Problem 32 - Algebra 2

Question

The game of euchre (YOO ker) is played using only the 9s, 10s, jacks, queens, kings, and aces from a standard deck of cards. Find the probability of being dealt a 5-card hand containing all four suits.

Step-by-Step Solution

Answer

The probability of being dealt a 5-card hand containing all four suits is approximately 0.61.

Step 1: Calculate the Total Number of Cards Used in Euchre

Euchre uses only the cards 9, 10, J, Q, K, A from each of the four suits. Since each suit has 6 cards, we have a total of \(6 \times 4 = 24\) cards in a euchre deck.

Step 2: Determine the Total Number of 5-Card Hands

To find the total possible 5-card hands from the euchre deck, use the combination formula \(\binom{n}{k}\), which gives the number of ways to choose \(k\) items from \(n\) items without regard to order. Here \(n = 24\) and \(k = 5\), so the total number of hands is \(\binom{24}{5} = 42,504\).

Step 3: Find the Number of Hands with All Four Suits Present

To have all four suits present in a 5-card hand, choose one card from each suit. For the first four cards, pick one card from each of the four suits, which can be done in \(6 \times 6 \times 6 \times 6 = 1,296\) ways. The fifth card can be any card from the remaining 20 cards (5 cards per suit except the one already chosen). Thus, the number of ways is \(1,296 \times 20 = 25,920\).

Step 4: Calculate the Probability

The probability is the number of favorable 5-card hands divided by the total number of possible 5-card hands. Thus, the probability \( P \) is given by \( P = \frac{25,920}{42,504} \). Simplifying this fraction gives \( P \approx 0.61\).

Key Concepts

Card Combinations

Imagine you're playing euchre with a deck that includes 9s, 10s, jacks, queens, kings, and aces from all four suits. The fascinating aspect of euchre is that it doesn't use the full 52-card standard deck but rather a smaller, specialized one. This creates unique challenges when you want to draw a hand.
A 5-card hand means choosing 5 cards out of the available 24. This is where card combinations come into play! A combination helps us figure out how many ways we can select a subset (in this case, a hand) from a larger pool and is denoted using the notation \(\binom{n}{k}\).

• "n" is the total number of items available, which is 24 in euchre.
• "k" is the number of items to choose, here 5 for the hand.

It doesn't matter how you pick them—the order is irrelevant, emphasizing the essence of combinations. By calculating \(\binom{24}{5}\), we determine there are 42,504 unique 5-card hands possible. This sets the stage for computing probabilities of specific hands, like one that contains all four suits.

Binomial Coefficient

At the heart of calculating card combinations is a mathematical marvel known as the binomial coefficient. This coefficient tells you how many ways you can choose a specific number of items from a larger set without regard to order. It's like figuring out how many ways you can build a team from a group of people.
The binomial coefficient is represented by the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where "!" signifies a factorial, meaning you multiply a series of descending natural numbers. So for euchre:

• \(24!\) ("24 factorial") means multiplying every whole number from 24 down to 1.
• \(5!\) means multiplying every whole number from 5 down to 1.
• \(19!\) comes from \(n-k\), meaning multiply from 19 down to 1.

Using this formula for \(\binom{24}{5}\) gives us those 42,504 possible hands. Each step helps encapsulate how many different ways are possible, rooted in the randomness and possibilities that fuel card games and probability theory.

Conditional Probability

Conditional probability paints a vivid picture of probability under specific conditions. It doesn't just ask, "What's the chance of picking a card?" but "What's the chance of picking a card given that certain cards have been picked?"
In our euchre scenario, we're specifically looking at the likelihood of drawing a hand with all four suits. First, ensure each of the first four cards comes from a different suit—a task that multiplies each suit's choices: \(6 \times 6 \times 6 \times 6 = 1,296\) ways to make sure that a card from each suit is included.

• The initial choices constrain the conditions for the fifth card—selecting from the remaining pool of 20.
• This specific planning translates into \(1,296 \times 20 = 25,920\)—ways to achieve our target 5-card hand under the given condition.

The conditional aspect shines as we calculate the probability of these qualifying hands against all possible hands we've calculated before. The resulting probability is \(P = \frac{25,920}{42,504} \approx 0.61\), indicating that not all hands are equally likely, but this is the fraction that satisfies our conditional craving for an all-suited hand. It's a subtle, intricate dance of chance that mirrors countless real-world scenarios where certain conditions guide outcomes.