Problem 13 - A First Course in Probability

Question

Suppose that an ordinary deck of 52 cards (which contains \(4 \text { aces })\) is randomly divided into 4 hands of 13 cards each. We are interested in determining \(p,\) the probability that each hand has an ace. Let \(E_{i}\) be the event that the \(i\) th hand has exactly one ace. Determine \(p=\) \(P\left(E_{1} E_{2} E_{3} E_{4}\right)\) by using the multiplication rule.

Step-by-Step Solution

Answer

To find the probability \(p\) that each hand has exactly one ace in a standard deck of 52 cards divided into 4 hands of 13 cards each, we calculated the total number of ways to distribute the cards and the number of favorable outcomes. Using the multiplication rule, the probability \(p\) is given by: \[ p = \frac{C(48,12) * C(36,12) * C(24,12) * C(12,12)}{C(52,13) * C(39,13) * C(26,13) * C(13,13)} \] Where \(C(n,k)\) is the combination formula, representing the number of ways to choose k items from a set of n items without repetition. Calculate the values of the combinations using this formula and divide the number of favorable outcomes by the total number of ways to distribute the cards to get the probability \(p\).

Step 1: Calculate the total number of ways to distribute the cards

Let's first calculate the total number of ways to divide the 52 cards into 4 hands, each containing 13 cards. This can be done using the combination formula, which is given by: \[ C(n,k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \] So for each hand, there are 13 cards to be chosen out of 52. We perform this process 4 times for all the hands. The total number of ways to distribute the cards is then: \[ C(52,13) * C(39,13) * C(26,13) * C(13,13) \]

Step 2: Calculate the number of favorable outcomes

To find the number of favorable outcomes, first, let's distribute the aces. There are 4 aces in the deck, and we want to ensure that one ace goes to each hand. There's only one way to do this. Now let's continue to distribute the rest of the cards. After assigning an ace to each hand, we have 48 non-ace cards left in the deck, and we need to choose 12 more cards for each of the remaining hands. Similar to step 1, we find the number of ways to distribute these cards as follows: \[ C(48,12) * C(36,12) * C(24,12) * C(12,12) \]

Step 3: Find the probability p

Finally, let's find the probability p by dividing the number of favorable outcomes by the total number of ways to distribute the cards: \[ p = \frac{\text{number of favorable outcomes}}{\text{total number of ways to distribute the cards}} = \frac{C(48,12) * C(36,12) * C(24,12) * C(12,12)}{C(52,13) * C(39,13) * C(26,13) * C(13,13)} \] Now you can calculate the value of p using the formula above to find the probability that each hand has exactly one ace.

Key Concepts

Multiplication Rule in Probability

Understanding the multiplication rule in probability is essential when dealing with independent events. Imagine flipping a coin and rolling a die. What is the probability of getting heads and rolling a four? Here, we use the multiplication rule: the probability of both events occurring is the product of their individual probabilities.

In our card problem, we're looking at a slightly more complex scenario. Rather than simple independent events, we're distributing cards into hands, where the outcome of one affects the others. However, the multiplication rule still applies when we calculate the probability that each hand has an ace. We first assess the probability of an ace being in one hand and then multiply these probabilities for all hands considering the changing conditions after each distribution. This approach is a powerful tool for calculating the likelihood of multiple outcomes occurring in sequence.

Combinatorics

Combinatorics is the branch of mathematics that studies counting, both as a means and an end in obtaining results, and certain properties of finite structures. It's the math behind how many ways different events can occur, and it's steeped in the study of combinations and permutations.

The card problem requires us to determine how many ways we can distribute cards into hands. Combinatorics provides the methods used in the steps, such as the combination formula, which tells us the number of ways to choose a subset of cards from a larger set, without regard for the order. Understanding combinatorics is essential to solve problems where the arrangement or selection of objects is central. For example, finding the number of five-card poker hands of a certain type involves combinatorics.

Factorial Notation

Factorial notation is a mathematical expression that defines the product of an integer and all the non-negative integers below it. Denoted by an exclamation mark (!), factorial notation is crucial in combinatorics. For example, 5! (read as 'five factorial') is equal to 5 x 4 x 3 x 2 x 1 = 120.

In probability, factorials appear in calculations of permutations and combinations. The formula for combinations, shown in the card problem as \( C(n,k) = \frac{n!}{k!(n-k)!} \), heavily relies on factorial notation to calculate the number of different combinations that can be made from a larger set. Factorials quickly grow to very large numbers, which is why they are often used to describe the huge number of ways elements can be arranged, like cards in a deck.

Probability Theory

Probability theory is the mathematical framework that allows us to analyze random phenomena and quantify the likelihood of various outcomes. It helps us make sense of the uncertainty and enables calculated predictions and informed decisions.

The formula used in the card problem to find the probability that each hand has an ace is a direct application of probability theory. By dividing the number of favorable outcomes by the total number of possible outcomes, we arrive at a probability that gives insight into the expected regularity of an event occurring. It's crucial to understand not just the calculations themselves, but also the principles that underlie probability theory, such as the concepts of independence, mutual exclusivity, and the total probability of all possible outcomes equaling one.