## forecast and fine

hi,
am i solving it right?
a coin is flipped twice with probability of 2/3 for heads.
you are supposed to forecast the number of heads with fine equals the absolute value of the mistake margin, and you wanna to minimize it.
so i need to minimize the expectation of fine that is the sum of |myguess-result|*p(result).
then i find the myguess which minimizes it. in this case i think it is one.
my question is, is there some elegant way to find the answer without assigning value. that is to find the derivative of this and find the min point.
problem is, it has abs in it which is not derivable. is there another way?
i can't manage to prove this expression to be monotonic either.

thanks