Let's assume I came up with a money management program that over the long run, beat the house.
(Please don't comment about how we can't make that assumption, etc, etc. I am asking you to make the assumption - just pretend for a second please.)
How many trials (dice rolls) would it take to show that the system is in fact a winning system?
Said another way - if I thought I had a winning system, and I simulated it 1000 times and it wins (more bank roll than starting bank roll) - would that be enough rolls to prove it? How about 2000 rolls? 5000?
I guess another way to say it, is given that the house edge with optimal strategy is 1.36 - 1.41% - how many rolls would it take with a 99% probability that bank roll is less than starting? If we get that number, could one use that number to then say - if I can show a positive bankroll after that number of rolls, there is only a 1% chance it isn't a winning system?
You also need to define "win" in specific terms. If your first bet won, your bankroll would be larger than when you started; would you stop betting at that point?
If the "system" is "always bet 1", then, if the bet is red on a single-zero roulette wheel and your starting bankroll is 100, the probability of being ahead at any point before losing the entire 100 is about 90%.
In fact, if the probability p of winning a single bet < 1/2, then, as your initial bankroll increases, the probability of being ahead B bets before losing the entire bankroll approaches ( (p / (1 - p) )B. In the single-zero roulette case, p = 9/19, so 1 - p = 10/19, p / (1 - p) = 9/10, and the probability of being ahead by 1 bet before busting approaches 9/10 as your initial bankroll increases, being ahead by 2 bets before busting approaches 81/100, being ahead by 3 bets before busting approaches 729/1000, and so on.
If I am crunching my numbers correctly, for the single-zero roulette even-money flat bet system, if you try it 100 times, you are expected to win 90 times, and the standard deviation is 3 (I am not sure if there is an "easy" way to calculate this; in 100 trials, assuming the probability of being ahead at some point before being behind by 100 is 0.9, then the probability of n out of the 100 trials being wins is (100)C(n) x 0.9n x 0.1100-n ; some Excel number crunching shows the mean is 90 and the standard deviation is 3), so, in this case, a 99% success rate would be +3 standard deviations.
Could you do the same for craps.
Assuming an overall house edge of 1.41,
What is the probability of being ahead (positive bank roll) after 1000 rolls? 5000? 10000?
I don't mean at any point - I mean at the end of the x number of rolls.
Thanks for the numbers.
I'm having a hard time following the last post.
The spreadsheet was for the probability of being ahead at any point right?
If you invented a money management system that defied the fundamental math of a game I'm not sure any number of trials would prove it, but I'm sure you'd be very famous with all the attention you'd attract in the mathematics world with your discovery. But if you had billions of quality simulations that showed something notable then you'd have my attention.
Your whole premise is flawed though. Money management just rearranges variance. You might be able to create a system that wins 99% of the time, but that 1% is going to ream you. The hard work has already been done: you can easily walk into a casino and do a roulette martingale tomorrow with less than a 1% chance of losing--- but God help you if it does.