Problem 42 - Calculus for Biology and Medicine

Question

You select 5 cards without replacement from a standard deck of 52 cards. What is the probability that you get four aces?

Step-by-Step Solution

Answer

The probability is \( \frac{1}{54,145} \).

Step 1: Identify the Total Number of Outcomes

Calculate the total number of ways to choose 5 cards from a deck of 52 cards. This is a combination problem that can be represented as \( \binom{52}{5} \). Compute this as \( \frac{52!}{5!(52-5)!} \).

Step 2: Calculate the Number of Favorable Outcomes

To have four aces in your hand, consider that there are 4 aces in the deck. We need to select all 4 aces, which can only happen in 1 way: \( \binom{4}{4} = 1 \). The fifth card can be any of the remaining 48 cards: \( \binom{48}{1} = 48 \). The total number of favorable outcomes is \( 1 \times 48 = 48 \).

Step 3: Find the Probability

Divide the number of favorable outcomes by the total number of outcomes to find the probability. This is given by \( \frac{48}{\binom{52}{5}} \). Calculate \( \binom{52}{5} = 2,598,960 \). So, the probability is \( \frac{48}{2,598,960} \).

Step 4: Simplify the Fraction

Simplify \( \frac{48}{2,598,960} \) to its simplest form. Both 48 and 2,598,960 can be divided by their greatest common divisor (GCD). Here, the GCD is 48, leading to \( \frac{1}{54,145} \).

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. Understanding combinatorics is essential when dealing with probability, especially in scenarios where you select items from a larger set without regard to order.

In the given problem, we use combinatorics to determine how many ways we can select 5 cards from a standard deck of 52 cards. Since the order of selecting cards does not matter here, we use combinations rather than permutations. The mathematical notation used for combinations is represented by \( \binom{n}{k} \), which reads as 'n choose k'.

To apply this to our scenario, we calculate \( \binom{52}{5} \) to find the total possible combinations for selecting 5 cards from a deck of 52. This is done using the formula:

• \( \binom{52}{5} = \frac{52!}{5!(52-5)!} \)

Here, the factorial symbol (!) represents the product of all positive integers up to a given number.

Card Games Mathematics

The study of card games within mathematics involves calculating probabilities and outcomes when dealing with a standard deck of 52 cards. Each card game holds a unique structure and strategy space characterized by the rules and sequence of play.

In our example, the goal is to select 5 cards and identify the probability of drawing exactly four aces. The strategic key is understanding that there are only four aces in a deck. Therefore, if we're to draw four aces, we must successfully choose all four from the available set.

Once all aces are chosen, the remaining card must be from the non-ace group. This is tackled by selecting one card from the remaining 48 cards in the deck. Mathematically, this step can be computed as:

• Choosing 4 aces: \( \binom{4}{4} = 1 \)
• Selecting an additional card from the remaining 48: \( \binom{48}{1} = 48 \)

Thus, the total number of favorable outcomes is the product of these two figures, resulting in 48 possibilities.

Probability Calculation

Probability is a measure of the likelihood that an event will occur, expressed as a fraction between 0 and 1. In the context of our problem, to find the probability of drawing four aces and one additional card, we must divide the number of favorable outcomes by the total number of possible outcomes.

In simpler terms, with our dataset already calculated:

• Total possible five-card combinations: 2,598,960
• Number of favorable outcomes for four aces: 48

The probability is then given by the fraction: \[ \frac{48}{2,598,960} \]

This fraction simplifies to \( \frac{1}{54,145} \) because both the numerator and the denominator share a greatest common divisor (GCD) of 48. This simplification is critical in making our probability easy to interpret. It shows that the chance of drawing four aces and one additional card in one try is highly improbable, reinforcing the challenge and excitement of card games in mathematical terms.