Expected Value of R squared Let $n$ be a fixed positive integer. Generate $n$ numbers $x_1, x_2, ..., x_n$ from the set $[0,1]$, with the probability distribution being the uniform one and the $x_i$ all being independent of each other. Now repeat this process to generate $y_1, ..., y_n$. If we let $X$ be a random variable which takes on $x_1, ..., x_n$ with probability $\frac{1}{n}$ each, and let $Y$ be a random variable which takes on $y_i$ whenever $X$ takes on a value of $x_i$. We can then compute the square of the correlation $R^2$ between $X$ and $Y$. What is the expected value of this $R^2$?
Another less rigorous phrasing of the problem is this: suppose we throw $n$ points at random on a graph spanning $[0,1] \times [0,1]$. What is the expected value of the $R^2$ of the line of best fit?
For instance, for $n=2$ the expected value is $1$ due to the $R^2$ value always being $1$. For $n=3$ one can numerically compute the expected value to be $\frac{1}{2}$. In general, it seems that the answer is $\frac{1}{n-1}$. I don't really have any idea how to do this problem in general; and even specific cases look nontrivial. Does anyone have any ideas? This looks like what should be a well-known result, but my searching didn't pick up on anything which looked useful.
This has applications in that when one is working with variables which are not expected to be very highly correlated, it is often difficult to tell when an $R^2$ value is significant. This result gives an idea of how big the $R^2$ needs to be for one to deduce there is some nontrivial correlation between two variables.
 A: This problem seems simple...but its not. For example, see here for a rather complex analysis for the prima facie simple case of ratios of normal rv and ratios of sums of uniforms.
In general, if your pairs are not from a bivariate gaussian, there is no nice formula for $E[R^2]$. 
Note:
$$R_n=\frac{n\sum x_iy_i-\sum x_i\sum y_i}{n^2s_Xs_Y}$$ 
This mess will have some distribution $f_{R_n}(r)$ that will be very sensitive to $n$.
I think your best bet is to simulate this (Monte Carlo) for $n\in [2....N]$ using a large number of trials (you can check convergence by running each simulation twice, with randomly chosen seeds and comparing these results to each other and to results from $n-1$).
Once you have this data, you can fit a curve to the it or some transformation thereof. Your general equation looks reasonable in terms of how the curve will look, since:
$$E[R^2_n] \xrightarrow{p} 0$$ for correlations between independent variables
Possible Solution
Since your variables are independent, I realized that we are really looking for the variance of the sample correlation (i.e., the square of the expected value of the standard error of the correlation coefficient (see p.6):
$$se_{R_n}=\sqrt{\frac{1-R^2}{n-2}}$$. However, you already know the true value of $R^2$, so you can increase the df in the denominator to get:
But: $R^2=0$ for independent variables, so it reduces to:
$$(se_{R_n})^2=\sigma^2_{R_n}=E[R^2_n]=\frac{1}{n-1}$$
There you have it...it matches your empirical results. As per Wolfies, I should note that this is an asymptotic result, but sums of uniform RVs generally exhibit good convergence properties ala CLT, so this may explain the good fit. 
For further reading, see @soakley's nice reference. I was able to pull the relevant page from JSTOR:

or, if you're really motivated, you can get this recent article (2005) about your exact problem. 
A: According to Kendall's Advanced Theory of Statistics (Exercise 16.17 in the 5th edition of Volume 1), Pitman (1937) showed the sample correlation coefficient $r$ has zero mean and variance or second moment of $$\sigma^2_{r}=E[r^2] = {1 \over {n-1}}$$ for any sample of size $n$ where $x$ and $y$ are independent continuous variates. 
Checking the reference, we find he shows $r^2$ has an approximate $\mathrm{Beta} \left( {1 \over 2}, {{n-2} \over {2}}\right)$ distribution. 
Reference: Pitman, E.J.G.. Significance tests which may be applied to samples from any population., v. 4, No. 1, II. The correlation coefficient test., v. 4, No. 2, $\it{Supp. J.R. Statist. Soc.},$ 1937.
A: I'm just copying the section from 
http://en.wikipedia.org/wiki/Coefficient_of_determination
I think it is what you are looking for.
A data set has n values marked $y_1...y_n$ (collectively known as $y_i$), each associated with a predicted (or modeled) value $f_1...f_n$ (known as $f_i$, or sometimes $ŷ_i$).
If $\bar{y}$ is the mean of the observed data:
$\bar{y}=\frac{1}{n}\sum_{i=1}^n y_i $
then the variability of the data set can be measured using three sums of squares formulas:
The total sum of squares (proportional to the variance of the data):
$SS_\text{tot}=\sum_i (y_i-\bar{y})^2,$
The regression sum of squares, also called the explained sum of squares:
$SS_\text{reg}=\sum_i (f_i -\bar{y})^2,$
The sum of squares of residuals, also called the residual sum of squares:
$SS_\text{res}=\sum_i (y_i - f_i)^2\,$
The notations $SS_\text{R}$ and $SS_\text{E}$ should be avoided, since in some texts their meaning is reversed to Residual sum of squares and Explained sum of squares, respectively.
The most general definition of the coefficient of determination is
$R^2 \equiv 1 - {SS_{\rm res}\over SS_{\rm tot}}.$
Note: I can't tell from the preview if it looks ok. I'll keep trying to make it look ok, or just follow the link.
If nothing else, look at the inset figure to the right. 
Here is the link to the graphic, with squares of data versus (difference of squared) $\bar{y}$ on the left compared to squares of data versus (difference of squared) fit line on right.
http://en.wikipedia.org/wiki/Coefficient_of_determination#mediaviewer/File:Coefficient_of_Determination.svg
