QZ's points are correct in a qualitative sense, but for VP 10 million hands
isn't very many in a statistical sense.
VP has very high variance, and this implies that it takes a truly huge
number of hands for the results to "tighten up" to this level of precision.
Assuming a game with a variance of 30, the standard deviation after
10 million hands is sqrt(30/10,000,000) = 0.001732. So, the standard
deviation is 0.17%, and 0.1% represents about 0.6 standard deviations.
This implies that the probability of being within 0.1% after 10 million
hands is only about 48%, which isn't even a concensus.
If we take "almost certainly" to mean a 95% probability that the results
will be within a tenth of a percent of expected return, then we need
two sigma to equal 0.1%, or:
N = 30/(0.0005)^2 = 120,000,000 hands.
If we use the "Ivory Standard" of (99 + 44/100)% sure, then we need
2.8 standard deviations, or:
N = 30*(2.8/0.001)^2 = 235,200,000 hands. At a rate of 1200 hands
per hour, this represents over 22 years of continuous play, without
a potty break. Youch!
So, following a single machine for its entire lifetime and measuring its
payback will not give results that are statistically significant beyond
"roughly speaking."
···
On Saturday 26 October 2002 11:53 pm, Quad Zilla wrote:
I don't think it's that difficult a concept. The paytable determines the %
payback on a VP machine. Each individual machine will eventually payback
THAT %, as long as it is in play.
Absolutely, positively FALSE. The payback by the machine over its
entire lifetime is governed by the same things that govern what each of
us gets over our lifetime. After, let's say 10,000,000 hands, the machine
will almost certainly, but not certainly, be within a tenth of a percent or
so of the expected return.