Calculating Variance(X + Y + 1) Not Duplicate. I know a question with similar data has been used here, but I am looking for something else. 
Two tire-quality experts examine stacks of tires and assign a quality rating to each tire on a 3-point scale. Let X denote the rating given by expert A and Y denote the rating given by B.
The following table gives the joint probability distribution for X and Y :

Calculate the Variance(X + Y + 1)
What I am confused about: is what to do with the 1 inside the variance. I know it is a random variable itself, but not sure how to apply the expected value formula on it: E(X) = sum(x * P(X)) for all x
I have found the formula for the Variance(X + Y), but that "1" is throwing me off.
Thanks
 A: $\newcommand{\Var}{\operatorname{Var}}$Recall the property $\Var(Z+c)=\Var(Z)$ for any constant $c$ and random variable $Z$ (i.e. adding a constant does not affect the variance). Hence $$\Var(X+Y+1)= \Var(X+Y).$$
A: If $\mathbb EZ^2<\infty$ then: $$\mathsf{Var}Z=\mathbb E(Z-\mathbb EZ)^2\tag1$$
If $c$ denotes a constant then $\mathbb E(Z+c)=\mathbb EZ+c$ and consequently $(Z+c)-\mathbb E(Z+c)=Z-\mathbb EZ$.
Application of $(1)$ results in:$$\mathsf{Var}(Z+1)=\mathbb E(Z-\mathbb EZ)^2=\mathsf{Var}Z$$
A: I agree with what the other two folks had said. I would only add two things.
First, for all practical purposes, 1 is not a random variable unless it has a probability distribution, as you assigned to the random pair $(X,Y).$
Second, the general formula for any real constants $a$ and $b$, and any random = variable $Z$, is 
$$
\text{var}(aZ + b) = a^2 \text{var}(Z).
$$
This, applied to the case $a=1$, $b=1$, and $Z = X+ Y$ (a sum of random scalars is a random scalar), implies $\text{var}(X+Y+1) = \text{var}(X+Y)$.
