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 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?