popper-logic-scientific-discovery

Our theorem has been derived by considering finite universes, and it
is indeed quite independent of the transition to infinite universes. It is
therefore independent of the formulae (1) and (2) of the preceding
appendix, that is to say, of the fact that in an infinite universe, we have
for any universal law a and any finite evidence e,
(2)
p(a) = p(a, e) = 0.
We may therefore legitimately use (1) for another derivation of (2);
and this can indeed be done, if we utilize an idea due to Dorothy
Wrinch and Harold Jeffreys.
As briefly indicated in the preceding appendix, 4 Wrinch and Jeffreys
observed that if we have an infinity of mutually incompatible or
exclusive explanatory theories, the sum of the probabilities of these
theories cannot exceed unity, so that almost all of these probabilities
must be zero, unless we can order the theories in a sequence, and
assign to each, as its probability, a value from a convergent sequence of
fractions whose sum does not exceed 1. For example, we may make the
following assignments: we may assign the value 1/2 to the first theory,
1/2 2 to the second, and, generally, 1/2 n to the nth. But we may also
assign to each of the first 25 theories the value 1/50, that is to say,
1/(2.25); to each of the next 100, say, the value 1/400, that is to
say, 1/(2 2 .100) and so on.
However we may construct the order of the theories and however we
may assign our probabilities to them, there will always be some greatest
probability value, P say (such as 1/2 in our first example, or 1/50), and
this value P will be assigned to at most n theories (where n is a finite
number, and n.P < 1). Each of these n theories to which the maximum
probability P has been assigned, has a dimension. Let D be the largest
dimension present among these n theories, and let a 1 be one of them,
with d(a 1) = D. Then, clearly, none of the theories with dimensions
greater than D will be among our n theories with maximum probability.
Let a 2 be a theory with a dimension greater than D, so that
d(a 2)>D = d(a 1). Then the assignment leads to:
4 Cf. appendix *vii, text to footnote 11.
appendix *viii 397