Yes, if there are two or more options and the score function depends only on the probability assigned to the correct outcome, then the only proper function is log. You can see that with the equation I gave
f0′ (x) = (k—x.f1′ (x))/(1-x)
for f0 = 0, it means
x.f1′(x) = -k
thus f1(x) = -k ln(x) + c (necessary condition)
Then you have to check that -k ln(x) + c indeed works for some k and c, that is left as an exercise for the reader ^^
Yes, if there are two or more options and the score function depends only on the probability assigned to the correct outcome, then the only proper function is log. You can see that with the equation I gave
f0′ (x) = (k—x.f1′ (x))/(1-x)
for f0 = 0, it means x.f1′(x) = -k thus f1(x) = -k ln(x) + c (necessary condition)
Then you have to check that -k ln(x) + c indeed works for some k and c, that is left as an exercise for the reader ^^