Originally posted by NemesioLet's start by stating Bayes' Theorem in plain English.
I've just read the Bayesian Theorem article on Wikipedia and don't even understand the chocolate
Hold my hand a bit and explain it to me and then explain what it means to be a Bayesian atheist.
I thank you in advance and am,
Originally posted by DoctorScribblesSorry I took so long to respond. I wanted to print out the post and look it over a few times and
Let's start by stating Bayes' Theorem in plain English.
Bayes' Theorem asserts that these two products have an equal value, for any propositions A and B:
1) The probability that A is true whenever B is true, times the probability that B is in fact true.
2) The probability that B is true whenever A is true, times the probability that A is i ...[text shortened]... right.
If that framework is clear, then we can go on to apply it to the cookie problem.
Originally posted by DoctorScribblesI'm too tired to review the problem's solution right now, but let me give a first blush impression
Suppose you are playing Hold'em against an odd opponent whom you
know with certainty raises before the flop only with pocket Aces,
pocket Queens, or Ace-Queen.
You are playing a hand and lo and behold, he raises before the flop.
Then, as he's shuffling his two cards he happens to expose one at
random - an Ace.
You hold pocket Kings and ...[text shortened]... shed Ace, what is the probability
that his other card is a Queen, leaving you the favorite?
Originally posted by NemesioOne problem with your analysis is that it does not account for all of the available information. Namely, it does not account for the fact that you saw a particular Ace.
I'm too tired to review the problem's solution right now, but let me give a first blush impression
and you can tell me why it's wrong.
He's flashed an Ace, so that means he has either Aces or Ace-Queen. So, here are the hands he
1) A-S + A-C
2) A-S + A-D
3) A-S + A-H
4) A-C + A-D
5) A-C + A-H
6) A-D + A-C
7) A-S + Q-S
8) A-S + Q-C as a slightly better than 3 to 1 favorite.
What's wrong with this analysis?
Originally posted by DoctorScribblesI should be asleep, but is this where the mistake is?
c) P(B|A): Here we compute the probability that he flashes an
Ace given that he holds Queens or Ace-Queen. If he has an Ace
and a Queen, an Ace is exposed 50% of the time. If he holds
Queens, an Ace is exposed 0% of the time. Weighting these by
the number of ways each type of hand could be dealt, we get
Originally posted by NemesioIf I have chosen the correct A, then that is not the problem, because A is assumed to be true with certainty when computing P(B|A). The problem might be in my choice of A.
I should be asleep, but is this where the mistake is?
We're not accounting for how often he flashes an Ace given that he has Ace-Ace.
Or am I totally lost?
Originally posted by PalynkaYes, that's correct!
Your mistake is in P(B|A). You use 28 in the denominator but you should have used 22 because it's the probabilities given A.