Topics Probability

Question Let's say there is some crime committed and that only 5% of similar crimes are committed by someone like Person A (based on demographics, personality type, previous criminal record, etc.). If the police later find evidence suggesting that Person A is the perpetrator of a crime and that there is only a 10% chance that the evidence could exist if Person A is innocent, then does that mean there is a 90% chance that Person A is guilty? Or do we have to factor in the fact that there was only a 5% probability that A was guilty before the evidence was found? Thanks!

Accepted:
May 24, 2012

Comments

What we're trying to get to is the probability, given all the evidence, that A is guilty. Let H be the hypothesis that A is guilty. You're supposing that our initial probability for H is 5%, i.e., .p(H) = .05. Then we get a piece of evidence – call it E – and the probability of E assuming that H is false is 10%, i.e., p(E/not-H) = .1. Your question: in light of E, how likely is H? What's p(H/E)?

We can't tell. We need another number: p(E/H). We need to know how likely the evidence is if A is guilty. And we can't infer that from p(E/not-H). Why not? Well, suppose the evidence is that the Oracle picked A's name out of a hat with 10 names, only one of which was A's. The chance of that if A is not guilty is 10%, but so is the chance if A is guilty (assuming Oracles don't really have special powers.) iI this case, the "evidence" is actually irrelevant.

The crucial question is this: what's the ratio of p(E/H) to p(E/not-H)? Intuitively, does H do a good job of explaining E? And knowing only one of p(E/H) or p(E/not-H) leaves that under-determined. So to answer your original question, p(E/H) = .1 does not imply p(H/E) = .9.

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Here's some technical detail. The relevant bit of math is something called Bayes' Theorem. Although the formula is a bit unintuitive, it's this:

p(H/E) = [p(H)p(E/H)]/[p(H)p(E/H) + p(not-H)p(E/not-H)]

So as you can see, p(H) does matter. Other things equal, the higher the prior probability the higher the probability given the evidence. But the ratio of p(E/H) to p(E/not-H) is the thing to watch. If p(E/not-H) = .1, as you've assumed, then the highest this ratio can possibly be is 10. As such things go, that's not very high. In fact, if the prior probability of H is .05, as you assumed, then with p(E/not-H) = .1, the highest value possible for the hypothesis given this bit of evidence is only about .345 - quite a bit lower than 90%. But even if the original probability that A did it was 50%, with p(E/not-H) = .1, we still can't get p(H/E) higher than about 84%.

What's going on? Intuitively, the idea is that if there's a 10% chance of seeing this evidence even if A isn't guilty, then it's too likely that the evidence is, so to speak, a false positive for it to count very heavily.