Mr. T said:
Sorry Fred, on belabouring this matter. But it is the matter of Variance and the size of your sample, 500.
What you are trying to show is that your House Edge for BS play is not affected by other players play. In other words you are trying to show that with 500 hands you can show that the HE remains the same.
Here is what I can show to you about flipping a coin 100 times and getting 50 heads and tail result.
The probability of getting exactly 50 of each is combin(100,50)*(1/2)100 = 7.96%.
In your case your sample is 500 so the probability that you are right in your statement is slightly higher than 7.96%. Not much to cheer about.
Smarter guys than me like Sonny and London Colin would know how to use the formula above to work out what is your exact % of probability of being correct. Have a good day.
T,
To show what the house edge is, or to measure the performances of various count systems, or to evaluate any measure that deals in hundredths of a percent does indeed require several million hands.
But to get a good idea whether the outcome of a chance event should fall at 50% or rather at say, 60% requires far less. Suppose someone were to claim that 3rd base's bad play will cause the other players to lose say, even 60% of the time. Then here's what a 500 hand sample will tell you if that sample actually produced only 50% losers
(in dollars wagered),as this one did.
The standard error of the sample was 12
.6 wins.
The number of actual losses was 264.
Losing 60% of the decisions
(in dollars wagered) would require 320 losses.
That's 56 losses, or 4
.4 standard deviations from the actual result.
My statistical "Z" score table says it's over 5000-to-1 that these 264 losses were lower than their true probability by at least 56 losses. My handling of the pushes may or may not be quite correct here, but either way, I'm pretty sure the indication will be just about the same. Perhaps Norm can comment on the effect of bad players down to the tenth of a percent.