Perhaps I should have asked instead, how many trials would it take to theoretically reach 1.9% with a 95% confidence interval?
Well alright. Now I can say "I don't know" instead of "your question sucks"
It's affected a lot by the big bets. The profit from thousands of small bets can be wiped out by a single lucky big bet (and vice versa).
And... I don't remember how to do a problem like that
It's complicated by the fact that with 1 trial, there is a 0% chance that the house will take within 1.89 to 1.91 percent.I think it can be assessed using Markov chains. Meni Rosenfeld's analysis of PPS pool bankruptcy probabilities (AoBPMRS, Appendix c) examines a similar problem and uses a Markov chain model. I haven't studied them, maybe someone else more familiar with Markov chains could do the analysis.
Markov chains are more useful when there's a relationship between the states of the system. In this case, it would be more like "losing 3 in a row changes your chances of winning the next one". Since we don't have that, you can use regular IID statistics.
The way I wrote the script that dooglus is using right now, is taking advantage of the fact that for a sum of random variables, you can just add their expected values, and add their variances. Square-root the resulting variance ot get the standard-deviation. The result will be a mean and std-dev, which you can use to compute a 3-sigma bounding box (or 2-sigma, if you want 95% confidence).
For binary systems like this (win or lose), you'd "normally" use different equations, but this construction is accurate as the number of trials gets large. SatoshiDice qualifies.