Got nerd-sniped hard by this. The basic question is simple; given a random person in the population, what are the values of x and y such that the chance that they will give a wrongful vote are 2/3? Obviously, we can see that y = ((2/3) - x)/(1 - x). A simple enough graph shows that the racist population would have to be surprisingly high to fulfill that condition; even a 20% die-hard "convict on race" population would still require the other 80% of people to similarly erroneously convict almost 60% of the time.

It becomes more interesting when we consider juries. Juries, as far as I know (not being American), require unanimous agreement to convict. A jury consists of 6-12 people. To find out if a jury will convict, we have to find the probability of every single person in that jury wrongfully convicting. The formula now is y = ((2/3)^(1/{6|12}) - x)/(1 - x). At this point, either x or y have to be very high for 2/3 of wrongful convictions to be black. Even when x = 66%, y still has to be above 80% for the 6-member jury to convict wrongfully that many times. This is, of course, assuming this is a perfectly mathematical scenario and we aren't dealing with real humans who will vote to convict because everyone else is doing so.

The error, then, is probably more due to humans being likely to fold their convictions under peer pressure than merely a high percentage of them being rabidly racist. If people adamantly voted their conscience in juries, and refused to be swayed by the deliberation of their fellow jurors, we would have many more hung juries and many fewer convictions.