Re: [Bug-gnubg] Alternative weights files

[Joseph Heled]

You are right, I was slightly off. It is 1/2 and not 2/3.

The single experiment is two matches (with reversed dice).
The possible outcomes are XX, (X wins both), OO and XO (which is the same as OX.
I assign a +1 for XX, zero for XO and -1 for OO

That way, under the null hypothesis P(XX) = 1/4, P(OO) = 1/4, P(XO) =
1/2 the mean is zero and the variance is 1/4*1^2 + 1/4*(-1)^2 = 1/2.
The variance of N matches is N* (1/2) and so the std is (N/2)^0.5

-Joseph

On 20 September 2012 08:35, Philippe Michel <address@hidden> wrote:
> On Wed, 19 Sep 2012, Joseph Heled wrote:
>
>> your test statistic is 117*-1 + 262*0 + 121*1 = 4
>> the standard deviation is (500 * (2/3))^0.5 = 18.25
>
>
> This doesn't seem right (why 2/3 ?) and this is more than the
> (1000 * 0.5 * (1 - 0.5))^0.5 = 15.8
> from the dumb rollout.
>
> On a second look I think it is :
>
> E(x) = (close to) 0
> E(x^2) = 121 + 117
>
> for a standard deviation of 15.4 or a little less.
>
> Duplicated dice would then be only a small improvement (at least in this
> example).