Suppose $\sum_i c_i = 0$ and $Z$ is random draw from the collection of unit vectors, i.e. it has Euclidean norm $\lVert Z \rVert = 1$. Of course $E\left[\sum_i c_iZ_i^2 \right]=0$. What can we say about $P\left(\sum_i c_i Z_i^2 > 0\right)$? I don't think this probability is always one half. Can anybody help to find the probability?
$\it{Example}:$ Consider $c_1 = 1$, $c_2 = 1$ and $c_3 = -2$. Then, we are interested in $P\left(Z_1^2 + Z_2^2 - 2 Z_3^2 > 0\right)$. It is not difficult to see that $\left\{z:z_1^2 + z_2^2 - 2 z_3^2 = 0, z_1^2 + z_2^2 + z_3^2 =1\right\} = \left\{z: z_1^2 + z_2^2 = \frac{2}{3}, z_3 =±\frac{1}{3}\sqrt{3}\right\}$
such that the set $\left\{z_1^2 + z_2^2 - 2 z_3^2 = 0, \lVert z\rVert=1\right\}$ consists of two circles with centres $(0,0,±\frac{1}{3}\sqrt{3})$, radius $\sqrt{\frac{2}{3}}$, and normal vector pointing in the $z_3$ direction. For this example, the question is then to find the surface of the unit sphere enclosed within the two circles, i.e. $P\left(Z_1^2 + Z_2^2 - 2 Z_3^2 > 0\right)=P\left(-\frac{1}{3}\sqrt{3}< Z_3<\frac{1}{3}\sqrt{3}\right)$.