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.
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.
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.
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.
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
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.
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…
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.
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?
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.
