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Playing Card Frequencies

A standard deck of 52 playing cards consists of 4 suits, with 13 kinds in each suit. In many card games, the kinds are ranked, and are often referred to as the ranks of the cards. In some games, the suits are also ranked.

Image from www.jfritz.com/cards

In the picture above, the four rows are the four suits. The clubs are all in the first row, followed by the spades, then the hearts, and last the diamonds. Among the 13 kinds, we find the numbers 2 through 10, and four other kinds. The A stands for ace, the J for jack, the Q for queen, and the K for king. The jack, queen, and king are often referred to as face cards. In many card games, a player has a number of cards, and this is referred to as his hand. In a few card games, the order in which the cards are received will matter, but more often a player receives all of the cards for his hand at one time. It should be noted that when cards make up a hand, the same card cannot appear twice. In other words, selections of cards to create a hand are done "without replacement" (that is, without returning the first card to the deck and replacing it with a second).

Determining Frequencies of Different Events

The number of ways any particular event can happen will depend upon the number of cards that form a hand. Let us begin with the simplest possible hand, a single card. (Most players would not even consider this a hand, since it has only one card, but mathematicians always include the extreme cases when creating their definitions.) The number of ways to have a heart is 13, since there are 13 hearts. The number of ways to not have a heart is 39, since there are 39 cards that are not hearts. The number of ways to have a queen is 4, since there are 4 queens. The number of ways to not have a queen is 48, since there are 48 cards that are not queens. The number of ways to have a face card is 12, since there are 3 kinds that are face cards, in four suits each. When a hand consists of multiple cards, the events are much more interesting. Suppose a hand consists of three cards. There are ${}_{52} C_3 = 22100$ ways to choose such a hand.

Let us consider a single suit, say hearts.

The number of ways to have three hearts is ${}_{13} C_3 = 286$, since any three hearts can be chosen from the 13 available hearts to make up the hand. The number of ways to have two hearts and one other card that is not a heart is $({}_{13} C_2)({}_{39} C_1) = 78 imes 39 = 3328$. In this computation, we chose 2 of the 13 available hearts, and one of the other 39 cards to make up the hand. The number of ways to have one heart and two other cards is $({}_{13} C_1)({}_{39} C_2) = 13 imes 741 = 9633$. We chose 1 of the 13 hearts, and two of the other 39 cards to make up the hand.

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The number of ways to have no hearts at all is ${}_{39} C_3 = 9139$, where we have chosen all 3 cards from the 39 that are not hearts. It can be verified that the four results above in fact add to 22100, the total number of ways that something can happen. The fact that we used hearts as the suit was irrelevant, the same frequencies would occur if the suit had been spades (or diamonds, or clubs).

We can do variations on that theme as well.

The number of ways to have two hearts and one diamond is $({}_{13} C_2)({}_{13} C_1) = 78 imes 13 = 1014$. Here, we chose 2 of the 13 hearts, and one of the 13 diamonds. The number of ways to have at least two hearts is $({}_{13} C_2)({}_{39} C_1) + {}_{13} C_3 = 78 imes 39 + 286 = 3328$. To do this problem, we had to break "at least two" into the cases "exactly two" and "exactly three". The number of ways to have at least one heart is ${}_{52} C_3 - {}_{39} C_3 = 22100 - 9139 = 12961$. Rather than to break this problem into three cases ("one heart", "two hearts", "three hearts"), we decided it would be easier to recognize "at least one" as the complement of "no hearts", so we subtracted the "no heart" case from the "anything" case.

We can also count various groupings of suits, without identifying specific suits.

The number of ways to have three cards in the same suit is $({}_4 C_1)({}_{13} C_3) = 4 imes 286 = 1144$. In this computation, we first chose which one suit of the 4 available would be represented, then chose the 3 cards out of the 13 available in that suit. The number of ways to have two cards of one suit and one of another is $({}_4 C_1)({}_{13} C_2)({}_3 C_1)({}_{13} C_1) = 4 imes 78 imes 3 imes 13 = 12168$. Here, we chose one of the 4 suits to be a pair, then 2 cards of the 13 in that suit, then one of the three remaining suits for the single card, then one of the 13 cards in that suit. The number of ways to have three cards of three different suits is $({}_4 C_3)({}_{13} C_1)^3 = 4 imes 13^3 = 8788$. Here, we chose the three suits all at once, then for each suit, identified which one of the 13 cards we would use. Again, we can verify that these three values do add to 22100. Listing all possibilities on a theme, then checking to see that all of the possible combinations have been accounted for, is a very effective way of avoiding errors in your computations.

Now let us consider a single kind, say queens.

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The number of ways to have three queens is ${}_4 C_3 = 4$, since any three queens can be chosen from the 4 available queens to make up the hand. The number of ways to have two queens and one other card that is not a queen is $({}_4 C_2)({}_{48} C_1) = 6 imes 48 = 288$. In this computation, we chose 2 of the 4 available queens, and one of the other 48 cards to make up the hand. The number of ways to have one queen and two other cards is $({}_4 C_1)({}_{48} C_2) = 4 imes 1128 = 4512$. We chose 1 of the 4 queens, and two of the other 48 cards to make up the hand. The number of ways to have no queens at all is ${}_{48} C_3 = 17296$, where we have chosen all 3 cards from the 48 that are not queens.

These four values also add to 22100.

We can also consider a single kind, without identifying the specific card being sought. The number of ways to have three of the same kind (often simply called "three of a kind" by card players) is $({}_{13} C_1)({}_4 C_3) = 13 imes 4 = 52$. The computation involved identifying the particular kind, then choosing 3 of the 4 cards of that kind. The number of ways to have two of one kind and one of another (known as "one pair" to card players) is $({}_{13} C_1)({}_4 C_2)({}_{12} C_1)({}_4 C_1) = 13 imes 6 imes 12 imes 4 = 3744$. Here, we determined which kind would be the pair, then chose 2 of the 4 of that kind, then which of the 12 remaining kinds would be the other card, and which of the 4 suits for that card. The number of ways to have three cards of all different kinds is $({}_{13} C_3)({}_4 C_1)^3 = 286 imes 4^3 = 18304$. We first chose 3 of the 13 kinds, then for each of those 3 cards, identified which one of the 4 suits would be represented. And as before, we can verify that these three values add up to 22100.

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