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I've been studying critical thinking and come across what looks like a paradox. Let's say we have the following argument:

P1) If a person is A, then it's likely that that person is also B.

P2) This person is A.


C) This person is likely to be B.

Here, "likely" is interpreted as having a probability greater than 0.5. Using conditional probability we can write P1 as

Pr(B|A) > 0.5

This argument is deductively valid. If our premises are true we can confidently say that the person is more likely to be B than not.

Now, let's consider another argument:

P1) If a person is C, then it's likely that that person is also B.

P2) This person is C.


C) This person is likely to be B.

Just like with the first argument we can write P1 as

Pr(B|C) > 0.5

And again, taking this argument in isolation and assuming the premises are true, we can say that the person in question is more likely to be B than not.

Finally, let's assume that we have both arguments and their second premises refer to the same person. Is it likely that that person is B? On the surface it seems to be likely - we have two arguments both of which confirm it. At the same time, what I now need to show is that

Pr(B|A,C) > 0.5

Using Bayes theorem we can write it as follows:

Pr(B|A,C) = Pr(B ∧ A ∧ C) / Pr(A ∧ C)

I believe, those values cannot be expressed using Pr(B|C) and Pr(B|A), so we can't be confident that Pr(B|A,C) > 0.5. I have found an article on Cross Validated which confirms my intuition.

This looks paradoxical because we have two arguments each of which confirms likelihood of the person being B, but still, when we combine them they do not reinforce each other. My only explanation of this paradox is that, as humans, we tend to put significance on value 0.5 which separates "likely" from "unlikely" while in mathematics it's just an ordinary number between 0 and 1.

So my questions are:

  1. Is my intuition about impossibility of estimating Pr(B|A,C) from Pr(B|A) and Pr(B|C) correct?
  2. If it's correct why does this situation look paradoxical?
  • The crucial question is how independent P1 and P2 are. If they are independent then the two inputs would "reinforce" each other, see Bayesian confirmation. Although it can never be verified with certainty where this sort of reasoning is typically used the assumption of independence is often reasonable. – Conifold Mar 17 '18 at 3:04
  • Thank you for the link, @Conifold. I considered an assumption of independence but it didn't get me too far. If I use step-by-step addition of evidence, then I still have to estimate the effect of C when I already know that A is true. This pretty much leads me to the same problem of calculating Pr(B|A,C). Assumption of independence simplified the result but not to a degree where it was obvious that A and C reinforce each other. Also, it's not clear from Bayesian confirmation article if new evidence which independently confirms a hypothesis also confirms it in the presence of other evidence. – Vlad Mar 26 '18 at 12:06
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Your intuition is not always correct because the areas AC and ABC (see Venn diagram image attached) would be double counted if to find the P(A,C) you simply added them. If AC is particularly small and ABC particularly large, then as ABC is used both in P(B|A) and P(B|C), P(B|A,C) can be <0.5 . E.g. If A=0.3, C=0.3, BC=0, BA= 0, BCA =0.4 then clearly your intuition does not hold in this case as BCA can outweigh A and C individually but not combined

More detailed maths below

Consider a 3-circle venn diagram picture attached

It is easier to see a case where your intuition is justified, e.g. if the intersection of the 3 circles (ABC) has value 1. Hence P(B|A,C) =1 , P(B|A)=1 etc. Your intuition is justified in most cases: if the regions on the venn diagram are A=a,B=b,C=c,AB=d,AC=e, BC=f,ABC=g we get the following. (I would strongly recommend drawing the venn diagram to follow the maths

*a+e< d+g **c+e< f+g (these are just restatements of P(B|A)>0.5 and P(B|C)>0.5 You want to know if ***a+e+c < f + g + d , i.e is P(B|A,C) > 0.5

If e=g or e>g then from (adding * and **) a + 2e + c < d+ 2g + f , then a+e+c < f + g + d as if e>g then more is being subtracted from a+ 2e +c than d + 2g +f

If ef and a>d, It is possible for your intuition not to hold in this case

  • 1
    Maybe I did not express my question clearly, but the answer I expected could not be "true" in some cases and "false" in others. What I meant was that, in general, it's not possible to say that Pr(B|A,C) is greater or less than 0.5 given estimates of Pr(B|A) and Pr(B|C). However, by providing examples in which Pr(B|A,C) can be less or greater than 0.5 you essentially answered my question, confirming impossibility of estimating Pr(B|A,C). Thank you for your help, @Ethan. – Vlad Mar 26 '18 at 10:56
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You have five people, and properties A, B and C.

Your people respectively have properties AB, AB, BC, BC, AC.

P(B|A) = 2/3

P(B|C) = 2/3

P(B|A,C) = 0

  • You misunderstand the premises. It's not the case that that we can only have two properties out of A B and C. – Canyon Mar 30 '18 at 1:07
  • I disagree. The question was about cases where A is evidence for B and C is evidence for B and the intuitive (but invalid) assumption that evidence can be combined to give that A,C is evidence for B. I gave a self-explanatory counterexample. – Veedrac Mar 30 '18 at 1:33

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