But if I have something like this
p(x,y,z)= (1/3,1/3,1/3)
then I have entropy 1.585. But now I could have p(y|x) = 1/2 or p(y|x) = 1 as long as p(y|z)=1-p(y|x) the overall probability p(x,y,z) stays the same. So I have the same entropy. But in the case of p(y|x)=1 I can only use two...