Hello Jonas, > > The scores were > > A vs C 3-1 > > B vs C 0-4 > > With respect to the 5 % level this refutes the hypothesis that > > A is not stronger than B. > > > > To prove this you have to find upper bounds for > > (1-p)^3 * [1-p + 4p] * p^4, where p is in the interval [0,1]. > > This maximum is clearly below 0.05. > > I do not completely agree with this calculation.
> What you have done could be seen as multiplying p-values. Forbidden. Why? Probabilities of independent events can be multiplied. In total I did not only multiply, but built an expression which had both multiplication and summation. > Specifically, you have computed the probability under p of having both > A-C 3-1 or better for A > B-C 0-4 or worse for B > The problem is that you are comparing A to B. So you have to take all > the cases where the apparent difference in strength is at least as big. > This notably includes the case: > A-C 4-0 > B-C 1-3 You are correct. > In fact in this case that's the only part to be added. Right. And I had included that term in my computation a year (or so) ago. > We still get a p-value less than 0.04, it's still significant. Correct. Ingo. -- GRATIS: Spider-Man 1-3 sowie 300 weitere Videos! Jetzt freischalten! http://portal.gmx.net/de/go/maxdome _______________________________________________ Computer-go mailing list [email protected] http://dvandva.org/cgi-bin/mailman/listinfo/computer-go
