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  • A is probable because of B.
  • B is probable because of C.
  • C is true, therefore A is probable.

What type of argument or inference is this?

closed as unclear what you're asking by Joseph Weissman Jan 24 '16 at 2:41

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The question isn't particularly clear about what you mean by "what kind" of argument, however Modal Logic primarily deals with the possibility and necessity of statements like the one in the question. Probability is a different option that is used in various systems (such as multi-valued logics that assign each proposition a probability value such as 0.7, with computations from those).

In modal logic, your argument would be written as follows:

B ⇒ ⋄A

C ⇒ ⋄B

C

⊢ ⋄A

  • Well.... but... it may be possible to write the argument like that, but it is worth pointing out that the argument you have written is not valid. – Bumble Nov 27 '15 at 21:53
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Your use of "because of" is slightly odd. If we write it using the conventional notation of conditional probabilities, then it becomes P(A|B) is high, P(B|C) is high, therefore P(A|C) is high. In the general case this is not true, so an argument of this form is not guaranteed to be cogent. It may be, for example, that the truth of C renders A highly improbable. One way to understand this is to think in terms of worlds in which the propositions are true. It may be that most C worlds are B worlds and most B worlds are A worlds, but few C-and-B worlds are A worlds.

Try this for an example: If Alice spends lots of money on expensive luxuries she'll probably become poorer. If Alice wins the lottery, she'll probably spend lots of money on expensive luxuries. One cannot conclude that if Alice wins the lottery she'll probably become poorer.

Another way to express this kind of argument is using default reasoning. By default, if B then A; by default, if C then B. C, therefore A. In many cases this will work, but not in general, because C may be a defeating condition for the conditional if B then A, i.e. it may be the case that if B and C then not A.

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