Suppose that $X_1,X_2,\ldots,X_n$ is a random sample from a distribution with probability function:
$$p_X(x)=\begin{cases} 1/2 &x=-1,1 \\ 0 & \text{otherwise} \end{cases}$$
Now if we define the average
$\bar{X}= \frac{1}{n} \sum X_i$, I need to show that that for an odd $n$ the probability function for $\bar{X}$ is:
$$p_{\bar{X}} (x)= \frac{\binom{n}{\frac{n}{2} (x+1)}}{2^n}$$ for $x=\pm 1/n,\pm 3/n,\ldots,\pm 1$, $0$ otherwise.
Since $P[\sum X_i=k] =P [\bar{X}=k/n ]$, I thought it would be easier to first derive the probability function for the sum. Since they are still $2^n$ n-tuples and each one is equilikely, I need to count the ones whose sum is $k$ for:
$k=\pm 1,\pm 3,\ldots \pm n-2,\pm n$ (since even numbers are not a feasible combination)
The problem here though is that I do not immediately recognize how I could count all these tuples. Could you please help me with this? If it is easier to proceed another way, I am of course all ears.