[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.