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A bridge deck has 52 cards with 13 cards in each of four suits: spades, hearts, diamonds, and clubs. A hand of 13 cards is dealt from a shuffled deck. Find the probability that the hand has (a) a distribution of suits 4,4,3,2 (for example, four spades, four hearts, three diamonds, two clubs). (b) a distribution of suits 5,3,3,2 .

Short Answer

Expert verified
(a) 0.0113 (b) 0.0092

Step by step solution

01

Understand the Setup

We need to calculate the probability of being dealt a 13-card hand with specified distributions of suits from a standard deck of 52 cards. The deck and hand are defined by the problem statement.
02

Calculate Total Number of Hands

The total number of different 13-card hands possible from a 52-card deck can be calculated using combinations: \[\binom{52}{13} = 635,013,559,600\]
03

Step 3a: Calculate Suit Distribution 4,4,3,2

To find a specific distribution of suits (4,4,3,2), select 4 spades from 13, 4 hearts from 13, 3 diamonds from 13, and 2 clubs from 13. The number of ways to do this is:\[\binom{13}{4} \times \binom{13}{4} \times \binom{13}{3} \times \binom{13}{2} = 7,195,680\]
04

Step 4a: Calculate Probability for 4,4,3,2

The probability of getting a 4,4,3,2 distribution is the number of favorable hands divided by the total possible hands.\[P(4,4,3,2) = \frac{7,195,680}{635,013,559,600} \approx 0.0113\]
05

Step 3b: Calculate Suit Distribution 5,3,3,2

To find a specific distribution of suits (5,3,3,2), select 5 cards from one suit, 3 from another, 3 from the next, and finally 2 from the last suit. Calculate the number of ways:\[\binom{13}{5} \times \binom{13}{3} \times \binom{13}{3} \times \binom{13}{2} = 5,825,136\]
06

Step 4b: Calculate Probability for 5,3,3,2

The probability of getting a 5,3,3,2 distribution is the number of favorable hands divided by the total possible hands.\[P(5,3,3,2) = \frac{5,825,136}{635,013,559,600} \approx 0.0092\]

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Key Concepts

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

Combinatorics
Combinatorics involves studying the arrangement of objects in specific sequences or combinations. In card games, combinatorics helps us calculate the number of possible hands or distributions. When you have a set number of items, combinatorics can tell you how many different ways you can pick these items without caring about the order.

For example, in calculating card hands, you often use combinations denoted by \( \binom{n}{k} \). This represents the number of ways to choose \( k \) items from \( n \) items without regard to order. It can be calculated using the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
  • \( n! \) (n factorial) means multiplying all positive integers up to \( n \).
  • Combinations do not consider the sequence of cards, just the group.
Using combinations allows us to find the number of ways certain hands can appear in deck games like bridge, poker, and others. It’s essential in calculating probabilities for card distributions.
Card Suits
A typical deck of cards is divided into four suits: spades, hearts, diamonds, and clubs. Each of these suits contains 13 unique cards, ranging from Ace to King. Understanding the basic structure of a deck is crucial when calculating probabilities involving card games.

Understanding card suits involves recognizing:
  • The suit itself: Each represents a different category, and in games like bridge, they can have varying levels of significance or power.
  • The count of each suit in a deck: 13 cards per suit, resulting in 52 cards in total in a standard deck.
In probability problems, such as determining the likelihood of specific suit distributions, knowing the number of cards in each suit is critical. When a hand is dealt, understanding suits helps compute the potential distributions possible.
Probability Distributions
Probability distributions describe how likely different outcomes are when an experiment or random event occurs. In card dealing, we often look at probability distributions to describe how cards (or particular characteristics like suits) are drawn.For example, when dealing hands in a card game, you want to calculate the probability of receiving certain distributions such as the 4,4,3,2 and 5,3,3,2 suit distribution mentioned in the exercise. To do this:
  • Calculate the total number of possible outcomes (all possible card hands).
  • Find the number of favorable outcomes (hands that match the desired suit distribution).
  • Divide the favorable outcomes by the total number of outcomes to find the probability.
The formula for probability is:\[P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]In these kinds of problems, understanding how to work with these distributions is key to solving complex probability questions. Distributions help you see how likely different configurations of deals are within a given framework.

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Most popular questions from this chapter

You are presented with four different dice. The first one has two sides marked 0 and four sides marked \(4 .\) The second one has a 3 on every side. The third one has a 2 on four sides and a 6 on two sides, and the fourth one has a 1 on three sides and a 5 on three sides. You allow your friend to pick any of the four dice he wishes. Then you pick one of the remaining three and you each roll your die. The person with the largest number showing wins a dollar. Show that you can choose your die so that you have probability \(2 / 3\) of winning no matter which die your friend picks. (See Tenney and Foster. \(\left.^{8}\right)\)

A number \(U\) is chosen at random in the interval \([0,1] .\) Find the probability that (a) \(R=U^{2}<1 / 4\). (b) \(S=U(1-U)<1 / 4\) (c) \(T=U /(1-U)<1 / 4\)

On the average, only 1 person in 1000 has a particular rare blood type. (a) Find the probability that, in a city of 10,000 people, no one has this blood type. (b) How many people would have to be tested to give a probability greater than \(1 / 2\) of finding at least one person with this blood type?

A worker for the Department of Fish and Game is assigned the job of estimating the number of trout in a certain lake of modest size. She proceeds as follows: She catches 100 trout, tags each of them, and puts them back in the lake. One month later, she catches 100 more trout, and notes that 10 of them have tags. (a) Without doing any fancy calculations, give a rough estimate of the number of trout in the lake. (b) Let \(N\) be the number of trout in the lake. Find an expression, in terms of \(N,\) for the probability that the worker would catch 10 tagged trout out of the 100 trout that she caught the second time. (c) Find the value of \(N\) which maximizes the expression in part (b). This value is called the maximum likelihood estimate for the unknown quantity N. Hint: Consider the ratio of the expressions for successive values of \(N\).

A manufactured lot of brass turnbuckles has \(S\) items of which \(D\) are defective. A sample of \(s\) items is drawn without replacement. Let \(X\) be a random variable that gives the number of defective items in the sample. Let \(p(d)=P(X=d)\) (a) Show that $$ p(d)=\frac{\left(\begin{array}{l} D \\ d \end{array}\right)\left(\begin{array}{l} S-D \\ s-d \end{array}\right)}{\left(\begin{array}{l} S \\ s \end{array}\right)} $$ Thus, \(\mathrm{X}\) is hypergeometric. (b) Prove the following identity, known as Euler's formula: $$ \sum_{d=0}^{\min (D, s)}\left(\begin{array}{l} D \\ d \end{array}\right)\left(\begin{array}{l} S-D \\ s-d \end{array}\right)=\left(\begin{array}{l} S \\ s \end{array}\right) $$

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