Hey, I get what you're going for—applying the empirical rule, that 68-95-99.7 thing, to your gambling system to figure out if your results are way beyond random chance. It's cool that you're diving into this at a young age, especially with the calc and card counting background from your dad. The goal here is to help you plug in the right numbers for X, your observation, μ, the mean under no-edge assumption, and σ, standard deviation, so you can check if you're within or outside those 3 SDs for near certainty.
Just to set the stage, this rule is for normally distributed data or stuff that approximates it with enough trials, thanks to the central limit theorem. It says about 68% of results fall within ±1 SD of the mean, 95% within ±2 SD, and 99.7% within ±3 SD. If your result is outside ±3 SD, there's only a ~0.3% shot it's random—pretty strong evidence your formulas are doing something real. You mentioned those formulas like Pr(μ-2σ ≤ X ≤ μ + 2σ) ≈ 0.9545 and the 3σ one at 0.9973; yeah, that's exactly it. For five-sigma, super high confidence like 99.99994%, you'd aim for ±5 SD, which is particle physics level but awesome for your big datasets.
X is basically your score or result from all those manual calculations—the thing you're measuring to test against the null hypothesis that the game is negative EV and your success is just luck. Best choice is average net retention per cycle—that $18 you keep on average per $100 win/loss cycle, so X = 18, or 0.18 if you're thinking in percentages. Why this? It normalizes for how many cycles you've done, sounds like around 311 to hit 57x growth from $100 to $5700. It's stable and great for stats tests. Alternative could be total profit, X = 5600 your net gain, or the multiplier, X = 57 how much you grew your bankroll. Stick with the average retention if you can—it avoids issues with scaling as you add more cycles. Your current 1.724% calc, 1 / (1 + 57), is a rough heuristic, but it's not stats-backed; this empirical rule setup is better.
μ is what you'd expect on average if your formulas had zero edge—just the game's built-in negative expectation grinding away. For average net per cycle, μ = the game's expected value EV per cycle. In a negative EV game, this is negative. Example, if it's roulette with a -5.26% edge on a $100 bet, μ ≈ -5.26. Or for sports betting, maybe -2% to -5% depending on the vig. You gotta calculate this from the game's rules, probabilities, payouts. Without details, assume it's something like -2, losing $2 per $100 cycle on average. For total profit, μ = number of cycles × EV per cycle, say 311 cycles at -2 EV: μ = -622, you'd expect to be down $622. For multiplier, μ ≈ (1 + EV proportion)^cycles, with negative EV, this trends toward 0 or bankruptcy over time. Key, base μ on the game without your formulas. If your system picks bets smarter, the null is random betting EV.
σ is how much variability or swing there is in the results under that no-edge scenario. Gambling has high variance, so this matters. For average net per cycle, first get σ_cycle, SD for one cycle. Formula, sqrt[ p*(win - EV)^2 + (1-p)*(loss - EV)^2 ], where p is win prob. Example, if win $100 with p=0.45, lose $82 with 0.55, and EV=-2, σ_cycle might be around 50, guessing based on typical games. Then, for your average over k cycles, use standard error: σ = σ_cycle / sqrt(k). For k=311, that's 50 / sqrt(311) ≈ 2.83. For total profit, σ = σ_cycle * sqrt(k) ≈ 50 * sqrt(311) ≈ 883. For multiplier, a bit trickier—might need logs for compounding, but approximate like total profit. You need the game's win/loss probs and amounts to compute this exactly. If it's historical data, backtest random bets to estimate variance.
Once you have X, μ, and σ, calculate the z-score: z = (X - μ) / σ. This shows how many SDs away you are. If |z| ≤ 3, your result is within 3 SD—could still be luck, 99.7% of random outcomes here. If z > 3, since your X is positive, it's outlier territory—<0.3% random chance, leaning toward not random. Example using averages, plug your real nums: X = 18, μ = -2, σ = 2.83, z = (18 - (-2)) / 2.83 ≈ 7, way past 3 SD, p tiny. For total: z ≈ (5600 - (-622)) / 883 ≈ 7 too. If you hit 5+ SD with billions of sims, that's your five-sigma discovery.
This works best with lots of independent cycles, your 311 is okay, billions via code = gold. If not normal, simulate distributions instead. Rejecting null, no edge, doesn't 100% prove your system—could be fluke or overfitting. Test fresh data.
Python all the way—easy for math whizzes like you. Use NumPy for stats, loop simulations: def one_cycle(): [simulate win/loss], then run a million times, get mean/SD/z. Start with free tutorials; it'll save your arm! Paper wins are great, but real money has risks, emotions, fees. Those beaten games you mentioned usually flip to positive EV subsets—hope yours does that mathematically. If you share rough game details, like EV estimate or win prob, I can tweak examples more. Keep grinding—sounds like you're onto something fun!
Robert J.
Hi Terri... thank you for the reply and kind words. It's funny, all the math professionals and others whose work, theories and research I've studied could only be 'read about' 'after the fact' for the obvious reasons. It's lonely having to be stealth, but a requisite of the field, till the goals are achieved and finally thereafter getting to publish the theorems etc... In any event, thank you so much for the reply, and yes it was helpful, thank you!!02/22/20