Now, I'm not amazing at math, but if we stipulate that bias is causing black drivers to be pulled over for less, meaning we're catching a less reasonably suspicious cross section of the black population by profiling them; if we stipulate that that less suspicious cross-section is being searched at double the rate of more suspicious white people, wouldn't that mean the likelihood of finding something should have a bigger discrepancy than 26%? Doesn't this actually show, within this sample size, the opposite of what is being implied(other than the raw disparity in how suspicion is treated)?
You mean like some kind of Bayesian caveat?
So if we have...
P(B) -> probability of being black in Ferguson
P(D) -> probability of any person having contraband
P(D | B) -> probability of contraband given that a person is black
P(B | D) -> probability of being black given that person has contraband
(and a similar scheme for white people with P(W))
So what we want to know is which of P(B | D) and P(W | D) is larger.
P(B | D) = P(D | B) * P(B) / P(D)
P(W | D) = P(D | W) * P(W) / P(D)
P(B) = .68
P(W) = .32
P(D | B) = .20
P(D | W) = .33
(Correct me if I'm wrong, I'm not totally sure about the numbers)
If we divide the first equation by the second, we can eliminate the need for knowing P(D), although I'm sure the exact number is around somewhere.
P(B | D) / P(W | D) = [P(D | B) * P(B)] / P(D | W) * P(W)]
So P(B | D) / P(W | D) = (.20 * .68) / (.33 * .32) = .136 / .105
This means that a person possessing contraband in Ferguson is 29% more likely to be black than white. Therefore, if black residents are being searched at a rate significantly higher than 29% more than white residents, the police force is most likely racially biased. We know that black residents are stopped 85% of the time, meaning white residents are stopped 15% of the time or less.
P(S | B) = .85
P(S | W) = .15
.85 / .15 = 5.67
.68 / .32 = 2.32
So for a randomly selected stop as it is, the suspect is 5.67 times as likely to be black, even though a randomly selected Ferguson resident is only 2.32 times as likely to be black. If we account for the likelihood of black residents having contraband, an appropriate ratio would be 2.99 black residents per white resident. Therefore, a black resident is searched 5.67/2.99*.32 (?) = 28% more often than they should be if the police weren't racially biased.