“Bayes' Theorem for Beginners: Formal Logic and ... - Richard Carrier

33% chance the parent is telling the truth, then the evidence is only contrary to
expectation by that much:
34
P(e|~h.b) = 1.00 – 0.33 = 0.67 (case with some support)
8. Stronger supporting evidence would generate a stronger factor. For example, if a
trusted colleague of yours, who has no connection to the student (and thus no interest in
the matter), attests to witnessing the incident, or there is a video recording of it, etc., you
may think there is only a 5% chance he would lie (or the video was forged or misleading
you, etc.), in which case there is only a 5% chance you would have the evidence you do
(missing homework, student’s story, and trusted colleague’s eyewitness report), which
means quite simply:
P(e|~h.b) = 0.05 (case with strong support)
9. So assuming the common case (no evidence or expectations of evidence beyond the
missing homework and the story, and using the prior probabilities concluded with above):
0.10 x 1.00
P(h|e.b) = ____________________________ = 0.10
[ 0.10 x 1.00 ] + [ 0.90 x 1.00 ]
= There is only a 10% chance the student is telling the truth (in this
case relying solely on your background evidence regarding both
student honesty and frequencies of dogs eating homework, i.e. how
often stories of dogs eating homework are typically true)
10. Assuming the case with small incongruity (some expectation that the student should
be able to present more evidence but can’t) but some support (parent’s testimony):
0.10 x 0.80
P(h|e.b) = ____________________________ = 0.117 = 0.12
[ 0.10 x 0.80 ] + [ 0.90 x 0.67 ]
= There is still only a 12% chance the student is telling the truth.