Quote from: organofcorti on April 03, 2013, 05:33:17 AM

Quote from: nimda on April 03, 2013, 04:16:24 AM

Quote from: dooglus on April 03, 2013, 03:35:52 AM

Quote from: Picklee on April 03, 2013, 01:58:34 AM

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).

The question should have one more number. It should be phrased "95% chance that the house edge will be within 1.89-1.91%" or similarly. The size of the acceptable error there will greatly influence the result.
And... I don't remember how to do a problem like that Cry

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.