Bayes' Theorem

[Rasputin]

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Bayes' Theorem relates the conditional probability of A given B to the conditional probability of B given A. Often, it is necessary to invoke Bayes' Theorem in order to transform a prior probability distribution into a posterior probability distribution.

However, in our case, A is the event that the second shooter finds a bullet and B is the event that the first shooter does not. While it is true that P(A|B)=1/4, nowhere do we need to relate this to P(B|A). In fact, we can just compute this directly from the definition of P(A|B) without using Bayes' Theorem at all.

Definition of Conditional Probability:
P(A|B) = P(A and B)/P(B)

Bayes' Theorem:
P(A|B) = P(B|A)*P(A)/P(B)

It is the first formula, not the second, which captures the essence of the (correct) answers given in this thread. So while it may just be a matter of terminology, this is technically not an example of the use of Bayes' Theorem. It is simply an exercise in the computation of conditional probabilities. I just wanted to point this out for the benefit of those who may want to look further into this topic.


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I am stealing jason1990's post from this thread (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=probability&Number=1541054 &page=0&view=collapsed&sb=5&o=14&fpart=all) to start this discussion since I have only barely heard of Bayes' Theorem and would like to know more.

If I read it correctly, the probability of A given that B has happened is equal to the probability of B given that A has happened multiplied by the probablility of A happening independently divided by the probability of B happening independently.

Okay, how does it apply to poker?

The probability of a given player getting an ace is 4/52+4/51 (.155).

The probability that there is an ace on the flop is 4/52+4/51+4/50 (.233).

The probability that there is an ace on the flop if there is an ace in someone's hand is 3/51+3/50+3/49 (.180)

So the probability that there is an ace in someone's hand given that there is an ace on the flop is the probability of there being an ace on the flop if there is an ace in someone's hand (.180) multiplied by the probability that there is an ace in someone's hand (.155) divided by the probability that there is an ace on the flop (.233)

.180*.155/.233 = .119

So the fact that an ace came on the flop changes the likelihood of an ace being in a given player's hand from 18% to 12%?

Do I have the gist of it there?

[jason1990]

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So the probability that there is an ace in someone's hand given that there is an ace on the flop is the probability of there being an ace on the flop if there is an ace in someone's hand (.180) multiplied by the probability that there is an ace in someone's hand (.155) divided by the probability that there is an ace on the flop (.233)

.180*.155/.233 = .119

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The reasoning here is correct and this is exactly the way to apply Bayes' Theorem. But your numbers are a little off.

First, does "an ace in someone's hand" mean "at least one person holds an ace" or are you fixing a particular someone and talking about the event that "this fixed someone holds an ace"? You seem to be using different interpretations in different parts of the problem. Second, when you say "an ace in someone's hand" or "an ace on the flop", do you mean "at least one ace" or "exactly one ace"? Again, you seem to be using different interpretations in different parts of your computations. Third, you cannot just add the probabilities like this

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The probability of a given player getting an ace is 4/52+4/51 (.155).

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because the events "the first card is an ace" and "the second card is an ace" are not disjoint. You must either use inclusion-exclusion or compute one minus the probability of not receiving an ace.

However, if you compute these probabilities correctly, then Bayes' Theorem will give you the correct answer. Incidentally, another way to compute the probability that someone holds an ace given that there is an ace on the flop (and this is the way most people compute it) is to pretend that the flop happened *before* the hole cards were dealt. Your computation with Bayes' Theorem would essentially amount to a proof that this more common way of computing the conditional probability works.