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Five cards are dealt from a standard 52 -card deck. What is the probability that we draw a. 3 aces and 2 kings? b. a "full house" (3 cards of one kind, 2 cards of another kind)?

Short Answer

Expert verified
a) \( \frac{1}{108290} \); b) \( \frac{1}{694} \).

Step by step solution

01

Calculate number of ways to draw 3 aces

There are 4 aces in a deck of 52 cards. To choose 3 aces, the combination formula \( \binom{n}{k} \) is used, where \( n \) is the total number of aces and \( k \) is the number of aces to choose: \( \binom{4}{3} = 4 \).
02

Calculate number of ways to draw 2 kings

There are 4 kings in a deck. To choose 2 kings from these 4, we use the combination formula again: \( \binom{4}{2} = 6 \).
03

Compute total ways to draw 3 aces and 2 kings

The number of ways to draw 3 aces and 2 kings is the product of the ways to choose the aces and the ways to choose the kings: \( 4 \times 6 = 24 \).
04

Calculate total ways to draw any 5 cards

The number of ways to draw any 5 cards from a 52-card deck is \( \binom{52}{5} = 2,598,960 \).
05

Calculate probability of drawing 3 aces and 2 kings

The probability is the ratio of the favorable outcomes to the total outcomes: \( \frac{24}{2,598,960} = \frac{1}{108290} \).
06

Calculate ways to form 3 cards of one kind

Choose a rank for the 3 cards and select 3 out of the 4 available cards of that rank. There are 13 possible ranks, and \( \binom{4}{3} = 4 \) ways to choose the cards: \( 13 \times 4 = 52 \).
07

Calculate ways to form 2 cards of another kind

Next, choose a different rank for the 2 cards and select 2 out of 4 available cards. There are 12 ranks left and \( \binom{4}{2} = 6 \) ways to choose the cards: \( 12 \times 6 = 72 \).
08

Compute total ways to draw a full house

Multiply the ways to choose the 3 cards by the ways to choose the 2 cards: \( 52 \times 72 = 3,744 \).
09

Calculate probability of drawing a full house

The probability is the ratio of favorable ways to the total 5-card combinations: \( \frac{3,744}{2,598,960} = \frac{1}{694} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects. It's essential for calculating probabilities in card games.

For instance, when determining how many ways we can select three aces from four available aces in a deck, we use combinatorics. The formula used for these calculations is the combination formula notated as \( \binom{n}{k} \). Here, \( n \) represents the total number of items to select from, and \( k \) denotes the number of items to select.

Using this formula, we calculate how many ways we can pick a set number of cards from a full deck. For example, choosing 3 aces from 4 aces is calculated as \( \binom{4}{3} = 4 \). This applies similarly when calculating the probability of selecting 2 kings from a set of 4, which is \( \binom{4}{2} = 6 \).

These calculations help in understanding the basis of probability in card games and other forms of combinatorial probability computations.
Deck of Cards
A standard deck of playing cards consists of 52 cards divided into 4 suits: hearts, clubs, diamonds, and spades. Each suit has 13 ranks, ranging from the number 2 to 10, and then the face cards: jack, queen, king, and ace.

Knowing the structure of a deck is crucial when calculating probabilities. Each deck ensures equal representation of suits and ranks, making probability calculations straightforward. For example, when checking the probability of drawing a specific set of cards like 3 aces and 2 kings, knowing the composition of the deck beforehand is vital.

Understanding the deck's structure allows us to calculate scenarios such as drawing any 5 cards from the deck using \( \binom{52}{5} \), which results in 2,598,960 possible outcomes. These basics enable students to dive deeper into more complex probability and combinatorics problems related to card games.
Full House
In card games like poker, a "full house" is a hand consisting of three cards of one rank and two cards of another rank. Calculating the probability of drawing a full house involves understanding not only the concept but also the combinatorial arrangements.

The process begins by selecting a rank for the three cards. There are 13 ranks to choose from, and once a rank is chosen, there are \( \binom{4}{3} = 4 \) ways to select 3 cards from it. This gives a total of \( 13 \times 4 = 52 \) ways.

After selecting the three cards, a different rank is chosen for the pair. With 12 ranks left and \( \binom{4}{2} = 6 \) ways to select 2 cards, this results in \( 12 \times 6 = 72 \) possible selections.

Finally, multiplying the ways of choosing the triplet and the pair, \( 52 \times 72 = 3,744 \) gives the total number of full house combinations. The probability is then found by dividing this number by the total possible 5-card hands from the deck, \( 2,598,960 \), resulting in \( \frac{1}{694} \). Understanding how to compute such probabilities is essential for mastering strategic decision-making in card games.

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Most popular questions from this chapter

The following game was played on a popular television show. The host showed a contestant three large curtains. Behind one of the curtains was a nice prize (maybe a new car) and behind the other two curtains were worthless prizes (duds). The contestant was asked to choose one curtain. If the curtains are identified by their prizes, they could be labeled \(G, D_{1},\) and \(D_{2}\) (Good Prize, Dud1, and Dud2). Thus, the sample space for the contestants choice is \(S=\left\\{G, D_{1}, D_{2}\right\\}\). a. If the contestant has no idea which curtains hide the various prizes and selects a curtain at random, assign reasonable probabilities to the simple events and calculate the probability that the contestant selects the curtain hiding the nice prize. b. Before showing the contestant what was behind the curtain initially chosen, the game show host would open one of the curtains and show the contestant one of the duds (he could always do this because he knew the curtain hiding the good prize). He then offered the contestant the option of changing from the curtain initially selected to the other remaining unopened curtain. Which strategy maximizes the contestant's probability of winning the good prize: stay with the initial choice or switch to the other curtain? In answering the following sequence of questions, you will discover that, perhaps surprisingly, this question can be answered by considering only the sample space above and using the probabilities that you assigned to answer part (a). i. If the contestant choses to stay with her initial choice, she wins the good prize if and only if she initially chose curtain \(G\). If she stays with her initial choice, what is the probability that she wins the good prize? ii. If the host shows her one of the duds and she switches to the other unopened curtain, what will be the result if she had initially selected \(G\) ? iii. Answer the question in part.(ii) if she had initially selected one of the duds. iv. If the contestant switches from her initial choice (as the result of being shown one of the duds), what is the probability that the contestant wins the good prize? v. Which strategy maximizes the contestant's probability of winning the good prize: stay with the initial choice or switch to the other curtain?

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