Section 2.1 – Card Hands

In previous examples, we've used cards to come up with examples of counting problems. We can take those methods further to calculate the probability of getting specific hands in games like blackjack and poker. If you aren't familiar with calculating probabilities, I've put a quick explanation in Appendix A.

Blackjack

To answer questions involving different card hands, it helps to focus on the specific choices that you have to make. For example, if you're dealt two cards then you would ask yourself, "How many different ways can I be dealt the first card? How many ways can I be dealt the second card?" Once you have those numbers, you can get the total number of combinations by multiplying the values together using the Multiplication Rule.

A Different Approach

With a blackjack, the rules are pretty simple and you only start off with two cards so it was easy just to count the number of cards that fell into each category. With poker hands, the situations are going to be a little more complicated. The procedure for counting those hands will usually involve two steps: first, chose a value for a card and then chose a suit for it. I'm not suggesting that you use this method in every situation. In Example 1, for example, it was clearly easier just to count the actual cards and say, "There are a total of 16 ten's, jack's, queen's and king's." What I want you to see here is how the method works in a simple case so that you have it in your toolbox when we get to more complicated problems.

Suppose I wanted to find (total ways to choose a ten point card). Using the method of choosing a value then choosing a suit, the process would look like this:

(number of ten point card choices) = (ways to choose the card's value) · (ways to choose the card's suit)

There are a total of 4 types of ten point cards. We're only thinking about the numbers at this point, not the suits, so we have to choose one of 10, jack, king or queen. That makes

$$\textrm{ways to choose the cards value} = \begin{pmatrix}\textrm{4 types of 10 point cards} \\ \textrm{choose 1 of them} \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \frac{4!}{(4-1)! \cdot 1!} = 4$$

Now, I'll apply the same reasoning to the number of choices for the card's suit.

$$\textrm{ways to choose the card's suit} = \begin{pmatrix}\textrm{4 suits} \\ \textrm{choose 1 of them} \end{pmatrix} = \begin{pmatrix}4\\1\end{pmatrix} = \frac{4!}{(4-1)! \cdot 1!} = 4$$

That makes the total number of ten card choices

number of ten card choices = (value choices) · (suit choices) = 4 · 4 = 16

This matches the result that we got in Example 1 from just counting the cards which confirms that the method is giving us correct results.

Poker

Poker hands give us some more interesting examples because the hands are bigger and have more "winning" combinations. For this kind of problem, we'll often have to use our, "pick a value then pick a suit" strategy.

To write out a poker hand, we'll often just right out the numbers followed by a symbol for the suit, e.g. 2H would be the two of hearts and 5D would be the five of diamonds. A whole hand would be the individual cards separated by dashes so 3H-3D-3S-6H-6D would be the three of hearts, the three of diamonds, the three of spades, the six of hearts and the six of diamonds. If the suits don't matter, I'll just leave out the letters, e.g. 3-3-3-5-5 would be a hand with three 3's and two 5's of any suit.

So what's the probability of being dealt a straight flush? There are 52 cards in a standard deck and, to form a hand, we need to choose 5 of them. Because the order of the cards in a poker hand doesn't matter, this is an example of a combination.

$$\textrm{total five card hands}=\begin{pmatrix}52\\5\end{pmatrix}=\frac{52!}{5!(47)!} = 2, 598, 960$$

That makes the probability of being dealt a straight flush

$$P(\textrm{straight flush}) = \frac{40}{2598960}= .00001539$$

or 0.001539%.