This is all scribbled in a notebook, so there's a good chance that it's wrong, but bear with me on the arithmetic.

As a first guess, let's agree that in expectation you earn $0.51 each time you flip the coin, so it should take you about 20 tries to reach $10. Let's do something better, though.

EDIT: The previous paragraph is totally wrong. Thanks, tome :)

Let's build a confidence interval with alpha = 0.01 so that (1 - alpha) = 0.99. First, we'll need some trials. For that, I wrote a program that flipped a weighted coin and played the game until it reached $10 using the rules that you described. I recorded the number of coin flips required in each of 15 trials:

298, 84, 268, 2712, 110, 66, 42, 128, 84, 48, 280, 80, 64, 42, 234

We'll need the sample mean, X_bar = 302.

Now, we'll compute the Z-score so that we can build an interval in which the true mean (mu) lies with 99% probability:

P(-z <= Z <= z) = 0.99

We know that Z = (302 - mu) / (sigma / sqrt(n)), where sigma (the standard deviation) = 650 and sqrt(n) = 4. I'm rounding. Therefore, Z = (302 - mu) / 168.

Now, let's look at the cumulative distribution function Phi(z) and note that if Phi(z) = 1 - (alpha / 2) = 0.995, then Phi(z) ~= 0.997, the approximate cutoff for the 3rd standard deviation. Thus, z ~= 3.

Thus, we have that P(-3 <= (X_bar - mu) / (sigma / sqrt(n)) <= 3) = 0.99, so P(X_bar - 504 <= mu <= X_bar + 504) = 0.99. Therefore, I am 99% confident that the true mean, mu, lies on [X_bar - 504, X_bar + 504] = [302 - 504, 302 + 504].

That's a really wide range, and seemingly completely unhelpful for the purposes of betting. More sample trials would teach us more and lead us to a smaller interval since we expect that within some large number of trials we will converge on mu.