What is the variance of a binomial distribution with -1 and 1? I am struggling to see how to solve the following problem:

I have n i.i.d bernoulli trials. The result can be -1 or 1. I can figure out the expected value of this = n(p-(1-p) but how do I know what the variance is if p is known?

I know that:
$$
{\rm Var}(X) = E[X^2] - E[X]^2
$$
I don't know where to go from here.
 A: When you work with variances, know these facts (in addition to the definition of variance):


*

*The variance of a sum of independent (or just uncorrelated) variables is the sum of their variances.

*The expectation of a sum of variables (independent or not) is the sum of their expectations.

*The expectation (of any discrete variable) is the sum of each possible value multiplied by its probability.
As an example, let's apply them to your case.  You may model the sum of your Bernoulli trials with a variable $X$ expressed as the sum of $n$ independent variables $X_i$.  Each $X_i$ takes on the value $1$ with probability $p$ and the value $-1$ with the probability $1-p$.  
Fact $(3)$ asserts $$\mathbb{E}(X_i) = p(1) + (1-p)(-1) = 2p-1$$ and $$\mathbb{E}(X_i^2) = p(1)^2 + (1-p)(-1)^2 = p + (1-p) = 1.$$
Fact $(2)$ asserts $$\eqalign{\mathbb{E}(X) &= \mathbb{E}(X_1+\cdots+X_n) = \mathbb{E}(X_1) + \cdots + \mathbb{E}(X_n) = n\mathbb{E}(X_1) \\&=n(2p-1).}$$
The definition of variance now tells you$$\operatorname{Var}(X_i) = \mathbb{E}(X_i^2) - \mathbb{E}(X_i)^2 = 1 - (2p-1)^2 = 4p(1-p).$$
Consequently, fact $(1)$ yields $$\eqalign{\operatorname{Var}(X) &= \operatorname{Var}(X_1+\cdots+X_n) = \operatorname{Var}(X_1) + \cdots + \operatorname{Var}(X_n)=n \operatorname{Var}(X_1) \\&= n(4p(1-p)).}$$

The appearance of the factor of $4$ might seem somewhat mysterious.  There's another extremely useful fact you might consider using to shortcut these considerations and identify the origin of that factor:
(4) The variance of a shifted, rescaled variable is the square of the scale factor times the variance. In mathematical symbols, $$\operatorname{Var}(\sigma X + \mu) = \sigma^2 \operatorname{Var}(X)$$ no matter what values the numbers $\sigma$ and $\mu$ might have.
This applies by noting that your $X_i$ can all be expressed as $2Y_i-1$ where $Y_i$ is a true Bernoulli (that is, $0-1$) variable.  It's simple to show this: since $2\times 1-1=1$ and $2\times 0 - 1=-1$, the values of $1$ and $0$ taken on by $Y_i$ become $1$ and $-1$, respectively, for $X_i$.  The probabilities are unchanged.
You might already know that $Y=Y_1+\cdots + Y_n$, the sum of $n$ independent Bernoulli variables with common probability $p$, is called a Binomial variable.  It has a Binomial distribution.  You can remember or look up its variance, which is $np(1-p)$.  Since $$X = X_1+\cdots+X_n = (2Y_1-1) + \cdots + (2Y_n-1) = 2(Y_1 + \cdots + Y_n) - n=2Y-n,$$ you can take $\sigma=2$ in applying fact $(4)$, which immediately tells you $$\operatorname{Var}(X) = 2^2 \operatorname{Var}(Y) = 4np(1-p).$$
A: if you take $x=2b-1$ where $b$ is bernoulli, then then $x$ is either 1 or -1, and mean and variance follow from Bernoulli $\mu_b=p$ and $\sigma^2_b=p(1-p)$, the extension to 'Binomial' is then just a sum (assuming independence), 
The mean of $x$ is then $\mu_x=2\mu_b-1=2p-1$, variance $\sigma_x^2=4\sigma_b^2=4p(1-p)$, 
The sum of $n$ independent such $x$'s has mean $n\cdot(2p-1)$ (your result) and the variance is $4\cdot n \cdot p(1-p)$
EDIT after your question in the comment
It is as @Pegah says:  $b$ is the ''usual'' Bernoulli with a success probability $p$ and outcomes 0 and 1.  It's expected value is $1 \times p + 0 \times (1-p)=p$ and the variance $p(1-p)$ (also from the definition. 
So $b$ is the known Bernoulli with outcome 0 and 1 and $x$ is just a linear function of it. The linear function is such that $x$ has outcomes -1 and 1, and the mean and variance can be obtained from the mean and variance of $b$. 
