zman said:
I have had over 15 sessions making 14 losses and 1 win.
Can i ask how much hands of BJ gives a good spread? Enough to show what the odds are for that BJ game, im gonna play vegas strip BJ.
Ok, lets say we want to estimate the expected value EV (negative) of a black jack hand and that we know the standard deviation SD a priori (it's ~1.16 for vegas strip BJ).
The result R after N hands flatbetting the amount x, is a normally distributed variable with mean N*EV*x and standard deviation sqrt(N)*SD*x.
Now, R/(N*x) has mean EV and standard deviation SD/sqrt(N).
If we want e.g. a 99 % confidence interval, we get the interval R/(N*x) 2.85*SD/sqrt(N)
So if we want get an estimate of EV with an error margin of 0.0001 (0.01 %) and a 99 % confidence level, we need to choose N such that
2.85*SD/sqrt(N) < 0.0001 => N > 1 092 963 600
If we only need an error margin of 0.001 (0.1 %) and a 95 % confidence level, we get
1.96*SD/sqrt(N) < 0.001 => N > 5 169 256
With the 10000 hands suggested, the standard deviation of R/(N*x) would be 1.16/100 = 0.0116, more than 1 %. The error margin on a 95 % confidence level is more than 2 %.
In betting units, the expected loss over 10000 hands with EV=-0.36% is 36 units, and the standard deviation is 100*1.16 = 116 units.