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Chapter 4: Problem 46

A poker hand consists of 5 cards dealt from a deck of 52 cards. Let \(X\) and \(Y\) be, respectively, the number of aces and kings in a poker hand. Find the joint distribution of \(X\) and \(Y\).

Short Answer

Expert verified
The joint distribution of \(X\) and \(Y\) is given by \(P(X = x, Y = y) = \frac{\binom{4}{x} \binom{4}{y} \binom{44}{5-x-y}}{\binom{52}{5}}\) for valid \(x\) and \(y\).

Step by step solution

01

Determine the Sample Space

A poker hand consists of 5 cards chosen from a standard deck of 52 cards. The number of possible 5-card hands can be calculated using the combination formula, which is \binom{n}{k}, where \(n\) is the total number of cards and \(k\) is the number of cards chosen. Therefore, \(\binom{52}{5}\) is the number of different poker hands.
02

Identify Possible Values for X and Y

There are 4 aces and 4 kings in the deck, so the value of \(X\) (number of aces) can be 0, 1, 2, 3, or 4, and similarly, the value of \(Y\) (number of kings) can be 0, 1, 2, 3, or 4. However, \(X + Y \leq 5\) because the total number of cards in a hand cannot exceed 5.
03

Calculate Probabilities for Joint Distribution

For each pair of values \( (x, y) \), where \(x\) represents the number of aces and \(y\) represents the number of kings, calculate the probability \(P(X = x, Y = y)\). This is done by selecting \(x\) aces, \(y\) kings, and the remaining \(5-x-y\) cards from non-ace, non-king cards:- Choose \(x\) aces from 4: \(\binom{4}{x}\)- Choose \(y\) kings from 4: \(\binom{4}{y}\)- Choose \(5-x-y\) cards from the remaining 44 cards: \(\binom{44}{5-x-y}\)The probability of \(P(X = x, Y = y)\) is determined by dividing the product by the total combinations: \[P(X = x, Y = y) = \frac{\binom{4}{x} \binom{4}{y} \binom{44}{5-x-y}}{\binom{52}{5}}\]
04

Compile Joint Distribution Table

Create a table where each row corresponds to a value of \(X\) and each column corresponds to a value of \(Y\). Using the formula derived in Step 3, calculate the probability for each combination of \(x\) and \(y\).

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Key Concepts

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

Probability
Probability is the field of mathematics that deals with the likelihood or chance of different outcomes occurring. In simple terms, it quantifies how likely an event is to happen. For instance, when dealing a poker hand, you might be interested in the likelihood of being dealt certain cards.

To find probabilities, one often divides the number of successful outcomes by the total number of possible outcomes. For example, the probability of drawing a specific card from a deck is 1 out of 52, as there are 52 unique cards. Probability values range between 0 and 1, where 0 means the event is impossible, and 1 means it is certain.
  • A probability of 0.5 indicates there's an equal chance for the event to occur or not occur.
  • Cumulative probability considers the likelihood of multiple, individual events occurring.
Understanding probability helps in areas like determining the chance of drawing aces or kings in a poker hand, as seen with joint distribution in our exercise.
Combinatorics
Combinatorics is a branch of mathematics focusing on counting, arranging, and analyzing possibilities. Combinatorial methods are especially useful in situations like figuring out the number of possible poker hands. Knowing how to count correctly is crucial in determining probabilities.

The combination formula \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]is employed to find out how many ways you can choose \(k\) items from \(n\) items without considering the order. This helps in calculating setups like poker hands where order doesn't matter. When dealing with a standard deck, if you want to know how many ways you can select 5 cards, you'd use: \[\binom{52}{5}\]

Understanding combinatorics ensures that every possibility is evaluated and taken into account, crucial for calculating joint distributions and probabilities in poker.
Poker Hand
A poker hand is a set of 5 cards drawn from a standard deck of 52. Each player is usually dealt one such hand in a game. Hence, understanding poker hands is critical in applying probabilities effectively.

In our exercise, we are especially concerned with hands containing specific numbers of aces and kings.
  • There are 4 aces and 4 kings in a complete deck.
  • The sum of aces and kings in any hand cannot exceed 5, since a poker hand only contains 5 cards.
By understanding these constraints, you can better evaluate the likelihood of receiving certain combinations, particularly for creating a joint probability distribution of aces and kings.
Random Variables
In probability theory, random variables are used to describe outcomes of certain random processes. They are variables whose possible values are numerical outcomes of a random phenomenon. In the context of poker, the number of aces or kings you receive in a hand can be thought of as random variables.

For our exercise, the random variables \(X\) and \(Y\) represent the number of aces and kings respectively in a poker hand. These can take values between 0 and 4, constrained by the maximum hand size of 5.
  • Random variables help quantify outcomes and allow for probability calculations.
  • They are essential for forming the joint distribution table, which helps visualize and calculate the probability of obtaining different combinations of aces and kings.
By understanding random variables, one can better grasp the underlying probability distributions and make informed predictions about poker hand outcomes.

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Most popular questions from this chapter

(Chung \(^{9}\) ) In London, half of the days have some rain. The weather forecaster is correct \(2 / 3\) of the time, i.e., the probability that it rains, given that she has predicted rain, and the probability that it does not rain, given that she has predicted that it won't rain, are both equal to \(2 / 3 .\) When rain is forecast, Mr. Pickwick takes his umbrella. When rain is not forecast, he takes it with probability \(1 / 3\). Find (a) the probability that Pickwick has no umbrella, given that it rains. (b) the probability that he brings his umbrella, given that it doesn't rain.

In Exercise 2.2 .12 you proved the following: If you take a stick of unit length and break it into three pieces, choosing the breaks at random (i.e., choosing two real numbers independently and uniformly from [0,1]\()\), then the probability that the three pieces form a triangle is \(1 / 4\). Consider now a similar experiment: First break the stick at random, then break the longer piece at random. Show that the two experiments are actually quite different, as follows: (a) Write a program which simulates both cases for a run of 1000 trials, prints out the proportion of successes for each run, and repeats this process ten times. (Call a trial a success if the three pieces do form a triangle.) Have your program pick \((x, y)\) at random in the unit square, and in each case use \(x\) and \(y\) to find the two breaks. For each experiment, have it plot \((x, y)\) if \((x, y)\) gives a success. (b) Show that in the second experiment the theoretical probability of success is actually \(2 \log 2-1\)

One coin in a collection of 65 has two heads. The rest are fair. If a coin, chosen at random from the lot and then tossed, turns up heads 6 times in a row, what is the probability that it is the two-headed coin?

Given that \(P(X=a)=r, P(\max (X, Y)=a)=s,\) and \(P(\min (X, Y)=a)=\) \(t,\) show that you can determine \(u=P(Y=a)\) in terms of \(r, s,\) and \(t .\)

You are given two urns each containing two biased coins. The coins in urn I come up heads with probability \(p_{1}\), and the coins in urn II come up heads with probability \(p_{2} \neq p_{1}\). You are given a choice of (a) choosing an urn at random and tossing the two coins in this urn or (b) choosing one coin from each urn and tossing these two coins. You win a prize if both coins turn up heads. Show that you are better off selecting choice (a).
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