Calling Out Cribs and Starrman

I've just read the Bayesian Theorem article on Wikipedia and don't even understand the chocolate
cookie example.

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,

Appreciatively yours,
Nemesio

Originally posted by Nemesio
I've just read the Bayesian Theorem article on Wikipedia and don't even understand the chocolate
cookie example.

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,

Appreciatively yours,
Nemesio

Originally posted by DoctorScribbles
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.

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 might have the best hand -- as long as he
doesn't hold pocket Aces.

Problem: Given that you've seen his flashed Ace, what is the probability
that his other card is a Queen, leaving you the favorite?





Bayes' Theorem comes to the rescue.

Define two propositions:
Proposition A: Your opponent was dealt Queens or Ace-Queen.
Proposition B: A random exposed card of this opponent is an Ace.

State the three known quantities:
a) P(A): We can compute the probability that the opponent holds
Queens or Ace-Queen by counting the ways he could hold Queens
or Ace-Queen and dividing it by the total possible hands he could
have (disregarding the flashed card). He could have Ace-Queen
16 ways (4 Aces, times 4 Queens), and he could have Queens 6
ways (4*3/2), and he could have a total of 28 (8*7/2) different
hands. Thus, P(A)=22/28.

b) P(B): Since the opponent's hand can contain only Aces or
Queens, and since he is equally likely to have recieved either
(still disregarding the flashed card), since there are four of each
in the deck, P(B)=1/2.

[Note that P(A) and P(B) must always be computed from a
position of ignorance with regard to any reavealed truth of A and B.
They are typically called prior probabilities, to be computed from a
set of information available prior to the truth of either A or B
being revealed (as B is revealed to be true in this scenario).]

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
P(B|A)=0*(6/28)+.5*(16/28)=8/28.


Finally, solve for the unknown:
d) P(A|B): This is the unknown that we are now ready to use
Bayes' Theorem to compute. According to the theorem:
P(A|B) = P(B|A)*P(A)/P(B)
= (8/28)*(22/28)/(1/2)
= 45%

Bayes' Theorem tells us that given all of our information, the
opponent has Queens or Ace-Queen with likelihood 45%,
which means that he holds Aces with probability 55%, which
means folding your Kings is the correct play.

It is the extra information, the flashed Ace, the revelation that
B is in fact true, that allows you to make the correct play in this
case, since pocket Kings would otherwise be on average a
favorite to his possible holdings in the absence of the extra
information.

Bayes' Theorem is all about revising probability estimates in
light of new information. The Monty Hall problem is another
good problem that you could try to work through using this method,
although that has other, more straightforward methods of solution.
The thing about being introduced to Bayes' Theorem is that at first
it seems like more trouble than it's worth when using it on toy
problems, since the final probability of interest can be solved
directly without too much trouble using non-Bayesian analysis.
The value of Bayes' is seen when the prior probabilities are much
easier to compute than the conditional probabilities, which is often
the case in real problems.

Originally posted by DoctorScribbles
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 Nemesio
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
could have:
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?

Nemesio

Originally posted by DoctorScribbles
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
P(B|A)=0*(6/28)+.5*(16/28)=8/28.

Originally posted by Nemesio
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?

Well, I'm a bit stumped for the moment, but my gut feeling is that as an empirical matter, Kings is definitely not a 3-to-1 favorite in this scenario, and before doing the Bayesian analysis, I would have guessed Kings to be a slight dog, so I think that analysis is right. I think there is something subtle in your distribution regarding your assumption that each holding is equally likely after the Ace flashes. I'll think it over a bit.

Yes. For one, he's more likely to have any one of your (1) through (6) as any one of your (7) through (22).

(This is the same idea as in the Monty Hall problem; Monty can't open the door with the prize; and you can't ever see a Queen exposed, but there are two cards you could have seen in each hand when he has Aces. Your equally-likely assumption about your remaining distribution is akin to the faulty Monty Hall analysis of listing the remaining doors after one is opened, and concluding that you have a 50% chance of picking the right one.)

I'll work out the numbers and see if that reconciles it.

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.

Originally posted by Palynka
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.

Originally posted by DoctorScribbles
Yes, that's correct!

With the correct arithmetic, P(A|B)=57%, which matches the number given by the revised seven-way distribution, leaving Kings a slight favorite.

Originally posted by Palynka
I'm just here waiting for the Bayesian atheist part. 😉

Originally posted by DoctorScribbles
I'll let Starrman handle the theological application.