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Blitzstein, Introduction to Probability (2019 2 ed), p 58, Example 2.3.10 (Six-fingered man).

A crime has been committed in a certain country. The perpetrator is one (and only one) of the $n$ men who live in the country. Initially, these n men are all deemed equally likely to be the perpetrator. And the eyewitness then reports that the crime was committed by a man with six fingers on his right hand. Let $p_{0}$ be the probability that an innocent man has six fingers on his right hand, and $p_{1}$ be the probability that the perpetrator has six fingers on his right hand, with $p_0$ < $p_1$. (We may have $p_1 < 1$ since eyewitnesses are not 100% reliable.)

Rugen lives in the country. He is found to have six fingers on his right hand.

I skip part (a).

(b) Now suppose that all n men who live in the country have their hands checked, and Rugen is the only one with six fingers on his right hand. Given this information, what is the probability that Rugen is the perpetrator?

Here's the textbook's solution.

Let $R$ be the event that Rugen is guilty and $M$ be the event that he has six fingers on his right hand.

I skip the solution for part (a).

Solituion to this

I was wondering how did we arrive at this?

R.jan
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    There are surely many books with that title, so you should give the author(s) too. – Hans Lundmark May 18 '21 at 09:22
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    @HansLundmark Ohh I updated it. – R.jan May 18 '21 at 09:47
  • For $P(M,N|R)$, we assume Rugen is the perpetrator. The example states that the perpetrator has probability $p_1$ of having 6 fingers, so $P(M|R) = p_1$. Then the other $n-1$ men are innocent, and for each of them the probability is $1-p_0$ of not having 6 fingers. Then we just multiply these probabilities to get $P(M,N|R)$ (assuming, as is plausible, the appropriate form of independence of fingeredness across individuals). – Joe Blitzstein Jul 28 '21 at 07:04

2 Answers2

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The calculation of $P(M,N\mid R)$ has a dizzying number of counterfactuals to sort out. I also have difficulty understanding what the authors meant by it.

How can the event $M, N$ (the event that Rugen has six fingers and no other man in the country does) be dependent on whether Rugen committed a crime? After all, Rugen and the other men were all born with their particular numbers of fingers long before the crime was even contemplated.

But the thing that really trips me up is the use of $p_1.$ It is supposed to represent the unreliability of witness statements, but that's not a good model for unreliability of witness statements.

In reality, there is some probability $p_T$ that a witness seeing a six-fingered man would correctly say he had six fingers, and some probability $p_F$ that a witness seeing a five-fingered many would incorrectly say he had six fingers. If these probabilities were equal then witness statements would have no value at all. It is reasonable to assume they are not equal, in fact that $p_T > p_F.$

Now the question is what $p_1$ means. The book says it is the probability that the perpetrator has six fingers. The evidence actually presented is that a witness said the perpetrator has six fingers. From some set of prior probabilities including $p_T$, $p_F$, and something about the distribution of six-fingered men in the population -- perhaps the probability that the next baby born has six fingers, perhaps the actual fact that the population has one six-fingered man and $n-1$ others -- we might be able to compute $p_1.$ But the book is using $p_1$ as if it were something that could be estimated independently of anything we know about the frequency of six-fingeredness. It smells to me.

David K
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  • I agree with your criticisms. I'd say this is a textbook example of a bad textbook example. – user159517 May 18 '21 at 12:31
  • I agree this example is tricky, but I also think it's an instructive one to think about. I'm glad at least you found it thought-provoking even though you don't like how some of the probabilities were formulated.

    Regarding "How can the event , ... be dependent on whether Rugen committed a crime? After all, Rugen and the other men were all born with their particular numbers of fingers long before the crime was even contemplated.": the events $M$ and $R$ are very dependent; that's about probability not causality. See Ex. 2.2.2 in the book and various Kahneman-Tversky examples.

     – Joe Blitzstein Jul 28 '21 at 06:38
  • @JoeBlitzstein I agree this is about probability, not causality. And the first equation in the alleged solution of (b) is just straightforward Bayes. But (among other things) the calculation of the second equation (expressing the probabilities in terms of $p_0$ and $p_1$) appears to make a number of assumptions of independence of the events that Rugen has six fingers and that Bob has six fingers, given that Rugen is guilty (or not). Where do those assumptions come from? I suspect a hidden (and in this case unjustified) causal model. – David K Jul 28 '21 at 13:32
  • @DavidK There is an implicit independence assumption, as in many probability problems phrased in words rather than equations. It's just very hard for written English to have full mathematical precision without the statement of any problem becoming very verbose and lawyerlike. I think the independence assumption is plausible and natural though: unless Rugen and Bob are brothers or something like that, Rugen's fingeredness doesn't seem informative about Bob's (given that Rugen is guilty; unconditionally, there is some dependence since if Rugen is exonerated then Bob is more likely to be guilty). – Joe Blitzstein Jul 28 '21 at 17:18
  • @JoeBlitzstein The implicit independence assumption is suitable in problems such as rolling dice (where there is no causal connection that would contradict independence) and in male and female births in a family (which may not be independent but at least in some conceivable world could be independent by a causal argument, and we choose to assume they are). In this problem, however, we have a crime and an unreliable witness statement mixed in. But perhaps it is better if I put it this way: how would you estimate $p_1$? – David K Jul 28 '21 at 17:51
  • @DavidK Of course this problem is a vast simplification of a complicated real world application about eyewitness testimony, forensic evidence, genetics etc. That's to make it simple enough to be suitable as an example of Bayes' rule, but challenging enough to be thought-provoking rather than routine plug-and-chug. Causal inference and estimation of parameters are extremely important areas of statistics but not the topic of this problem. – Joe Blitzstein Jul 28 '21 at 21:16
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Bayes rule states: $$P(A) = \frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)} $$

We have:

  1. P(R) = Rugen is guilty = $\frac{1}{n}$ since equally likely
  2. $P(R^c)$ = Someone else is guilty = $1 - \frac{1}{n}$ since equally likely
  3. $P(M,N|R) =$ 1 six-fingered man is guilty and all other $n-1$ five-fingered men are innocent = $p_1(1-p_0)^{n-1}$ since M and N are independent
  4. $P(M,N|R^c) =$ 1 five-fingered man is guilty, 1 six-fingered man is innocent and all other $n-2$ five-fingered man are innocent = $p_0(1-p_1)(1-p_0)^{n - 2}$.
Tankut Dogrul
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    thanks, @tankut: Yes but from a beginner's perspective how did we calculate 3)P(M, N|R) and how could we represent this with a probability tree? – R.jan May 18 '21 at 09:45