Statistics Probability Combinatorics

Simple Card Game Problems

In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. So we adjust our permutations formula to reduce it by how many ways the objects could be in order because we aren't interested in their order any more:. In other words choosing 3 balls out of 16, or choosing 13 balls out of 16 have the same number of combinations. We can also use Pascal's Triangle to find the values.

Go down to row "n" the top row is 0 , and then along "r" places and the value there is our answer. Here is an extract showing row Let us say there are five flavors of icecream: And just to be clear: Order does not matter, and we can repeat! Now, I can't describe directly to you how to calculate this, but I can show you a special technique that lets you work it out. Think about the ice cream being in boxes, we could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and we will have 3 scoops of chocolate!

So it is like we are ordering a robot to get our ice cream, but it doesn't change anything, we still get what we want. We can write this down as arrow means move , circle means scoop.

From Probability to Combinatorics and Number Theory

OK, so instead of worrying about different flavors, we have a simpler question: Notice that there are always 3 circles 3 scoops of ice cream and 4 arrows we need to move 4 times to go from the 1st to 5th container. In other words it is now like the pool balls question, but with slightly changed numbers. And we can write it like this:. But knowing how these formulas work is only half the battle. Figuring out how to interpret a real world situation can be quite hard. Hide Ads About Ads. Combinations and Permutations What's the Difference? So, in Mathematics we use more precise language: When the order doesn't matter, it is a Combination.

When the order does matter it is a Permutation. So, we should really call this a "Permutation Lock"!

Combinatorics, Probability and Computing - Wikipedia

Note that the remaining cards cannnot be of the same kind as the X cards of a kind. 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. 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? 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:. 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:. There are ten possible sequences in total:. Now, since we are completely disregarding the suits i. 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.

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! 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.

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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. 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. Sign in or sign up and post using a HubPages Network account. Comments are not for promoting your articles or other sites.

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? Other product and company names shown may be trademarks of their respective owners.

HubPages and Hubbers authors may earn revenue on this page based on affiliate relationships and advertisements with partners including Amazon, Google, and others. To provide a better website experience, owlcation. Please choose which areas of our service you consent to our doing so. For more information on managing or withdrawing consents and how we handle data, visit our Privacy Policy at: Cracking Probability and Combinatorics: Card Game Problems Updated on September 14, Ryan K Y Lai more. 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 n C r 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.

Combinations and Permutations

Here is a table with all the cards colours are missing because of formatting constraints, but the first two columns should be red: 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: 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.

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: 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: You can check by listing the four favourable cards: 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. The second one is only slightly harder, and with the above theorem in mind, it should be a piece of cake as well. 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: 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.

Permutations

It's advisable to know both methods so that you can check your answer by employing the other one. The second card is more restrictive, however. It must correspond to the suit of the previous card. We can also use combinatorics to solve this question.