Absolute (i.e., not just relative) probabilities in chi-squared test for proportions Why do the absolute probabilities affect the p-value of a chi-square test so much? For example:  
> successes.1 = c(400, 500)
> successes.2 = c(40, 50)
> trials = c(1000, 1000)
> test.1 = prop.test(successes.1, trials)
> test.2 = prop.test(successes.2, trials)
> test.1

    2-sample test for equality of proportions with continuity correction

data:  successes.1 out of trials
X-squared = 19.8, df = 1, p-value = 8.598e-06
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.14438565226 -0.05561434774
sample estimates:
prop 1 prop 2 
   0.4    0.5 

> test.2

    2-sample test for equality of proportions with continuity correction

data:  successes.2 out of trials
X-squared = 0.9424, df = 1, p-value = 0.3317
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.029165387766  0.009165387766
sample estimates:
prop 1 prop 2 
  0.04   0.05 

 A: Your data are distributed as a binomial.  The variance of a binomial is a function of the mean and/or the probability of success (which is the same thing).  Specifically, the variance is:
$$
{\rm Var}(X) = Np(1-p)
$$
where $p$ is the (absolute) probability of success.  Because the probabilities differ between your two examples, the standard error of $\hat p$ differs.  And thus the statistical power / the $p$-value of the test differs.  
A: Consider the variance of a binomial proportion - $\text{Var}\hat p = p(1-p)/n$. This is because $\hat p = x/n$ where $x$ is the observed count, which under a binomial model has variance $np(1-p)$. Basic properties of variance mean that the variance of the sample proportion is then that variance divided by $n^2$.
This means the standard error of the proportion is $\sqrt{p(1-p)/n}$.
When $p$ is near $\frac{_1}{^2}$, this is about $0.5/\sqrt{n}$. With $n=1000$, that's about 0.0158
When $p$ is very small, its about $\sqrt{p}/\sqrt{n}$. So if $p$ is near 0.05, the standard error is about 0.007 - much smaller, a bit less than half the size.
But notice that the difference in proportions is one tenth as big (we've gone from 0.5 - 0.4 = 0.1 down to 0.01). As a fraction of the gap between proportions, the standard errors are relatively large.
Here's a plot with the proportions on the log scale (so the relative shift in proportion is the same size), with the endpoints of a 95% interval for each proportion drawn in:

In the first case, the two proportions are some distance (in terms of standard errors) apart* while in the second case the two proportions are not - the overlap is substantial.
* I am glossing over some issues - e.g. it's really the standard error of the difference in proportion that matters.
