AK1089 comments on Underspecified Probabilities: A Thought Experiment

[AK1089]

These are the probabilities of each state:

State	Probability
$A\cap B\cap C$	$x$
$A\cap B\cap \neg C$	$0.3-x$
$A\cap \neg B\cap C$	$0.15-x$
$A\cap \neg B\cap \neg C$	$x$
$\neg A\cap B\cap C$	$0.15-x$
$\neg A\cap B\cap \neg C$	$0.15+x$
$\neg A\cap \neg B\cap C$	$0.05+x$
$\neg A\cap \neg B\cap \neg C$	$0.2-x$

with $x$ being the probability of all three parts of a component being fine. (Obviously, $x\le 0.15$, because $P(\neg A\cap B\cap C)\ge 0$.)

This is not enough information to solve for x, of course, but note that $P(A\cap B\cap C)=P(A\cap \neg B\cap \neg C)$. Note also that $P\left(A\right)P\left(B\right)\approx P(A\cap B)$ and $P\left(A\right)P\left(C\right)\approx P(A\cap C)$ (ie A is not strongly correlated or anti-correlated with $B$ or $C$). However, $P\left(B\right)P\left(C\right)>P(B\cap C)$ by quite a long way: $B$ is fairly strongly anti-correlated with $C$.

Now here’s the estimation bit, I suppose: given that $A$ holds, we’d probably expect a similar distribution of probabilities across values of $B$ and $C$, given that $A$ is not (strongly) correlated with $B$ or $C$. So $P(B\cap C|A)\approx P(B\cap C)$ etc. This resolves to $x\approx P\left(A\right)\cdot P(B\cap C)=0.45\cdot 0.15=0.0675$.

State	Probability
$A\cap B\cap C$	$0.0675$
$A\cap B\cap \neg C$	$0.2325$
$A\cap \neg B\cap C$	$0.0825$
$A\cap \neg B\cap \neg C$	$0.0675$
$\neg A\cap B\cap C$	$0.0825$
$\neg A\cap B\cap \neg C$	$0.2175$
$\neg A\cap \neg B\cap C$	$0.1175$
$\neg A\cap \neg B\cap \neg C$	$0.1325$

This seems… not super unreasonable? At least, it appears slightly better than going for the most basic method, which is $P(A\cap B\cap C)\le 0.15$, so split the difference and say it’s $0.075$ or thereabouts.

The key assumption here is that “if $A$ is pretty much uncorrelated with $B$ and $C$, it’s probably uncorrelated with the conjunction $B\cap C$. This is not strictly-always true as a matter of probability theory, but we’re making assumptions on incomplete information based on a real-world scenario, so I’d say this skewing our guess by a factor of 10% from the most naive approach is probably helpful-on-net.

This means in expectation, we guess the in-house machine to produce $0.0675\cdot 1000000=67500$ good widgets. I’d take that many from the Super Reliable Vendor if offered, but if they were offering less than that I’d roll the dice with the Worryingly Inconsistent In-House Machine. That is, I’m indifferent at $x=0.0675$.