Interesting variant of binomial distribution. Suppose i had $n$ Bernoulli trials with $X_{i}=1$ if the $i$th trial is a success and $X_{i}=-1$ if it is a failure each with probability $\frac{1}{2}$. 
Then the difference between the number of successes and failures can be represented by the random variable $Y=|\sum_{i=1}^{n} X_{i}|$ what is the best way to calculate $\mathbb{E}(Y)$? 
One thing i was thinking is $|\# \mathrm{Successes}-\# \mathrm{Failures}|=n-2\#\mathrm{successes}$ and so $\mathbb{E}(Y)=\mathbb{E}(2|\frac{n}{2}-\#\mathrm{Successes}|)$ which we could relate to the variance of $X=\sum_{i=1}^{n} X_{i}'$? where $X_{i}'$ is an indicator random variable equal to $1$ if $X_{i}$ is a success and $0$ otherwise. Am i along the right lines? 

Let $S_{2n}$ be the number of successes in $2n$ trials. Then there are $2n - S_{2n}$ failures, with a difference of $|S_{2n} - (2n - S_{2n})| = |2S_{2n} - 2n|$. Now let $0 <k \leq 2n$ be an even number. Then
\begin{align*}
\mathbb{P}(|2S_{2n} - 2n| = k) &= \mathbb{P}(2S_{2n} - 2n = k) + \mathbb{P}(2S_{2n} - 2n = -k) \\
&= \mathbb{P}\left(S_{2n} = n + \frac{k}{2}\right) + \mathbb{P}\left(S_{2n} = n - \frac{k}{2}\right) \\
&= \frac{1}{2^{2n}} \left(\binom{2n}{n + \frac{k}{2}} + \binom{2n}{n - \frac{k}{2}}\right) \\
&= \frac{1}{2^{2n-1}}\binom{2n}{n + \frac{k}{2}}.
\end{align*}
If $k$ is odd, then it is easy to check that $\mathbb{P}(|2S_{2n} - 2n| = k) = 0$. We have 
\begin{align*}
\mathbb{E}(Y_{2n}) &= \sum_{k=2}^{2n}k\,\mathbb{P}(|2S_{2n} - 2n| = k)\\
&= \frac{1}{2^{2n-2}}\sum_{k=1}^n k \, \binom{2n}{n + k} \\
&= \frac{n+1}{2^{2n-1}}\binom{2n}{n+1} \\
&= \frac{n+1}{2^{2n-1}} \cdot \frac{(2n)!}{(n+1)!(n-1)!} \\
&= \frac{n+1}{2^{2n-1}} \cdot \frac{(2n)!}{n!n!} \cdot \frac{n}{n+1} \\
&= \boxed{\frac{2n}{2^{2n}} \binom{2n}{n}}.
\end{align*}
Similarly, let $S_{2n+1}$ be the number of successes in $2n+1$ trials, for a difference of $|S_{2n+1} - (2n+1 - S_{2n+1})| = |2S_{2n+1} - 2n - 1|$. Let $1 \leq k \leq 2n+1$ be an odd number in the form $k = 2k' + 1$. Then
\begin{align*}
\mathbb{P}(|2S_{2n+1} - 2n - 1| = k) &= \mathbb{P}(2S_{2n+1} - 2n - 1 = k) + \mathbb{P}(2S_{2n+1} - 2n - 1 = -k) \\
&= \mathbb{P}\left(S_{2n+1} = n + \frac{k+1}{2}\right) + \mathbb{P}\left(S_{2n+1} = n + \frac{1-k}{2}\right) \\
&= \frac{1}{2^{2n+1}} \left(\binom{2n+1}{n + \frac{k+1}{2}} + \binom{2n+1}{n + \frac{1-k}{2}}\right) \\
&= \frac{1}{2^{2n}}\binom{2n+1}{n + \frac{k+1}{2}}.
\end{align*}
Writing $k = 2k' + 1$, the probability becomes
$$\mathbb{P}(|2S_{2n} - 2n - 1| = k) = \frac{1}{2^{2n}}\binom{2n+1}{n + k' + 1}.$$
Even differences are not possible, so we have
\begin{align*}
E(Y_{2n+1}) &= \sum_{k=1}^{2n+1} k \, \mathbb{P} (|2S_{2n+1} - 2n - 1| = k) \\
&= \frac{1}{2^{2n}}\sum_{k'=0}^n (2k'+1) \,\binom{2n+1}{n + k' + 1} \\
&= \frac{n+1}{2^{2n}}\binom{2n+1}{n+1} \\
&= \frac{n+1}{2^{2n}}\cdot \frac{2n+1}{n+1} \cdot \frac{2n!}{n!n!} \\
&= \boxed{\frac{2n+1}{2^{2n}} \binom{2n}{n}}.
\end{align*}
Stirling's approximation states that
$$n! \sim \sqrt{2\pi n}n^n e^{-n},$$
so
$$\binom{2n}{n} \sim \frac{2^{2n}}{\sqrt{\pi n}}.$$
This gives us
$$E(Y_{2n}) \sim \frac{2n}{\sqrt{\pi n}} = \frac{2}{\sqrt{\pi}} \sqrt{n}.$$
Thus the expected difference is $\boxed{\mathcal{O}(\sqrt{n})}.$
EDIT 1: My previous answer didn't have the right conditional expectation, so I just computed general probabilities and used the definition of expectation.
EDIT 2: Added asymptotics.
