For better or worse, traditional probability problems tend to involve gambling problems, such as die games and card games, perhaps because they are the most commonplace examples of truly equiprobable sample spaces. A middle (junior secondary) school student first trying her hand at probability will be confronted with simple questions like 'What is the probability of getting a 7?' Yet by the last days of high school and the early days of university, the going gets rough.
Mathematics and statistics textbooks are of varying quality. Some provide useful examples and explanations; others do not. However, few if any of them offer a systematic analysis of the various question types you will actually see in an exam. So when students, particularly those less gifted at mathematics, are faced with new question types they've never seen before, they find themselves in a perilous situation.
This is why I'm writing this. The purpose of this article - and its subsequent installments, if the demand is great enough for me to continue - is to help you apply the principles of combinatorics and probability to word problems, in this case card game questions. I assume you already know the basic principles - factorials, permutations vs. combinations, conditional probability, and so on. If you've forgotten everything or haven't learnt these yet, scroll down to the bottom of the page, where you'll find a link to a statistics book on Amazon covering these topics. Problems involving the Rule of Total Probability and Bayes' theorem will be marked with a *, so you may skip them if you have not learnt these aspects of probability.
Even if you aren't a student of mathematics or statistics, don't leave yet! The better part of this article is devoted to the chances of getting different poker hands. Thus, if you're a big fan of card games, you may well be interested in the 'Poker Problems' section - scroll down and feel free to skip the technicalities.
There are two points to note before we start:
- I will be focusing on probability. If you want to know the combinatorics part, look at the numerators of the probabilities.
- I will be using both the nCr and the binomial coefficient notations, whichever is more convenient for typographical reasons. To see how the notation you use corresponds to the ones I use, refer to the following equation:
Understanding the Standard Pack
Before we proceed to discuss card game problems, we need to make sure you understand what a pack of cards (or a deck of cards, depending on where you're from) is like. If you're already familiar with playing cards, you may skip this section.
The standard pack consists of 52 cards, divided into four suits: hearts, tiles (or diamonds), clubs and spades. Among them, hearts and tiles (diamonds) are red, while clubs and spades are black. Each suit has ten numbered cards - A (representing 1), 2, 3, 4, 5, 6, 7, 8, 9 and 10 - and three face cards, Jack (J), Queen (Q) and King (K). The face value is known as the kind. Here is a table with all the cards (colours are missing because of formatting constraints, but the first two columns should be red):
|Kind \ Suit||♥ (Hearts)||♦ (Diamonds)||♠ (Spades)||♣ (Clubs)|
Ace of Hearts
Ace of Diamonds
Ace of Spades
Ace of Clubs
1 of Hearts
1 of Diamonds
1 of Spades
1 of Clubs
2 of Hearts
2 of Diamonds
2 of Spades
2 of Clubs
3 of Hearts
3 of Diamonds
3 of Spades
3 of Clubs
4 of Hearts
4 of Diamonds
4 of Spades
4 of Clubs
5 of Hearts
5 of Diamonds
5 of Spades
5 of Clubs
6 of Hearts
6 of Diamonds
6 of Spades
6 of Clubs
7 of Hearts
7 of Diamonds
7 of Spades
7 of Clubs
8 of Hearts
8 of Diamonds
8 of Spades
8 of Clubs
9 of Hearts
9 of Diamonds
9 of Spades
9 of Clubs
10 of Hearts
10 of Diamonds
10 of Spades
10 of Clubs
Jack of Hearts
Jack of Diamonds
Jack of Spades
Jack of Clubs
Queen of Hearts
Queen of Diamonds
Queen of Spades
Queen of Clubs
King of Hearts
King of Diamonds
King of Spades
King of Clubs
From the above table, we notice the following:
- The sample space has 52 possible outcomes (sample points).
- The sample space can be partitioned in two ways: kind and suit.
A lot of elementary probability problems are based on the above properties.
Simple Card Game Problems
Card games are an excellent opportunity to test a student's understanding of set theory and probability concepts such as union, intersection and complement. In this section, we will only go through probability problems, but the combinatorics problems follow the same principles (just like at the numerators of the fractions).
Before we begin, let me remind you of this theorem (the non-generalised form of the Additive Law of Probability), which will pop up constantly in our card game problems:
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In short, this means the probability of A or B (a disjunction, indicated by the union operator) is the sum of the probabilities of A and B (a conjunction, indicated by the intersection operator). Remember the last part! (There is a complex, generalised form of this theorem, but this is rarely used in card game questions, so we won't discuss it.)
Here is a set of simple card game questions and their answers:
- If we draw a card from a standard pack, what is the probability that we will get a red card with face value smaller than 5 but greater than 2?
Firstly, we enumerate the number of possible face values: 3, 4. There are two types of red cards (diamonds and hearts), so the there are altogether 2 × 2 = 4 possible values. You can check by listing the four favourable cards: 3♥, 4♥ 3♦, 4♦. Then the resulting probability = 4 / 52 = 1 / 13.
- If we draw one card from a standard pack, what is the probability that it is red and 7? How about red or 7?
The first one is easy. There are only two cards that are both red and 7 (7♥, 7♦). The probability is thus 2 / 52 = 1 / 26.
The second one is only slightly harder, and with the above theorem in mind, it should be a piece of cake as well. P(red ∪ 7) = P(red) + P(7) - P(red ∩ 7) = 1 / 2 + 1 / 13 - 1 / 26 = 7 / 13. An alternative method is to count the number of cards that satisfy the constraints. We count the number of red cards, add the number of cards marked 7 and subtract the number of cards which are both: 13 × 2 + 4 - 2 = 28. Then the required probability is 28 / 52 = 7 / 13.
- If we draw two cards from a standard pack, what is the probability that they are of the same suit?
When it comes to drawing two cards from a pack (as with many other probability word problems), there are usually two possible ways to approach the problem: Multiplying the probabilities together using the Multiplicative Law of Probability, or using combinatorics. We will look at both, though the latter option is usually better when it comes to more complex problems, which we'll see below. It's advisable to know both methods so that you can check your answer by employing the other one.
By the first method, the first card can be whatever we want, so the probability is 52 / 52. The second card is more restrictive, however. It must correspond to the suit of the previous card. There are 51 cards left, 12 of which are favourable, so the probability that we'll get two cards of the same suit is (52 / 52) × (12 / 51) = 4 / 17.
We can also use combinatorics to solve this question. Whenever we pick n cards from a pack (assuming the order is not important), there are 52Cn possible choices. Our denominator is thus 52C2 = 1326.
As for the numerator, we first choose the suit, then choose two cards out of that suit. (This line of thought will be used quite often in the next section, so you'd better remember it well.) Our numerator is 4 × 13C2 = 312. Putting it all together, our probability is 312 / 1326 = 4 / 17, confirming our previous answer.
Poker problems are very common in probability, and are harder than the simple question types mentioned above. The most common type of poker question involves choosing a five cards from the pack and asking the student to find the probability of a certain arrangement, called a poker hand. The most common arrangements are discussed in this section.
A word of caution before we continue: When it comes to poker problems, it's always advisable to use combinatorics. There are two main reasons:
- Doing this by multiplying probabilities is a nightmare.
- You'll probably get tested on the combinatorics involved anyway. (In the situation that you do, just take the numerators of the probabilities we've discussed here, if order is not important.)
X of a Kind
X of a Kind problems are self-explanatory - if you have X of a kind, then you have X cards of the same kind on your hand. There are usually two of these: three of a kind and four of a kind. Note that the remaining cards cannnot be of the same kind as the X cards of a kind. For example, 4♠ 4♥ 4♦ 5♦ 4♣ is not considered three of a kind because the last card is not a three of a kind because of the last card. It is, however, a four of a kind.
How do we find the probability of getting a X of a kind? Let's first look at 4 of a kind, which is more simple (as we'll see below). A four of a kind is defined as a hand where there are four cards of the same kind. We employ the same method used for the third question above. First, we choose our kind, then we choose four cards from that kind, and finally we choose the remaining card. There's no real choosing in the second step, since we're choosing four cards from four. The resulting probability:
See why it's a bad idea to gamble?
Three of a kind is slightly more complicated. The last two cannot be of the same kind, or we'll get a different hand called a full house, which will be discussed below. So this is our game plan: Choose three different kinds, pick three cards from one kind and one card from the other two.
Now, there are three ways of doing this. At first glance, they all seem to be correct, but they result in three different values! Obviously, only one of them is true, so which?
I have the answers below, so please don't scroll down until you've thought it over.
The three approaches differ in the way they choose the three kinds.
- The first one chooses the three kinds separately. We are choosing three distinct kinds. If you multiply the three elements where we chose kinds, we get a number equivalent to 13P3.This leads to double counting. For example, A♠ A♥ A♦ 3♦ 4♣ and A♠ A♥ A♦ 4♣ 3♦ are treated as two.
- The second one chooses all three suits together. Thus, the suit chosen to be the 'three of a kind' and the two remaining cards are not distinguished. The probability is thus lower than it should be. For example, A♠ A♥ A 3♦ 4♣ and 3♠ 3♥ 3 A♦ 4♣ are not distinguished and regarded as one and the same.
- The third one is just right. The kind involved in 'three of a kind' and the other two kinds are distinguished.
Remember that if we choose the three sets in three separate steps, we are distinguishing between them. If we choose all of them in the same steps, we aren't distinguishing between any. In this question, the middle ground is the right choice.
Above, we described three of a kind and four of a kind. How about two of a kind? In fact, two of a kind is known as a pair. We can have one pair or two pairs in a hand.
Having gone through three of a kind, one pair and two pairs need no additional explanation, so I will only present the formulae here and leave the explanation as an exercise to the reader. Just note that, like the two hands above, the remaining cards must belong to different kinds.
A hybrid of one pair and three of a kind is full house. Three cards are of a kind and the two remaining cards are of another. Again, you are invited to explain the formula yourself:
Straight, Flush and Straight Flush
The three remaining hands are straight, flush and straight flush (a cross of the two):
- Straight means the five cards are in consecutive order, but not all are in the same suit.
- Flush means the five cards are all in the same suit, but not in consecutive order.
- Straight flush means the five cards are both in consecutive order and in the same suit.
We can start by discussing the probability of flush ∪ straight flush, which is a simple probability. First, we pick the suit, then we pick five cards from it - simple enough:
Straight are only slightly harder. When computing the probability of a straight, we need to note the following order:
A 1 2 3 4 5 6 7 8 9 10 J Q K A
Thus A 1 2 3 4 and 10 J Q K A are both permissible sequences, but Q K A 1 2 is not. There are ten possible sequences in total:
Now, since we are completely disregarding the suits (i.e. there are no constraints), the number of possible suit permutations is 45. The leads us to what is probably our easiest probability yet:
The probability of a straight flush should be obvious at this point. Since there are 4 suits and 10 possible sequences, there are 40 hands classified as straight flush. We can now derive the probabilities of straight and flush, too.
A Final Word
In this article, we have only covered combinations. This is because order is not important in a card game. However, you may still come across permutation-related problems from card to time. They usually requires you to choose cards from the deck without replacement. If you see these questions, don't worry. They're most likely simple permutation questions which you can handle with your statistics prowess.
For example, in the case where you are asked about the number of possible permutations of a particular poker hand, simply multiply the number of combinations by 5!. In fact, you can redo the above probabilities by multiplying the numerators by 5! and replacing 32C5 with 32P5 in the denominator. The probabilities will remain unchanged.
The number of possible card game questions is numerous, and to cover all of them in a single article is impossible. However, the questions I've showed you constitute the most common types of problems in probability exercises and exams. If you have a question, feel free to ask in the comments. Other readers and I may be able to help you out. If you liked this article, consider sharing it on social media and voting on the poll below so I know what article to write next. Thanks!
Note: John E Freund's Mathematical Statistics
John E Freund's book is an excellent introductory statistics book that explains the basics of probability in lucid and accessible prose. If you had difficulty understanding what I've written above, you are encouraged to read the first two chapters of this book before coming back.
You are also encouraged to try out the exercises in the book after reading my articles. The theory questions really get you thinking about statistics ideas and concepts, while application problems - the ones you'll most likely see in your exams - allow you to gain hands-on experience with a wide range of question types. You may buy the book by following the link below if needed. (There's a catch - answers are only provided for odd-numbered questions - but this is unfortunately true of the vast majority of college-level textbooks.)
A Quick Poll
Daniell on July 17, 2020:
I think there is a mistake in the second paragraph of 'Final Word'. 32C5 and 32P5 should read 52C5 and 52P5 for the denominator.
Keith Keithy on March 14, 2019:
The advanced probability questions from the mathematical statistics and data analysis, by John A. rice. my email is email@example.com please give advice or links to resources if you can, your article has opened my eyes thank you
Adam Cartisano on July 10, 2018:
Draw the top 10 cards of a standard deck of playing cards. What is the probability of drawing exactly 3 hearts and exactly 2 face cards if a card which is both a heart and a face card can only count as one or the other?