You can definitely break even with N=13, assuming the unrealistic even distribution between 0 and 1, perhaps a bit earlier with a more sophisticated strategy.
Example strategy: Choose randomly the first time, the same side in all further throws. If the other side comes up before the 6th throw stop (6th throw because that's the earliest your expected winnings for the next throw don't increase by stopping if the other side comes up). Otherwise continue to N.
The probability of the same side coming up n times is 2/(n+1), the probability of the other side coming up exactly the n+1th time 2/(n^2 +3n +2).
Expected winnings each throw: 1st: -1$, 2nd: -1/3 $, 3rd: 0$ 4th: 1/10 $, 5th: 2/15 $, 6th-Nth: 1/7 $
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