Section 5.6: More on Probability
- Page ID
- 215612
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- Use sample spaces and compute probabilities for drawing a card from a standard deck of cards
In Section 5.1, we explored the fundamental concepts of probability. We worked with simple scenarios like coin flips, rolling a single die, spinner wheels, and drawing colored marbles from bags. These examples helped us understand the core principles without getting bogged down in complexity. We could easily list all outcomes, count them, and calculate probabilities.
Now it's time to apply these foundational principles to situations you encounter in everyday life, specifically, dice games and card games. Whether you're playing Monopoly with family, enjoying poker with friends, or simply trying to understand the odds in a casino, dice and cards are universal tools of chance that have entertained humanity for centuries.
Why Dice and Cards?
Why focus on dice and cards specifically? Here are several compelling reasons:
- Accessibility - Nearly everyone has played with dice or cards at some point. These aren't abstract mathematical objects, they're tangible items you can hold and experiment with.
- Fair and Random - Both dice and standard playing cards are designed to be fair. Each face of a die is equally likely to appear, and each card in a shuffled deck is equally likely to be drawn. This makes them perfect for studying probability.
- Rich Mathematical Structure - Despite their simplicity, dice and cards generate complex probability problems that illustrate important concepts like independent events, conditional probability, and counting techniques.
- Cultural Relevance - Dice and card games appear across cultures and throughout history. Understanding their mathematics connects you to a rich tradition of games, gambling, and probability theory itself (which was invented partly to analyze gambling games).
- Practical Application - Many board games use dice, and card games are popular worldwide. Understanding the probabilities involved makes you a more informed player and helps you develop strategic thinking.
The good news is that all the principles you've already learned still apply. We're not learning new rules of probability, instead we're simply applying the same rules to richer, more interesting scenarios. Think of it like learning to cook. First, you learned basic techniques: how to chop, sauté, and season. Now we're moving from simple recipes (scrambled eggs) to more complex dishes (soufflés). The techniques are the same; we're just combining them in more sophisticated ways.
When analyzing the probability of rolling two dice, it is important to distinguish between the possible sums and the actual sample space of the experiment. While many applications focus on the sum of the two dice showing face up, which produces eleven possible results (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), these eleven outcomes are not equally likely to occur. Each sum does not have a probability of 1/11 because some sums can be achieved in more ways than others. For instance, there are significantly more ways to obtain a sum of 7 than a sum of 2 or 12.
To properly analyze the probabilities associated with rolling two dice, we must first identify the complete sample space of the experiment. Each die has 6 possible outcomes (1, 2, 3, 4, 5, 6), and since the dice are rolled independently, we can apply the Fundamental Counting Principle. The first die contributes 6 possible outcomes, and the second die contributes 6 possible outcomes. Therefore, the total number of equally likely outcomes in the sample space is 6 × 6 = 36.
These 36 outcomes represent all possible ordered pairs (first die, second die) that can occur when rolling two distinguishable dice. To clearly illustrate this complete sample space and demonstrate why certain sums are more likely than others, we will organize these 36 outcomes in a systematic table format.
| Die #2: 1 | Die #2: 2 | Die #2: 3 | Die #2: 4 | Die #2: 5 | Die #2: 6 | |
|---|---|---|---|---|---|---|
| Die #1: 1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
| Die #1: 2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
| Die #1: 3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
| Die #1: 4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
| Die #1: 5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
| Die #1: 6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
- The rows represent the outcomes of the first die (1 through 6)
- The columns represent the outcomes of the second die (1 through 6)
- Each cell in the grid contains an ordered pair (first die, second die)
To find any specific outcome, locate the row corresponding to the first die's value and the column corresponding to the second die's value. The intersection of that row and column shows the ordered pair representing that outcome.
A pair of single six-sided die is rolled. Answer the following questions.
- What is the probability a sum of \(7\) is rolled?
- What is the probability a sum of \(3\) is rolled?
- What is the probability a sum of \(8\) is rolled?
- What is the probability a sum of \(13\) is rolled?
- What is the probability a sum of at least \(11\) is rolled?
- What is the probability a sum of at most \(5\) is rolled?
- What is the probability there is at least one '\(4\)' on either or both dies rolled?
- What is the probability that both dice are the same?
✅ Solution:
So, we look at the table for the outcomes of rolling two dice.
- There are six outcomes that make a sum of \(7\). They are \((1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)\). Thus, the probability is \(\frac{6}{36}=\frac{1}{6}\).
- There are two outcomes that make a sum of \(3\). They are \((1, 2), (2, 1)\). Thus, the probability is \(\frac{2}{36}=\frac{1}{18}\).
- There are five outcomes that make a sum of \(8\). They are \((2, 6), (6, 2), (3, 5), (5, 3), (4, 4)\). Thus, the probability is \(\frac{5}{36}\).
- There are no (zero) outcomes that make a sum of \(13\). Thus, the probability is \(\frac{0}{36}=0\).
- At least \(11\) means \(11\) or more. There are three outcomes that make a sum of at least \(11\). They are \((5, 6), (6, 5), (6, 6)\). Thus, the probability is \(\frac{3}{36}=\frac{1}{12}\).
- At most \(5\) means \(5\) or less. There are ten outcomes that make a sum of at most \(5\). They are \((1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 2) (1, 4), (4, 1), (2, 3), (3, 2)\). Thus, the probability is \(\frac{10}{36}=\frac{5}{18}\).
- There are eleven outcomes that make at least one '\(4\)' on either or both dies. They are \(\text(1, 4), (2, 4), (3, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (4, 4)\). Thus, the probability is \(\frac{11}{36}\).
- There are six outcomes that make both dice the same. They are \((1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\). Thus, the probability is \(\frac{6}{36}=\frac{1}{6}\).
Now that we understand the basics of probability, let's explore one of the most common and useful applications: calculating probabilities using a standard deck of playing cards. Card problems appear frequently in probability because they provide a perfect example of a well-defined sample space with known outcomes.
A standard deck of playing cards contains exactly 52 different cards, making it an ideal tool for probability calculations. The deck is organized using two classification systems that work together to create each unique card. First, there are 4 different suits, which are the symbols that appear on each card: hearts (♥), diamonds (♦), clubs (♣), and spades (♠). Second, there are 13 different ranks, which represent the values or names of the cards. When we combine each of the 4 suits with each of the 13 ranks, we get our complete set of 52 distinct cards (4 × 13 = 52).
Understanding this structure is essential for solving card probability problems because it helps us count favorable outcomes and determine the total sample space. The systematic organization of suits and ranks also allows us to identify patterns and relationships between different types of cards, which becomes crucial when calculating probabilities for various events involving card draws. Below is the sample space for a standard deck of playing cards.
For each of the following questions, draw a single card from a standard deck of playing cards.
- What is the probability a heart is drawn?
- What is the probability a red card is drawn?
- What is the probability a '\(7\)' is drawn?
- What is the probability a '\(5\)' of clubs is drawn?
- What is the probability a '\(9\)' of moons is drawn?
- What is the probability a '\(6\)' or a queen is drawn?
- What is the probability an ace or spades is drawn?
- What is the probability a '\(4\)' or a black card is drawn?
✅ Solution:
So, we look at the sample space for the \(52\) outcomes of a standard deck of playing cards.
- There are \(13\) hearts. Thus, the probability is \(\frac{13}{52}=\frac{1}{4}\).
- There are \(26\) red cards. Thus, the probability is \(\frac{26}{52}=\frac{1}{2}\).
- There are four '\(7\)'s. Thus, the probability is \(\frac{4}{52}=\frac{1}{13}\).
- There is only one '\(5\)' of clubs. Thus, the probability is \(\frac{1}{52}=\).
- There are no suits called moons. Thus, the probability is \(\frac{0}{52}=0\).
- There are four '\(6\)'s and four queens. Thus, the probability is \(\frac{8}{52}=\frac{2}{13}\).
- There are \(4\) aces and there are \(12\) spades not including the ace of spades already counted. Thus, the probability is \(\frac{16}{52}=\frac{4}{13}\).
- There are four '\(4\)'s and there are \(24\) black cards not including the '\(4\)'s already counted. Thus, the probability is \(\frac{28}{52}=\frac{7}{13}\).
Ancient Beginnings
The exact origin of playing cards remains somewhat mysterious, but historians generally agree that playing cards originated in China during the Tang Dynasty (618-907 CE). These early cards were likely derived from ancient Chinese "money cards" or "domino cards" and were used for both games and fortune-telling.
From China, playing cards spread along trade routes, reaching:
- Persia and Arabia (by the 11th century), where they evolved into new forms
- Egypt (by the 12th-13th centuries), where Mamluk cards featured suits of cups, coins, swords, and polo sticks
- Europe (by the late 14th century), arriving through trade with the Islamic world and Mediterranean commerce
Evolution in Europe
When playing cards reached Europe in the 1370s, they quickly became popular among all social classes. Different regions developed their own suit systems:
- Italian suits: Cups, Coins, Swords, Batons
- German suits: Hearts, Bells, Leaves, Acorns
- Spanish suits: Cups, Coins, Swords, Clubs
- Swiss suits: Shields, Roses, Acorns, Bells
The French Standard
The playing cards we use today—the French deck—emerged in the late 15th century in France. French cardmakers made several innovations:
- Simplified suit symbols: They created the four suits we know today:
- Hearts (♥) - representing the clergy and love
- Diamonds (♦) - representing merchants and wealth
- Clubs (♣) - representing peasants and agriculture
- Spades (♠) - representing the military and nobility
- Standardized court cards: Kings, Queens, and Jacks (originally called Knaves)
- Efficient design: The French suits were easier and cheaper to reproduce using stencils, giving them an economic advantage
- Reversible face cards: By the 19th century, cards were printed with images that looked right-side-up from either direction
Why This Design Succeeded
The French deck became the international standard for several reasons:
- Economic efficiency: Simple suit symbols reduced production costs
- Clarity: The four distinct suit symbols were easy to distinguish
- Colonial spread: French and British colonization spread this deck worldwide
- Printing advances: The design worked well with mass production techniques
- Gaming versatility: The structure supported countless game variations
The English Contribution
When the French deck reached England in the late 15th century, the English made their own modifications:
- Changed the French term "Valet" to "Jack" (or "Knave")
- Added indices (the small numbers and suit symbols in the corners) in the 19th century
- Standardized the court card designs we recognize today
- Introduced the Joker around 1860 (originally for the game of Euchre)
The Anglo-American deck is essentially the French deck with English naming conventions and design refinements.
Connection to Probability Theory
Historically, card games (along with dice games) motivated the development of probability theory:
- Gerolamo Cardano (1501-1576) studied dice and card games, writing the first book on probability
- Blaise Pascal and Pierre de Fermat (1654) developed probability theory while analyzing gambling problems
- Christian Huygens (1657) published the first probability textbook, inspired by games of chance
- Modern combinatorics grew partly from analyzing card and dice probabilities
When you study card probability, you're following in the footsteps of mathematical giants who were themselves fascinated by these same questions.
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A pair of single six-sided die is rolled. Answer the following questions.
- What is the probability a sum of \(2\) is rolled?
- What is the probability a sum of \(11\) is rolled?
- What is the probability a sum of \(9\) is rolled?
- What is the probability a sum of \(1\) is rolled?
- What is the probability a sum of at least \(10\) is rolled?
- What is the probability a sum of at most \(3\) is rolled?
- What is the probability there is at least one '\(5\)' on either or both dies rolled?
- What is the probability that both dice are different?
- What is the probability that the sum is even?
- What is the probability that the sum is a prime number? (A prime number is a number, other than 1, that is divisible by 1 and itself).
- For each of the following questions, draw a single card from a standard deck of playing cards.
- What is the probability a black card is drawn?
- What is the probability a diamond is drawn?
- What is the probability a '\(9\)' is drawn?
- What is the probability a '\(5\)' of spades is drawn?
- What is the probability a '\(3\)' or a jack is drawn?
- What is the probability an '\(8\)' of cups is drawn?
- What is the probability a king of diamonds is drawn?
- What is the probability a red card or a heart is drawn?
- What is the probability a '\(10\)' or a red card is drawn?
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A pair of single five-sided die with outcomes {1, 2, 3, 4, 5} is rolled. Answer the following questions.
- What is the probability a sum of \(6\) is rolled?
- What is the probability a sum of \(10\) is rolled?
- What is the probability a sum of at least \(4\) is rolled?
- What is the probability a sum of \(12\) is rolled?
- Answers
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- a) \(\frac{1}{36}\); b); \(\frac{2}{36}=frac{1}{18}\); c) \(\frac{4}{36}=frac{1}{9}\); d) \(\frac{0}{36}=0\); e); \(\frac{6}{36}=frac{1}{6}\); f) \(\frac{3}{36}=frac{1}{12}\); g) \(\frac{11}{36}\); h) \(\frac{30}{36}=frac{5}{6}\); i) \(\frac{18}{36}=frac{1}{2}\); j) \(\frac{15}{36}=frac{5}{12}\).
- a) \(\frac{26}{52}=frac{1}{2}\); b); \(\frac{13}{52}=frac{1}{4}\); c) \(\frac{4}{52}=frac{1}{13}\); d) \(\frac{1}{52}\); e); \(\frac{8}{52}=frac{2}{13}\); f) \(\frac{0}{52}=0\); g) \(\frac{1}{52}\); h) \(\frac{39}{52}=frac{3}{4}\); i) \(\frac{28}{52}=frac{7}{13}\).
- a) \(\frac{5}{25}=frac{1}{5}\); b); \(\frac{1}{25}\); c) \(\frac{22}{25}\); d) \(\frac{0}{52}=0\).