That in turns implies that P(V|~G)*P(~G) > 0.7576. And for this to be the case you have to have the following probability, P(V|~G) > 0.8111.

Should read, P(V|G) < 0.7576 and for this to be the case it must be that P(V|~G) < 0.06.

I was using P(V) to mean the probability that Mr. Puckett was there.

Not in your initial calculations. In your initial calculations you used 0.05. Now you want to backpedal and try to claim you used 1? Please, your calculation was,

P(G|VIC) = 1.0 * 1:15 / .05 = 20:15 = 4:3

And your 1:15 is wrong too. It should be 1:14.

And the correct calculation is,

Odds(G|V) = Odds(G) *[Pr(V|G)/Pr(V|~G)].

Why you have 0.05 in the denominator in your initial calculation is not clear to me.

The numbers add up.

No, they don’t. Your numbers, taken in conjunction violate the theorem of total probability. Thus either P(V) is too low or P(G) is too high. There is no room for doubt about this. Your numbers just don’t add up and as such your claims of guilt beyond a reasonable doubt are now in doubt.

Your mistake is using the probability of 1 man’s guilt when I’m talking about the probability that any California man would be in the vicinity.

I made no mistake. I merely applied the theorem of total probability to your set up. To be specific the theorem of total probability is for disjoint events B and ~B,

P(A) = P(A|B)*P(B) + P(A|~B)*P(~B).

Now you want to change your supposedly inviolate numbers because you’ve been caught in a mistake.

Using your numbers,

0.05 = x*0.066 + P(V|~G)*P(~G)

Now for this equality to hold it must be the case that 0.05 – x*0.066 > 0. That in turns implies that P(V|~G)*P(~G) > 0.7576. And for this to be the case you have to have the following probability, P(V|~G) > 0.8111.

I’m sorry, your made up out of cloth priors stink. I think I’d rather leave them to an expert which is not you.

Oh, and regarding this,

P is the likelihood….

You do know the difference between probability and likelihood don’t you?

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If P(V) means the probability that any California man would be in the vicinity, then you get:

P(V) = .05

P(V|G) = 1.0
P(G) = 1/16M

P(V|NG) = .04999994
P(NG) = (16M – 1) / 16M

The numbers add up. Your mistake is using the probability of 1 man’s guilt when I’m talking about the probability that any California man would be in the vicinity. I was most certainly not talking about the “probability” that Mr. Puckett was in the vicinity, which I assumed to be 1.0.

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I was using P(V) to mean the probability that Mr. Puckett was there. It was presumptuous of me to assume a 100% chance, despite the fact that he seems to have admitted to being there and the police have strong evidence placing him there. There’s always a chance everyone is wrong.

He apparently admitted that he was there, which he probably would not have done if it wasn’t true. Let’s say there’s a 1/20 chance, if he was innocent, that he was wrong about being there (despite the other evidence showing he was there).

Let’s drop P(V) to .95. We still end up with 618 * P : 1 odds of guilt, which are beyond a reasonable doubt, IMO.

P is the likelihood that the person who raped and murdered Diana Sylvester was later arrested for any felony (not just any sex crime, because I made a mistake as to the information contained in the DB) and had his DNA collected as a result. That suggests P should be higher than previously thought. A P of .5 seems reasonable. Someone who would commit a rape/murder seems like the kind of screw up to end up with a felony record, even if the later felonies weren’t sex crimes. Even a P of .3 would produce odds of about half a percent that Mr. Puckett is innocent.

My numbers add up. He is guilty beyond a reasonable doubt, to my subjective and personal understanding of what “beyond a reasonable doubt” means.

You have a problem with your calculations.

You state that P(V) = 0.05 however that is too low based on your other probabilities you use. This result follows from the theorem of total probability which holds, in this case,

P(V) = P(G)P(V|G) + P(~G)P(V|~G).

However you also have asserted that P(V|G) = 1 and that P(G) = 0.066. Hence you are asserting that

P(~G)P(V|~G) = -0.016 which clearly cannot happen since both P(~G) & P(V|~G) are in the closed set [0,1].

This highlights a bigger problem with Daryl’s posts. These numbers aren’t just pulled out of one’s ear. They have to make sense. Eliciting prior probabilities in Bayesian analysis is not as trivial an exercise as Daryl makes it out to be.

No it’s not. It’s a totally crappy non-point, and you were right to miss it on the first go. The odds of a single test yielding a false positive are 1 in 1.1 million. The odds of 338,000 tests collectively yielding a false positive are (1 x 338,000) in 1.1 million. Which is indeed the same as (1 x 338,000 / 338,000 = 1) in (1.1 million / 338,000).

I think this is not quite correct and this is also a problem in the original article. The chance of one or more hits is close to 1 in 4. However, that is merely the chance of a hit, it says nothing about guilt of the person’s whose name pops out. For that you need to use Bayes theorem.

I urge everyone still following this discussion to go here and read the article. It is about how to determine if a woman has breast cancer given that a test came back positive for breast cancer. Here is the nut of the problem,

1% of women at age forty who participate in routine screening have breast cancer. 80% of women with breast cancer will get positive mammographies. 9.6% of women without breast cancer will also get positive mammographies. A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?

The problem is very similar to the one we are dealing with here, but with the exception that all probabilities are known. One might be tempted to look at such test results and say, “Crap I (my wife) has breast cancer.” However that would be wrong if you are looking at the 80% number. The actual probability that the woman with a positive test has breast cancer is 7.8%! Ten times lower than what most people think is the actual answer.

Probabilistic reasoning is often counter intuitive. Look at the Monty Hall problem. People are still arguing about that one and there are mathematical proofs out there as well as different computer codes that show “switching” is always a better strategy than “sticking”.

Simply because the answer is counter intuitive or doesn’t strike you as right doesn’t mean it is wrong. There are other paradoxes in statistics too. Simpson’s Paradox, the Allais Paradox. the Ellsberg Paradox, and then there is the work of Daniel Kahneman about how people routinely flub probabilitistic/decision making under uncertainty and behave “irrationally”–usually a violation of Bayes theorem or some other theorem in probability theory.

Practically it does make a difference.

Not really. For a criminalist, DNA is a means to identification, nothing more. Its a tool used to exclude people from consideration. A five loci match cannot exclude as many people as a 13 loci match…which is why criminalists continue to use other methods to refine their identification, and why I find the discussion rather specious to begin with.

As I mentioned a few days ago, a criminalist with a sample DNA fragment is going to run a database query just as he/she would if they had a surveillance photo of a suspect. Or a partial fingerprint. If, and only if, there is a possible match, will it be further investigated, and only at that time does the reliability and the confidence of the possible identification come into question.

Which is indeed the same as (1 x 338,000 / 338,000 = 1) in (1.1 million / 338,000).

But 1 in (1.1 million/338,000) is the same as 1 divided by (1.1 million/338,000) — which I think still ends up being 338,000 divided by 1.1 million. And not the other way around.

Correct me if I’m wrong.

Maybe I’m missing something.

Shouldn’t they have said “That’s the same as dividing 338,000 by 1.1 million” and not the other way around?

After all, they end up with a roughly 1 in 3 chance.