Sure, I just couldn’t be bothered. If you ignore the complicating factor of having black balls in the white urn and vice-versa (i.e. we look at the problem where the urn is 100% white or 100% black). then the maths is easier, and we can quickly see that this always holds whatever variables we put in.
In fact, it also works for many different scoring algorithms. If our scoring algorithm is f(x) then it the proof works whenever xf(x)+(1-x)f(1-x) is increasing on [0.5,1].As far as I can tell this happens all of the time that f(x) is increasing on [0,1].
Sure, I just couldn’t be bothered. If you ignore the complicating factor of having black balls in the white urn and vice-versa (i.e. we look at the problem where the urn is 100% white or 100% black). then the maths is easier, and we can quickly see that this always holds whatever variables we put in.
In fact, it also works for many different scoring algorithms. If our scoring algorithm is f(x) then it the proof works whenever xf(x)+(1-x)f(1-x) is increasing on [0.5,1].As far as I can tell this happens all of the time that f(x) is increasing on [0,1].