My knowledge of probability is basic, and I understand the point of Bayesian interpretation most roughly. The following is part of this paper. It is about how p can be rational for person 1 and not-p can be rational for person 2 even when they share all their evidence. So the case is about two detectives and two body of evidence Q and R. one detective has Q, the other has R. Q supports p and R equally supports not-p. Now when they share their evidence with one another, how can Q+R supports p?
In particular, in the following scheme I don't understand where n came from, and when the numbers 2 and 4 came from.
Now, as it happens, the only plausible non-categorical notion of evidence around is the Bayesian one, according to which, recall, Q is evidence for P iff P is more probable conditional on Q than it is on its own, that is, iff Pr(P | Q) > Pr(P ). The natural way to extend this definition to one of strong evidential support is to add to it the requirement that Pr(P | Q) be high, where “high” needs to be filled in (though, patently, we will not want it to mean that Pr(P | Q) = 1; surely Q can be strong evidence for P if it fails to make P entirely certain).22 To appreciate, then, why Goldman’s negative verdict on whether the detectives in the example can rationally stick to their beliefs after sharing their evidence is not nearly as obviously correct as his wording—“Surely not”—makes it appear,23 consider the class of probability models defined by the following schema (n ∈ N+):
P Q R Pr P Q R Pr
T T T 1/(2n + 4) F T T 0
T T F n/(2n + 4) F T F 1/(2n + 4)
T F T 0 F F T (n + 1)/(2n + 4)
T F F 1 F F F 0
As is easy to check, on each model satisfying this schema it holds that (i) Pr(P | Q) = (n + 1)/(n + 2) > (n + 2)/(2n + 4) = Pr(P ); (ii) Pr(¬P | R) = (n + 1)/(n + 2) > (n + 2)/(2n + 4) = Pr(¬P ); and (iii) Pr(P | Q ∧ R) = 1 > (n + 2)/(2n + 4) = Pr(P ). So, by taking n sufficiently large, we have on any acceptable reading of “high” in the definition of “strong evidential support” that Q is strong evidence for P, R is strong evidence for ¬P, yet the conjunction of Q and R is again strong evidence for P. Furthermore, it is easy to see that, on its own, Q is equally strong evidence for P as R, also on its own, is evidence for ¬P (and thus against P). It follows that one detective may have what in the eyes of both of them is strong evidence incriminating Lefty, the other what is also in the eyes of both (equally) strong evidence incriminating Righty, yet after pooling their evidence the former detective has become even more confident—in fact, subjectively certain—that Lefty committed the crime while the latter has become more confident—subjectively certain—that Righty is the perpetrator. If this is not immediately clear, just use the above schema to (partially) define the detectives’ degrees-of-belief functions by choosing an appropriate value for n and making the following substitutions: for one detective’s degrees-of-belief function, substitute “Lefty committed the crime” for P, the house-maiden’s testimony for Q, and the gardener’s testimony for R; for the other detective’s degrees-of-belief function, substitute “Righty committed the crime” for P, the gardener’s testimony for Q, and the house-maiden’s testimony for R. It hence appears that the mechanical principle that equal but opposing forces cancel out has no analogue in the only formally precise theory of evidence we presently have.