Finding combinations involving only face cards

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.

How many 5 card hands with at least 3 black cards Combination

The advanced probability questions from the mathematical statistics and data analysis, by John A. 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. HubPages Inc, a part of Maven Inc. As a user in the EEA, your approval is needed on a few things. To provide a better website experience, owlcation. Please choose which areas of our service you consent to our doing so. 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.

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.

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. 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. 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. 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.

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Whenever we pick n cards from a pack assuming the order is not important , there are 52 C n possible choices. 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. Poker Problems Poker problems are very common in probability, and are harder than the simple question types mentioned above. 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. See why it's a bad idea to gamble? 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.

Answer to Puzzle # N Face Up Cards In A Dark Room

We are choosing three distinct kinds. If you multiply the three elements where we chose kinds, we get a number equivalent to 13 P 3.


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This leads to double counting. 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. The third one is just right. The kind involved in 'three of a kind' and the other two kinds are distinguished. Pairs Above, we described three of a kind and four of a kind. 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.

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. 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. 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.

Answer to Puzzle #36: N Face Up Cards In A Dark Room

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. 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. 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. Whenever we pick n cards from a pack assuming the order is not important , there are 52 C n possible choices. 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.

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Poker Problems Poker problems are very common in probability, and are harder than the simple question types mentioned above. There are two main reasons: Doing this by multiplying probabilities is a nightmare. You'll probably get tested on the combinatorics involved anyway.

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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. See why it's a bad idea to gamble?


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  • 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.