Subtraction of Independent random variables

Question

let X be a random variable for a six sided die 

$X(\Omega)=\{1,2,3,4,5,6\}$

$\sigma^2_X = 2.916$

let Y be a random variable of a four sided die

$Y(\Omega) = \{1,2,3,4\}$

$\sigma^2_Y = 1.25$

let $Z = X - Y$

Now $\sigma^2_Z = \sigma^2_X  + \sigma^2_Y$

$\sigma^2_Z = 4.166$

but when i try to calculate without applying the identity.

$Z(\Omega) =\{-3,-2,-1,0,1,2,3,4,5 \}$

$P(-3) = P(5) = \frac{1}{24}$

$P(-2) = P(4) = \frac{2}{24}$

$P(-1) = P(3) = \frac{3}{24}$

$P(2) = P(1) = P(0) = \frac{4}{24}$

$E[Z] = 1$

$E[Z^2] = 4.5$

$\sigma^2_Z = 4.5 - 1 = 3.5$

Where am i going wrong?

Answer 1

$\begin{aligned}\mathbb{E}Z^{2} & =\frac{1}{24}\left(\left(-3\right)^{2}+5^{2}\right)+\frac{2}{24}\left(\left(-2\right)^{2}+4^{2}\right)+\frac{3}{24}\left(\left(-1\right)^{2}+3^{2}\right)+\frac{4}{24}\left(0^{2}+1^{2}+2^{2}\right)\\ & =\frac{34}{24}+\frac{40}{24}+\frac{30}{24}+\frac{20}{24}\\ & =\frac{124}{24}=\frac{31}{6} \end{aligned} $

So that $\mathbb EZ^2-(\mathbb EZ)^2=\frac{31}6-1=\frac{25}6\approx4.166$.

Answer 2

Alternatively: $$\begin{align}Var[Z]=E[(Z-E[Z])^2] = &[(-3-1)^2+(5-1)^2]\cdot \frac1{24}+\\ &[(-2-1)^2+(4-1)^2]\cdot \frac2{24}+\\ &[(-1-1)^2+(3-1)^2]\cdot \frac3{24}+\\ &[(0-1)^2+(1-1)^2+(2-1)^2]\cdot \frac4{24}=\\ =&\frac{32+36+24+8}{24}=\\ =&\frac{100}{24}=\\ =&4.166...\end{align}$$