I was recently given the following statistics: On a particular highway, 18% of drivers are black, 63% of drivers searched by the police are black. So, a black driver is 7.7 times more likely to be searched.

I'm having some difficulty deriving the figure 7.7 myself. Am I correct to interpret the 63% as the conditional probability of being black given that you have been searched by the police as being 0.63? Thus, 0.37 as the probability of not being black given that you have been searched by the police.

Furthermore I'm not sure how we would arrive at the probability of being searched given that you are black (in order to make the comparison and arrive at 7.7). Or is this even the correct probabilistic interpretation of the sentence?

(if anyone is interested, this data is from an economics paper by Knowles, Persico and Todd, 2001)

  • 2
    $\begingroup$ It would be helpful to reproduce a full quotation from the text, since the way that the claim is phrased can make a difference. $\endgroup$ – Sycorax Dec 7 '18 at 15:46
  • $\begingroup$ I honestly don't know what "7.7 times more likely" mean. e.g. what does it mean when people say "'A' happens 7.7 times more likely than 'B'" and you find out that P(B) = 0.9? $\endgroup$ – gota Dec 7 '18 at 15:57
  • 1
    $\begingroup$ @gota The odds ratio is the only probability metric that works to describe any $2 \times 2$ table association, but it's so frequently reported as a risk ratio that, when it's possible to do so, a risk ratio is reported instead. However, it should be noted that the probabilities quoted here aren't risks. $\endgroup$ – AdamO Dec 7 '18 at 16:07

82% are "not black" and comprise the other 37% who are searched. So, in an abuse of statistical notation,

P(Search | black) / P(Search | not black) = = P(Black | search)/P(Black) / (P(not black | search) / P(not black) ) = (0.63 / 0.18 ) / (0.37 / 0.82) = 3.5 / 0.45 = 7.7

  • $\begingroup$ That's very odd... In your calculation P(Search | black) = 3.5... whaaat?! $\endgroup$ – gota Dec 7 '18 at 16:00
  • $\begingroup$ @gota you're right. It's a reverse conditional. The point is that P(search) cancels out of the ratio when applying Bayes rule. I made this semi-explicit. $\endgroup$ – AdamO Dec 7 '18 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.