Bayes Theorem has nothing to do with this problem.
The easiest way to solve this problem is this. What is the probability of drawing 3 cards of different ranks.? This is a problem in conditional probabilities.
You have a deck of cards consisting of 52 cards, with 4 suits, each containing 13 cards, and 13 ranks, each containing 4 cards.
You draw one card at random from the deck. What is the probability that it has a rank? Obviously 1.
You draw a second card at random from the deck. What is the probability that it is of a rank different from the first card? There are 51 ways to draw a second card, but only 48 ways to draw a card of a different rank. So the probability of drawing a second card of a different rank is 48/51.
You draw a third card at random from the deck. What is the probability that it is of a rank different from the ranks of both the first and second cards drawn. There are 50 ways to draw a third card, but only 44 ways to draw a card of a third rank. So that probability is 44/50.
Thus the probability of drawing three cards, all of different rank, is
[MATH]1 * \dfrac{48}{51} * \dfrac{44}{50} = \dfrac{16}{17} * \dfrac{22}{25} = \dfrac{352}{425}[/MATH]
The probability of drawing 2 or 3 cards of the same rank is
[MATH]1 - \dfrac{352}{425} = \dfrac{73}{425} \approx 17.17647\%.[/MATH]
You can get there another way. How many ways can you pick three cards from fifty-two (ignoring the order in which they are drawn)?
[MATH]\dbinom{52}{3} = \dfrac{52 * 51 * 50}{ 3 * 2} = 22100.[/MATH]
How many ways can you draw 2 cards from 4 of the same rank, say the ace? Just 6. They could be spades-hearts, spades-diamonds, spade clubs, hearts-diamonds, heart-clubs, or diamonds-clubs. How many ways can you draw 1 card from the 48 cards of different rank? 48 obviously. So the number of ways to pick 3 cards with 2 aces is 6 * 48 = 288.
So the probability of drawing exactly two aces is 288/22100.
What is the probability of drawing 3 aces? By similar logic, we get 4/22100.
So the probability of drawing two or three aces is
[MATH]\dfrac{4}{22100} + \dfrac{288}{22100} = \dfrac{292}{22100} = \dfrac{4 * 73}{4 * 5525} = \dfrac{73}{5525}.[/MATH]
But we need to multiply that by 13 to get the probability of two or three of ANY rank.
[MATH]13 * \dfrac{73}{5525} = 13 * \dfrac{73}{13 * 425} = \dfrac{73}{425}.[/MATH]
But that is precisely what we got the other way. In neither case did we use Bayes Theorem.
The relevance of Bayes Theorem arises from knowledge. Poincarre said that probability is a measure of our ignorance. As we know more, probabilities change. Bayes Theorem tells us how to compute the change exactly. But it is simply common sense. If you already hold 2 aces. your chance of drawing an ace will be less than if you hold no aces. That is the meaning of Bayes Theorem. The practical use of the exact computations specified in Bayes Theorem is admittedly probably close to zero because few of us can do that kind of arithmetic in our heads. I certainly can’t.
If you need an ace to make three of a kind playing seven card stud, your prospects get worse if someone else is dealt an ace. You now have only one ace you could possibly get as opposed to two. That is Bayes Theorem in qualitative form. The people who are telling you that it has no relevance are pig ignorant.
