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In Probabilistic Reasoning For Intelligent Systems by Judea Pearl, in chapter 2, the three prisoner paradox was handled.

One scenario was raised where $P(I_{B}^{'}|G_A)=q$ is assumed; here $G_A$ is the event of prisoner $A$ (the one that asks the guard) being convicted and $I_B^{'}$ is the event of the guard reporting that prisoner $B$ will be released. Here we consider the guard's bias towards reporting $B$'s release to be an unknown ($q$).

$P(G_A|I_B^{'})$ = $\frac {q} {1+q}$.

We can even assume a probability density function $f(q)$ for the random variable $q$ that varies between $0$ to $1$, in which case -

$P(G_A|I_B^{'})$
= $\int_{0}^{1} {\frac {q} {1+q} f(q|I_B^{'}) dq}$
= $\frac {\int_0^1 {q f(q) dq}} {1 + \int_0^1 {q f(q) dq}}$

I do not understand the transformation from second to third line here.

Note - $f(q)$, being a probability density function for $q$ varying between $0$ - $1$, I assume $\int_{0}^{1} {f(q)} dq$ = $1$.

Anirban Chakraborty
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    Looks like this. What about $f(q)=\mathbf{1}_{q\in[0,1]}$? – metamorphy Aug 04 '22 at 04:57
  • @metamorphy I should have verified it with an example first. Now that you mention this, I think I will verify the second and third expressions (that I referred to in my query) for equality, numerically, with a few candidates for $f(q)$ that satisfy $\int_0^1 {f(q) dq}$ = 1. $f(q)$ = 1 is one such candidate and true it does not satisfy the equality. Hence, we already know it cannot be generally correct.

    Did I understand your point correctly?

     – Anirban Chakraborty Aug 04 '22 at 17:14
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    Exactly. The (counter-)example $f(q)=\frac23(1+q)$ is even easier to check, since all the integrands are polynomials: $I=\int_0^1 qf(q),dq=\frac59$, so $\int_0^1\frac{qf(q)}{1+q},dq=\frac13\neq\frac{I}{1+I}=\frac5{14}$. – metamorphy Aug 04 '22 at 17:39

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