I would like to make two points. First, in the context of my Post #85, the probability that life on Earth emerged as the product of an intelligence can
never exceed the probability that before life existed on Earth, there existed an intelligent entity that attempted to create life on Earth. Second, observing white swans
does increase the chance that there are no black swans.
To address the first point, recall:
Quote:
Originally Posted by jason1990
A = "Life on Earth emerged from natural processes."
B = "Life on Earth emerged as the product of an intelligence."
C = "Life on Earth emerged."
...
D = "Before life existed on Earth, there existed an intelligent entity that attempted to create life on Earth."
In that post, I proved that
Quote:
Originally Posted by jason1990
P(B | C) = P(B & D | C) + P(B & (not D) | C)
= P(D | C)P(B | D & C) + P(not D | C)P(B | (not D) & C)
= P(D | C)P(B | D & C)
This means that, no matter what else we may assume, it will always be the case that P(B | C) ≤ P(D | C). Of course, this should be obvious, since B implies D.
To address the second point, let
X = "There are no black swans."
An = "We have seen n white swans and no black swans."
Let us assume that, for all n, P(X | A
n) > 0 and P(A
n+1 | A
n) < 1. We then have
P(X & An+1 | An) = P(X | An)P(An+1 | X & An)
= P(X | An).
On the other hand, we also have
P(X & An+1 | An) = P(An+1 | An)P(X | An+1 & An)
= P(An+1 | An)P(X | An+1).
Combining these gives
P(X | An) = P(An+1 | An)P(X | An+1) < P(X | An+1).
In other words, P(X | A
n) is an increasing function of n.