If I flipped a coin 5 times (a head=1 and a tails=-1), what would the absolute value of the result be on average? I.e., if I flipped five non-absolute values, then averaged them, then got the absolute value as the result, and I did that infinite times, what would be the average value of the result?
What mathematical method would I use to calculate this? I've only taken up to precalculus and I anticipate that calculating this would involve something I haven't learned yet but I don't know what it is.
Edit: I am asking this to understand how I can quantify how much lower sample size rather than an actual shift in the average for which there is a tendency can be expected to cause a deviation.

The distribution is binomial distributed and with that you can compute this manually.
If $X$ is the average of five coin flips (which I assume are fair) then
$$\begin{array}{}
P(X = -1) &=& \frac{1}{2^5}\\
P(X = -0.6) &=& \frac{5}{2^5}\\
P(X = -0.2) &=& \frac{10}{2^5}\\
P(X = 0.2)& =& \frac{10}{2^5}\\
P(X = 0.6) &=& \frac{5}{2^5}\\
P(X = 1) &=& \frac{1}{2^5}
\end{array}$$
And the expectation value for any function of the variable is
$$E[f(X)] = \sum_x P(X=x) \cdot f(x)$$
with $f(X) = |X|$ you can get your answer which should be close to the approximation below.

In the limit, you can estimate this with the half normal distribution, which has an expectation value of $\sigma \sqrt{\frac{2}{\pi}}$ and the distribution of the mean of $n$ coin flips is approximated by taking $\sigma = 1/ \sqrt{ n}$ giving
$$E[|X_n|] \approx \sqrt{\frac{2}{n\pi}}$$
which is equal to about $0.3568248$ in the case of $n = 5$, a bit less than 5% away from the exact answer.
