I many hours i have to play to see mathematical result in casino?
Do more hands lower fluctuation amount?
You cannot possibly live long enough to approach the long run.
However, the longer you play with an advantage the more likely you become to approximate the expected result.
varies by game and has been established by Brett Harris as N0 (N-Zero) where N is the number of hands required for your eV to exceed one standard deviation of variance. For a good single deck game this could be as short as 50 or 60 hours. For a mediocre play-all 8 deck game it could be as long as 600+ hours or more. Playing this number of hours gives you an approximately 84% chance of being even or ahead. For a 97.5% chance of being ahead (2 SD's) you'll need to play 4x the hours of N0...so...
The long run to guarantee a 97.5% chance of being ahead requires you to play anywhere from 200 hours for a great game on up to 2,500 or more hours for a very poor game. Divide that by the number of hours you expect to play in a given year and you can find out the long run for virtually any game in terms of months or years.
NO for a given game is simply its (Variance/EV)^2. So, if a game has a variance of 3.49 units per hand and a win rate of 2.5 units per 100 rounds, then NO would be ((3.49*100)/2.5)^2 or (349/2.5)^2 which equals (139.6)^2 or 19,489 rounds.
BJRM or CVCX will automatically generate the NO for a given game for any set of rules, bet spread, and penetration point.
Two bits I have picked up along the way which I enjoy.
As the number of trials increases, the absolute deviation tends to increase, while the percentage deviation decreases. Something to ponder.
And:
In the longrun we'll all be dead.
