# Statistics & profiling problem **URL:** \ **Category:** Factual Questions\ **Created:** [January 1, 2004, 4:12pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128 "2004-01-01T16:12:29Z")\ **Posts on this page:** 11\ **Page:** 4
**Author:** ![muttrox](https://avatars.discourse-cdn.com/v4/letter/m/a8b319/32.png) [@muttrox](https://boards.straightdope.com/u/muttrox)\ **Post date:** [January 8, 2004, 12:32pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/61 "2004-01-08T12:32:10Z")
I’m glad to see that after skipping most of this thread, I got the same answer as Omph and Lance. Guess that statistics degree stuck. Chances the black man is guilty = .25 (200/800) Chances the white man is guilty = .052 (450/8,550) .25/.052 = ~4.8, meaning the black man is 4.8 times more likely than the white man to be the guilty one. That works out to ~83% of the time, the black man is the guilty one. This is exactly the same reasoning as Lance and Omph used in a different form (and IMHO, much easier to understand). Note that all 4 numbers are used, so as we intuitively expect, changing any of the parameters will change the answer. This avoids the whole question of bad/bads… and it seems to me the OP was essentially assuming that one \*\* and exactly one \*\* of the suspects was guilty. ---
**Author:** ![Jurph](https://avatars.discourse-cdn.com/v4/letter/j/50afbb/32.png) [@Jurph](https://boards.straightdope.com/u/Jurph)\ **Post date:** [January 8, 2004, 2:59pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/62 "2004-01-08T14:59:30Z")
If I’m the cop? I chase they guy I can’t file a good description of later. Black guy with hair cut in a wild 1992 slanted box, with a bleached stripe in it, wearing bright blue Nike sneakers and an Iverson jersey? Let him run. I’m going after the 5’10" white guy with dirty blonde hair wearing jeans, beat-up sneakers, and a blank gray sweatshirt. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [January 8, 2004, 3:33pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/63 "2004-01-08T15:33:03Z")
Yes, it is assumed only one of them did it but, the other guy could be a bad guy by chance even if not involved. I am having great trouble accepting any of the conslusions presented here because they seem to lead to apparent contradictions. ---
**Author:** ![muttrox](https://avatars.discourse-cdn.com/v4/letter/m/a8b319/32.png) [@muttrox](https://boards.straightdope.com/u/muttrox)\ **Post date:** [January 8, 2004, 4:31pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/64 "2004-01-08T16:31:07Z")
Sailor, I don’t see the contradiction (I’ve gone back about 20 posts). What is your exact objection? The only thing I see you state is that probabilities should add up to one, which is satisfied. ---
**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\ **Post date:** [January 8, 2004, 7:15pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/65 "2004-01-08T19:15:04Z")
> [@](#): > > _Originally posted by sailor_ > **Yes, it is assumed only one of them did it but, the other guy could be a bad guy by chance even if not involved.** OK, this is a different assumption than I was using. I have been assuming that the officer knows that exactly one criminal (not zero, not two) is present at the scene. In your case, where **at least one criminal** is present, we can eliminate only the upper-left corner of the table ```auto Black innocent Black criminal White innocent 6840000 1710000 White criminal 360000 90000 ``` This gives us (this is just reiterating **Lance Turbo** ) a total of     1710000+360000+90000 = 2160000 possible pairings, in three categories:     1710000/2160000 = 79.2% have a white innocent and black criminal present;       360000/2160000 = 16.7% have a black innocent and white criminal present;         90000/2160000 = 4.2% have both a white criminal and black criminal present. For the final 4.2% of cases, we don’t have enough information to make an informed guess as to whom to chase. Like **muttrox** , I don’t know exactly what your objection is and so I’m not sure how to address it. ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [January 8, 2004, 8:48pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/66 "2004-01-08T20:48:09Z")
Now we’re getting somewhere. > [@](#): > > _Originally posted by sailor_ > **Yes, it is assumed only one of them did it but, the other guy could be a bad guy by chance even if not involved.** This is not an assumption that has been made by anyone else in this thread, but now that you have asserted it, we can get to the definitive answer. 1710000 cases bad black, good white 360000 cases good black, bad white 90000 cases bad black, bad white 2160000 cases in all. Period. Not even one more. Knowing that exactly one guy commited the crime allows us to divide the 90000 bad/bads further. 45000 guilty black guy, innocent bad white guy 45000 guilty white guy, innocent bad black guy Why can we divide this group exactly in half. Because… > [@](#): > > _Originally posted by sailor_ > **The bad guys are about equally bad…** So now we can compute the odds (1710000 + 45000)/2160000 = 13/16 = 81.25% black guy did it. (360000 + 45000)/2160000 = 3/16 = 18.75% white guy did it. These add up to unity, so it should be _obvious_ that it is correct. **sailor** , your 80% 20% results require the use of 2250000 for your denominator. This number is 1710000 + 360000 + 90000 + **90000**. That second 90000 is just wrong. No other way to look at it. A total of 1710000 + 90000 = 1800000 cases have a bad black man. A total of 360000 + 90000 = 450000 cases have a bad white man. Are two statements that have been agreed upon. However not every case with a bad black man has a guilty black man, and not every case with a bad white man has a guilty white man. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [January 8, 2004, 10:47pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/67 "2004-01-08T22:47:12Z")
Ok, let me ask you this. ```auto Problem #1 Problem #2 Problem #3 Cases where White is bad 2*10^8 20 p Cases where black is bad 8*10^8 80 q Cases where it could be either 100 10*10^8 r with equal probab. I say answer is Probabilty should be close to 20/80 50/50 ``` I say the answer in each case is P1 = (p+r) / (p+q+2r) P2 = (q+r) / (p+q+2r) which makes sense to me because those r cases have both bad buys and are counted on both sides. Forget about specific numbers. What are your answers for P1 and for P2 as a function of p, q and r in problem #3 above? ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [January 8, 2004, 11:09pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/68 "2004-01-08T23:09:02Z")
P1 = (p + r/2)/(p + q + r) P2 = (q + r/2)/(p + q + r) The total number of cases is clearly p + q + r. Why on earth would the total number of cases be p + q + 2r? ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [January 8, 2004, 11:31pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/69 "2004-01-08T23:31:29Z")
You are probably right. Not certainly right but probably right. 🙂 ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [January 8, 2004, 11:38pm UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/70 "2004-01-08T23:38:30Z")
Probably right is good enough for me on a probability problem. 😉 ---
**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\ **Post date:** [January 9, 2004, 12:03am UTC](https://boards.straightdope.com/t/statistics-profiling-problem/222128/71 "2004-01-09T00:03:55Z")
**sailor** , consider the case where p=r and q=0 (i.e., there are an equal number of cases where white is definitely bad and where either white or black is bad with equal probability; black is never definitely bad). In this case white is bad 3/4 of the time (half the time from the definite cases, and half of the other half of the cases). Your formula gives 2/3 instead. [Previous page](https://boards.straightdope.com/t/statistics-profiling-problem/222128.md?page=3)