Wednesday, February 12, 1997 10:05 PM
Paul R. Pudaite

In article <19970213020400.VAA22826@ladder01.news.aol.com>,
hittheflop@aol.com (HitTheFlop) wrote:

> My results after moving up to 10-20 HE have me encouraged.
> After 90 hours my win rate is $49/hr and standard deviation is
> $241/hr. This is a small sample and I would like to draw as
> much information out of it as is statistically reasonable. How 
> confident can I be that my true (longterm) win rate is at least 
> $30/hr? Is there someone with enough math to show me the 
> needed calculations? I'd like to be able to repeat the calculation
> when I have more data. Many thanks in advance.

Hi, Ed!  That's a mighty large hourly standard deviation you have there.
Are you sure you want to expose it?  :-)

Getting down to business, the standard error in the estimate of your
expected win rate is 241 / Sqrt(90) = $25.4/hr.  Your win rate of $49/hr
is thus .748 standard deviations above $30/hr.  Therefore, a classical
statistician would say you have a 77.3% confidence that your true win
rate is at least $30/hr.

However, let's take a Bayesian approach.  Suppose expected win rates in
the population of 10-20 hold'em players follow a normal distribution
with a mean of -$10/hour (due to the rake) and a standard deviation of
$15/hour.  This way, an expert who makes one big bet per hour is 2
standard deviations above the mean, the same requirement for Mensa
membership when considering IQ scores.

My Bayesian estimate of your win rate is +$5.25/hour with a standard
error of $12.92/hr.  That means that there's a 2.8% chance that your
expected win rate is at least $30/hr.

Paul R. Pudaite
pudaite@pipeline.com