--to answer Meeker, Yes, I assume all the N tests independent. --to answer Andy Latto, F_N(a,b,c,...) is NOT symmetric in its arguments because as I said we assume 0<a<b<c<...<z<1, or anyhow I made that assumption before writing the formulas. However, without that assumption, then it would be symmetric, in the sense that the correct formula would be got by 1. sort the N arguments into increasing order 2. apply my formula. This 2-step process of course produces a result which is a symmetric function. It is a piecewise polynomial function, and it is continuous, but the pieces are not symmetric polynomials. --to eyeball my F5 formula F_5(a,b,c,d,e) = [20bc^3-5b^4+30b^2d^2-60bcd^2+20b^3e-60bc^2e-60b^2de+120bcde-10c^3a+30cd^2a+30c^2ea-60cdea-10d^2a^2+20dea^2-5ea^3+a^4]a. one does observe interesting things such as all coefficients divide 120... I think one could prove that in the FN formula, all coefficients divide N!, the coefficient of a^N is +1, and the coefficient of abcd...z is +N!. Which is a long way short of complete understanding, but it is something.