Fw: Re: Fw: Re: [computer-go] Statistical Significance (was: SlugGov.s. Many Fac...

But we have a estimate of probability?

If if is it is 8/9 then ste standard deviation of them mean (= estimate of probabbility) is

SDevMean = sqrt(1/9*8/9) / sqrt(9) = 0,03.

So with about 66% probability desired probability is within 8/9 +- .03.

And with 99 it is 8/9 +- 0.09.

So with only nine samples we have measured the mean with about 10% accuracy.

So it is very unlikely that the program that won 8/9 is weaker than the other.

Petri Pitkänen

> From:: Compgo123@xxxxxxxxxxxxxxxxx
> To: computer-go@xxxxxxxxxxxxxxxxx
> Subject:: Re: Fw: Re: [computer-go] Statistical Significance (was: SlugGo v.s. Many Fac...
> Date: 09/09/2004

It's a good point to point to the binary (Bernoulli random number) process and the binomial distribution. I had the same feeling that what we discussed here were settled before the dawn of the 18th century.

However, you comment does not yet lead us to the promise land. The reason is that to calculate the variance of a binomial distribution one need to know the probability of one side winning. It's this probability we are trying to find out. If we know the variance, we can calculate the probability.

Daniel Liu

>Since this is a binary process - Program either wins or loses - we do not need to >measure variance. It is binomial distribution with well known properties. Measuring >that would be bit pointless as result is known beforehand

>Petri Pitkänen

Heart disease is Britains biggest killer. Join the British Heart Foundations Big Red Fightback: bhf.org.uk/fightback