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

The probability that each hand has an ace in a deck of 52 cards divided into 4 hands of 13 cards each is \(p = \frac{1}{2197}\).

Step 1: Find the probability of the first hand having an ace

The first hand can be formed from the original deck containing 52 cards. There are 4 aces available, so the probability of choosing an ace for the first hand is: \(P(E_{1}) = \frac{4}{52}\)

Step 2: Find the probability of the second hand having an ace

After the first hand takes one ace, there are 3 aces left in the remaining deck and 39 cards available to form the second hand (52-13=39). The probability of choosing an ace for the second hand is: \(P(E_{2} | E_{1}) = \frac{3}{39}\)

Step 3: Find the probability of the third hand having an ace

After the second hand takes another ace, there are 2 aces left in the remaining deck and 26 cards available to form the third hand (39-13=26). The probability of choosing an ace for the third hand is: \(P(E_{3} | E_{1}E_{2}) = \frac{2}{26}\)

Step 4: Find the probability of the fourth hand having an ace

Finally, after the third hand takes another ace, there is 1 ace left in the remaining deck and 13 cards available to form the fourth hand (26-13=13). The probability of choosing an ace for the fourth hand is: \(P(E_{4} | E_{1}E_{2}E_{3}) = \frac{1}{13}\)

Step 5: Use the multiplication rule to find the probability of all events

Using the multiplication rule, we can find the probability that all events occur by multiplying the probabilities of each event happening: \(p = P(E_{1}E_{2}E_{3}E_{4}) = P(E_{1}) \times P(E_{2} | E_{1}) \times P(E_{3} | E_{1}E_{2}) \times P(E_{4} | E_{1}E_{2}E_{3})\) So, \(p = \frac{4}{52} \times \frac{3}{39} \times \frac{2}{26} \times \frac{1}{13} \)

Step 6: Simplify and find the final probability

Now we can simplify the multiplication: \(p = \frac{4}{52} \times \frac{3}{39} \times \frac{2}{26} \times \frac{1}{13} = \frac{4}{52} \times \frac{3}{3 \times 13} \times \frac{2}{2 \times 13} \times \frac{1}{13}\) \(p = \frac{4}{4 \times 13} \times \frac{1}{13} \times \frac{1}{13} \times \frac{1}{13} = \frac{1}{2197}\) So, the probability that each hand has an ace is \(p = \frac{1}{2197}\).

Key Concepts

Multiplication Rule

The multiplication rule is a cornerstone of probability theory, especially when dealing with sequences of events. It tells us how to find the probability of multiple events happening in a sequence, considering that one event might affect the others.
We apply this rule by multiplying the probabilities of individual events. However, when events are dependent, as often seen in card games, each subsequent event's probability is conditional on the preceding ones.
For example, finding the probability that each hand in a card game has exactly one ace requires considering the events sequentially and using the multiplication rule to combine them.

• First, we identify the probability of the initial event occurring.
• Then, given that this event has occurred, we calculate the probability of the next event.
• We continue this process for all events.

In this exercise, using the multiplication rule allows us to logically follow through the series of probabilities for each hand drawn from the deck.

Conditional Probability

Conditional probability measures the likelihood of an event occurring, given that another event has already occurred. This concept is crucial when calculating probabilities in dependent settings, such as drawing cards from a deck.
In our exercise, after the first hand takes one ace, the probability of drawing another ace for the second hand changes as it depends on the first hand's outcome.
Simply put, the formula for conditional probability is:\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]Here, \( P(A | B) \) is the probability of event A occurring after event B has happened. It accounts for the reduced number of cards left in the deck after each ace is drawn.

• Initially, the probability of an event is calculated normally.
• The outcomes of subsequent events are recalibrated based on prior events.

This exercise highlights how the probability of each subsequent hand having an ace is recalculated considering the previous events.

Deck of Cards

A standard deck of cards is a familiar yet complex system for studying probability. It consists of 52 cards split into four suits: hearts, diamonds, clubs, and spades, each having 13 cards. Each suit contains one ace, four in total, making them key targets in many probability exercises.
When dividing this deck into hands, understanding the variations in possible outcomes becomes important. The setup takes advantage of the fixed structure of a deck of cards allowing us to predict probabilities more efficiently.
Key components of the deck:

• 52 total cards with 4 suits.
• Each suit contains one ace among other ranked cards.
• Card games often revolve around these physical constraints, influencing probability calculations.

The essence of this problem lies in correctly handling the structure of the deck, taking into account the decreasing number of cards and aces with each draw.

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations of objects. In probability theory, it helps us calculate the number of ways events can happen, essential for determining probabilities accurately.
In our exercise, combinatorics supports the arrangement and counting of ways to distribute cards into hands, particularly focusing on the drawn aces.
For instance, figuring out how a hand is likely to contain one ace involves:

• Counting the total number of ways to draw 13 cards from 52.
• Counting the successful outcomes where the arrangement results in one ace per hand.

Such combinations and distributions define the groundwork for accurately determining complex probabilities. In this case, combinatorics is disguised within the multiplication and conditional probabilities, ensuring each hand ends up with one ace.