's play a game," he says. "I've tampered with this quarter. It could come up all heads, all tails, or any value in between. I want you to predict a coin flip; if you get it right, I'll pay you $1, and if you're wrong, you pay me $3."
"Absolutely, on one condition," you reply. "We repeat this bet until I decide to stop or we finish N games."
What is the minimum value of N that lets you come out ahead on average?
Each game, you may choose heads or tails, or to end the sequence of bets with that coin. Assume that all heads:tails ratios are equally likely for the coin.
edit: since a couple people have gotten it, I'll link my solution: http://pastebin.com/XsEizNFL