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 <[email protected]> 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). _______________________________________________ Bug-gnubg mailing list [email protected] https://lists.gnu.org/mailman/listinfo/bug-gnubg
