Given that Rugen is the only one with six fingers on his right hand, what's the probability that Rugen's the perpetrator? 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).

I was wondering how did we arrive at this?
 A: 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.
A: 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:

*

*P(R) = Rugen is guilty = $\frac{1}{n}$ since equally likely

*$P(R^c)$ = Someone else is guilty = $1 - \frac{1}{n}$ since equally likely

*$P(M,N|R) =$ 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

*$P(M,N|R^c) =$ 1 five-fingered man is guilty, 1 six-fingered man is innocent and all other $n-2$ five-fingered man are innocent = $p_0(1-p_1)(1-p_0)^{n - 2}$.

