Suppose the following problem: A patient comes at the doctor and the doctor is sure the patient has either illness A, B or C. Statistically all three illnesses are equally likely to occur.
Note: It is absolutely known that the patient has one of these three illnesses, there is no such option as "other".
The doctor performs three independent tests. The tests result in the following probabilities:
$ P_1(A) = P_1(B) = P_1(C) = \frac{1}{3} $
$ P_2(A) = P_2(B) = 0.49 $ and $ P_2(C) = 0.02 $
$ P_3(A) = P_3(B) = 0.01 $ and $ P_3(C) = 0.98 $
As you can see, the test do not give a determined "winner". The main question is, can we calculate the probabilities of A, B and C from this tests?
I had a few thoughts on this. Firstly, test 1 does not give any information, so not using test 1 should not result in different probabilities.
Secondly, assume a test gives $P(X) = 0.90$ and another test gives $P(X) = 0.99$. This is a large difference and this should be reflected in the final outcome.
Note that just averaging the probabilities is not doing the job, since taking test 1 into account then results in a different outcome than not taking test 1 into account.
I thought about this solution, which I call multiplying, but there might be a proper name for it.
Define $M(X) = P_1(X) \cdot P_2(X) \cdot P_3(X)$. Then, the final outcome is: $P(A) = \frac{M(A)}{M(A) + M(B) + M(C)}$, and so on for $P(B)$ and $P(C)$.
Note that leaving test 1 out of the definition of $M(X)$ does not make any difference, a property which is desired. Also note that there is a huge difference in the final outcome if a tests outputs $P(X) = 0.99$ rather than $P(X) = 0.90$, which is also a desired property
I am very curious of your thoughts about it, please let me know!