A fun math puzzle whose solution will surprise you

First, pretend that the Mathematicians know there are an even number of white hats, and an even number of black hats. In that case it's easy for each one of them to guess what color hat they're wearing. So have the Mathematicians assume there are an even number of each type of hat. There's a 50% chance that the number of hats actually is even (non-trivial, but I'll leave out details for now), and therefore a 50% overall chance that the Mathematicians get every guess correct (and a 50% chance they get every guess wrong!)

This problem can be represented mathematically as a function f: X->Y, where:

X = {-1,1}^16 represents the set of hats the Mathematicians are actually wearing. Say -1 is a white hat and 1 is a black hat.

Y = {-1,0,1}^16 represents the guesses made by the Mathematicians, where -1 is a white hat, 1 is a black hat, and 0 is an "I don't know."

We can also write f(x_0,x_1,...,x_15) -> (y_0,y_1,...,y_15) . Then for the fixed set f(a_0,...,a_15) = (b_0,...,b_15), the function f is a correct guess if and only if

1. For all k, b_k = 0 or a_k .

2. For at least one k, b_k != 0.

Assume f(a_0,...,a_15) = (b_0,...,b_15) is a correct guess. There must be some b_k != 0 where b_k = a_k, so we can assume WLOG that b_0 = 1 = a_0. Now note that if f(1,a_1,...,a_15) = (b_0,b_1,...,b_15) and f(-1,a_1,...,a_15) = (c_0,c_1,...,c_15), then b_0 = c_0 . This is because mathematician 0 must make his guess based only on the colors of the other mathematicians' hats. Thus f(-1,a_1,...,a_15) is an incorrect guess.

Thus for every set (x_0,...,x_15) where f is a correct guess, we can come up with at least one set where f is an incorrect guess. Thus f cannot be correct more than 50% of the time.

...so, what am I missing?