Assume that I evaluate the probability that a murder suspect is guilty to 10%. Then I find a blood trace at the scene and only 1% of the population has this blood type. The suspect has this blood type and the test is accurate (say 100% accuracy for simplicity). What is now the probability that the suspect is guilty?

posterior = prior x likelihood / normalization = .1 x 1 / .01 = 10

What am I missing here?


2 Answers 2


Short answer: The denominator is missing a term. It's missing the "innocent and evidence" bit which is .009

Long answer:

You want to calculate the probability of guilty given that we have the evidence of the rare blood type. Assuming that if someone is guilty, we are always going to find his blood type as evidence then:

$P(Guilty|Evidence) = \frac{P(Evidence|Guilty) P(Guilty)}{P(Evidence)}$

$P(Guilty|Evidence) = \frac{P(Evidence|Guilty) P(Guilty)}{P(Guilty)P(Evidence|Guilty) + P(NotGuilty)P(Evidence|NotGuilty)}$

$P(Guilty|Evidence) = \frac{1 \times 0.10}{0.10 \times 1 + 0.90 \times 0.01}$

$P(Guilty|Evidence) = 0.917$

  • $\begingroup$ Can P(Guilty|Evidence) in this scenario be constant? I cant get my head around it: I would expect P(G|E) to be big if there are very few people within the range of the crime (lets say 100 people, so there would be only 1 with the given blood-type) and i would expect it to be rather smale if there are lots of people in range? $\endgroup$ Jan 28, 2019 at 11:04
  • $\begingroup$ Not sure I 100% understand what you mean by constant but strictly speaking, the only way to have P(Guilty|Evidence) constant is to never update it no matter how much more evidence you collect! Essentially, that would make the P(Guilty|Evidence)=P(Guilty) and in this example you’ll always have the prior 10% probability. We basically make use of the Bayes’ rule to incorporate data to our prior beliefs and update it if in the future we get more data $\endgroup$ Jan 28, 2019 at 12:00
  • $\begingroup$ Thanks for your reply. Lets say scenario1 has 100 People in the region (we would expect 1 to have bloodtype b1) and scenario2 has 10000 people in the region (we would expect 100 to have bloodtype b1). Is it possible that for both scenarious P(Guilty|Evidence) is the same value? In scenario1 the person we are suspecting might be pretty much the only one with bloodtype b1, so i would expect the 10% to raise. In scenario2 he is just one of 100 with bloodtype b1, which shouldn't raise the 10% that much? $\endgroup$ Jan 28, 2019 at 12:32
  • 1
    $\begingroup$ To answer your first question, the answer is yes. The probability above will always remain the same, no matter how many people are in the region. In the second part, you are imposing a new question and introducing additional information. The additional information is that there can be only one killer (it wasn't stated before) and the new question for modelling becomes P(he is the only killer | his blood type matches & n = n0). n0 can be 100 or 10000 but in the case of n0 = 100 your intuition will be completely right! the resulted probability will be higher compared to the case of n0 = 10000 $\endgroup$ Jan 28, 2019 at 23:42

You're calculating the posterior probability of a particular suspect being guilty, not a random person from the general population being guilty. Therefore, the normalisation term is not the probability of someone in the general population having the blood type, it's the probability of the suspect having that blood type given the prior probability that they're guilty. Since they must have the blood type if they're guilty, this is equal to $$\begin{align*} \mathbb{P}(\textrm{suspect has blood type}) =& \mathbb{P}(\textrm{suspect guilty})\\ &+ \mathbb{P}(\textrm{suspect innocent}) \mathbb{P}(\textrm{suspect has blood type if innocent})\\ =& 0.1 + 0.9 \times 0.01\\ =& 0.109, \end{align*}$$

which leads to a valid posterior probability.


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