Contradicting claims of probability distributions I am looking at the following distribution from here 

They proceed to make the following claims about this distribution 

1) Within each risk category, the proportion of defendants who reoffend is approximately the same regardless of race
  2) The overall recidivism rate for black defendants is higher than for white defendants (52 percent vs. 39 percent).

But are these two claims not contradicting? If the proportion of defendants who reoffend is approximately the same then wouldn't the overall recidivism rate be the same for both races?
 A: It doesn't say that the proportion of defendants who reoffend is the same for both races; that would indeed mean that the proportion of recidivism is the same.  It says among those classified as high-to-medium risk the recidivism rate is the same for both races, and also among those classified as low-risk, the proportion is the same for both races.  
However, the overall recidivism rate is higher for black defendants, and consequently, a higher proportion are classified as high-to-medium risk.  (By "consequently" I just mean that the statements in the previous paragraph couldn't be true otherwise.)  But this means that among blacks who do not reoffend, a higher proportion will have been classified as medium-to-high risk.  
Let's try an example, loosely based on the graphic.  Suppose that $\frac13$ of defendants classified as low risk reoffend, and $\frac23$ of defendants classified as medium-to-high risk reoffend, regardless of race.  Suppose that among $3000$ black defendants, $1800(60\%)$ are classified as medium-to-high risk, and the remaining $1200$ are classified as low risk.  $600$ of the medium-to-high people don't reoffend and $800$ of the low-risk people don't reoffend, so the proportion of non-recidivists classified as high-to-medium risk is $$\frac{600}{800+600}
\approx43\%$$
Also suppose that among $1800$ white defendants, only $600(\frac13)$ are classified as medium-to-high risk, and the remaining $1200$ are classified as low risk.  Then $200$ of the high-to-medium risk people don't reoffend, and $800$ of the low risk people don't reoffend.  The proportion of non-recidivists classified as medium-to-high risk is $$\frac{200}{200+800}=20\%$$
As the article points out, this is inescapable.  If you take any two groups A and B, where A has a higher rate of recidivism than B, if the classification algorithm produces the same rates of recidivism for low risk defendants in both groups and for high-to-medium offenders in both groups.
It's easy to prove this, under reasonable assumptions about the classifying scheme.  Suppose we have two groups, one with recidivism rate $a$ and the other with rate $b$, where $a>b$.  Suppose that the rate of recidivism among defendants classified as medium-to-high risk is $0<h<1$ and that the rate among those classified as low risk is $0<l<1$.  Also suppose $l<a,b<h.$  If there are $n$ defendants in the first group, and $k$ of them are classified medium-to-high risk, then in order for all these assumptions to be satisfied we must have$$
hk+l(n-k)=an\implies \frac{k}{n}=\frac{a-l}{h-l}$$ where $\frac{k}{n}$ is the fraction classified as medium-to-high risk.  Then the fraction classified as low risk must be $$1-\frac{a-l}{h-l}=\frac{h-a}{h-l}$$
The fraction of non-recidivists is $$\frac{a-l}{h-l}(1-h)+\frac{h-a}{h-l}(1-l)$$ and the proportion of those classified as medium-to-high risk is $$n_a=\frac{(a-l)(1-h)}{(a-l)(1-h)+(h-a)(1-l)}=\frac1{1+\frac{(h-a)(1-l)}{(a-l)(1-h)}}$$  Similarly for the second group we get  $$n_b=\frac1{1+\frac{(h-b)(1-l)}{(b-l)(1-h)}}$$  Now a simple calculation shows that $$\frac{h-a}{a-l}-\frac{h-b}{b-l}<0$$ so that $n_a>n_b$, as claimed.         
